#geometry-and-trigonometry | Mathematics | Page 228

weary drift Sep 16, 2019, 4:48 PM

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sure, ACD and ACE are a linear pair, so that's fine

surreal roost Sep 16, 2019, 5:13 PM

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I don't get it shouldn't it be 2.83?

📎 Screenshot_2.png

grand shard Sep 16, 2019, 5:16 PM

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hello i have question, how do i calculate the middle point of a circle?

the size of the circle is 35x35 (width: 35, height: 35)

worthy root Sep 16, 2019, 5:16 PM

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flonshed

grand shard Sep 16, 2019, 5:16 PM

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im kind of going coordinate-wise

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x, y

worthy root Sep 16, 2019, 5:33 PM

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pi*35^2

dark sparrow Sep 16, 2019, 5:54 PM

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@surreal roost 2.83 is not "rounded to the nearest tenth"

dreamy finch Sep 16, 2019, 6:20 PM

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@surreal roost nearest tenth = nearest multiple of 1/10, ie, 0.1

upper karma Sep 16, 2019, 6:38 PM

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What's your question

worthy root Sep 16, 2019, 6:52 PM

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https://cdn.discordapp.com/attachments/326138757474680852/623192804398596120/20190916_215121.jpg @upper karma

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<@&286206848099549185> i reposted it it more than am hpur. Help me . 2*

mighty gull Sep 16, 2019, 6:53 PM

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draw it out and do some principle of inclusion exclusion

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give me diagram first

upper karma Sep 16, 2019, 6:54 PM

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Wtf is c

worthy root Sep 16, 2019, 6:54 PM

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Wot ? @mighty gull

mighty gull Sep 16, 2019, 6:55 PM

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draw it out

upper karma Sep 16, 2019, 6:55 PM

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Yea

worthy root Sep 16, 2019, 6:55 PM

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Wait

upper karma Sep 16, 2019, 6:55 PM

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Literally

worthy root Sep 16, 2019, 6:55 PM

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I got it

upper karma Sep 16, 2019, 6:55 PM

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Drawl the angle

mighty gull Sep 16, 2019, 6:55 PM

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lmfao

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"i got it"

worthy root Sep 16, 2019, 6:55 PM

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Its 45-45-90 triangle

mighty gull Sep 16, 2019, 6:55 PM

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you just pinged helper

worthy root Sep 16, 2019, 6:55 PM

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I can do it i think

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I mean not the answer but a 45-45-90 triangle let me work over

mighty gull Sep 16, 2019, 6:57 PM

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please draw out your things and actually try

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b4 pinging helpers

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ffs

worthy root Sep 16, 2019, 6:57 PM

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I worked for an hour .

upper karma Sep 16, 2019, 6:57 PM

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Bro draw the angle

worthy root Sep 16, 2019, 6:57 PM

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Surfing on internet for answer

upper karma Sep 16, 2019, 6:57 PM

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Draw another angle

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Draw more

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Until you have a polygon

worthy root Sep 16, 2019, 6:58 PM

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Yes . Doing

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I am etting

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49(pi-\sqrt{2}/4

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Thats weong acc to my text

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Also people on internet say (49(pi-2)/4 is correct

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sadcat sadcat

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God(if u exist bless him) .

📎 2019_09_17_00.41.13_edit.jpg

south pebble Sep 16, 2019, 7:12 PM

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📎 Screenshot_20190916-121221_OneNote.jpg

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can someone help me solve this?

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im getting an answer thats not making sense at all

worthy root Sep 16, 2019, 7:13 PM

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isolate the sin(2x)

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,calc 150/2500

somber coyoteBOT Sep 16, 2019, 7:13 PM

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Result:

0.06

worthy root Sep 16, 2019, 7:14 PM

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,calc 0.06*9.81

somber coyoteBOT Sep 16, 2019, 7:14 PM

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Result:

0.5886

south pebble Sep 16, 2019, 7:14 PM

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arcsin right?

worthy root Sep 16, 2019, 7:14 PM

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Yes

south pebble Sep 16, 2019, 7:14 PM

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yes i did all that and im still not getting the right answer

worthy root Sep 16, 2019, 7:14 PM

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You can use calc ?

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,w calc arcsin(0.5886)

somber coyoteBOT Sep 16, 2019, 7:15 PM

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📎 wolf.png

Display results online and refine query

south pebble Sep 16, 2019, 7:15 PM

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then divide by 2?

worthy root Sep 16, 2019, 7:15 PM

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Amd divide hy 2

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Yes

south pebble Sep 16, 2019, 7:15 PM

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yup, still not getting the right answer for some reason..

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,calc 0.629326/2

somber coyoteBOT Sep 16, 2019, 7:16 PM

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Result:

0.314663

worthy root Sep 16, 2019, 7:16 PM

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Wdym by not getting right answer

south pebble Sep 16, 2019, 7:16 PM

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im supposed to be getting a larger angle

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maybe because its in radians?

worthy root Sep 16, 2019, 7:16 PM

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,w 150=(2500*sin(2x)/(9.81)

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Oh

somber coyoteBOT Sep 16, 2019, 7:16 PM

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📎 wolf.png

Display results online and refine query

worthy root Sep 16, 2019, 7:17 PM

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Then use peridicty identities

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@south pebble

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Do you know the peridoicty identities?

south pebble Sep 16, 2019, 7:18 PM

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err, not really no

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strange, i converted the answer to degrees using (180/pi)

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and its still not right..

worthy root Sep 16, 2019, 7:18 PM

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sin(x+2pi)=sin(x)

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This type identity? @south pebble

south pebble Sep 16, 2019, 7:19 PM

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are those also known as trig identities?

worthy root Sep 16, 2019, 7:20 PM

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Yes? Maybe

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f(n+xp)=n is peridic function

south pebble Sep 16, 2019, 7:22 PM

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i probably forgot

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i think im going to skip it for now

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thanks!

wind heart Sep 16, 2019, 7:24 PM

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📎 image0.jpg

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📎 image0.jpg

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I kinda need help with this question

weary drift Sep 16, 2019, 7:26 PM

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do you recall how to name a plane?

wind heart Sep 16, 2019, 7:27 PM

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it needs three points that are not on the same line

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so wait would it be like

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Plane C D A

weary drift Sep 16, 2019, 7:28 PM

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there you go

wind heart Sep 16, 2019, 7:28 PM

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oh cool

worthy root Sep 16, 2019, 7:31 PM

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Show that the minute hand of clock gains 5°30" on the hour hand in one minute .

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Is thera a mistake in the problem?

dark sparrow Sep 16, 2019, 7:34 PM

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yeah should be 5°30'

worthy root Sep 16, 2019, 7:34 PM

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Really man this textbook has hell.lot of mistakes

dark sparrow Sep 16, 2019, 7:34 PM

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not 5°30"

worthy root Sep 16, 2019, 7:34 PM

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Yes i know

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,w 1sec=xhr

somber coyoteBOT Sep 16, 2019, 7:39 PM

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Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Display results online and refine query

worthy root Sep 16, 2019, 7:39 PM

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sadcat

dark sparrow Sep 16, 2019, 7:39 PM

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1/3600

worthy root Sep 16, 2019, 7:39 PM

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Yes was just confirming

dark sparrow Sep 16, 2019, 7:39 PM

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,w convert 1 second to hours

somber coyoteBOT Sep 16, 2019, 7:39 PM

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📎 wolf.png

Display results online and refine query

worthy root Sep 16, 2019, 7:39 PM

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🤔

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1hr=60mins 1min=60sec

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1hr=3600secs

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📎 JPEG_20190917_011057.jpg

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I find the distance traveelid in 30sec

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That is

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36km/hr*30/3600hr

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3/10

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So i plugh this in arc lemght formula

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3/10=x

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The angle is just 18'

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But the textbook shows something else

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Woops

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I have to convert in degree

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Careless Krishna REEEE

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,calc 3/10*(180*7/22)

somber coyoteBOT Sep 16, 2019, 7:49 PM

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Result:

17.181818181818

worthy root Sep 16, 2019, 7:49 PM

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,calc 180*7

somber coyoteBOT Sep 16, 2019, 7:49 PM

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Result:

worthy root Sep 16, 2019, 7:49 PM

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,calc 0.18*60

somber coyoteBOT Sep 16, 2019, 7:49 PM

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Result:

10.8

worthy root Sep 16, 2019, 7:50 PM

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Ann how i convert this recurring decimal radiam to degree

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I mean i have to do it till mimute and seconds

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,w convert 3/10radian into degress

somber coyoteBOT Sep 16, 2019, 7:53 PM

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📎 wolf.png

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📎 wolf.png

Display results online and refine query

worthy root Sep 16, 2019, 7:53 PM

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,w convert (3/10)radian into degress

somber coyoteBOT Sep 16, 2019, 7:53 PM

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📎 wolf.png

Display results online and refine query

worthy root Sep 16, 2019, 7:53 PM

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They roumd thay off

vagrant karma Sep 16, 2019, 9:56 PM

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I need help

severe mortar Sep 16, 2019, 10:14 PM

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📎 image0.png

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I’m a little stuck

vagrant karma Sep 16, 2019, 10:15 PM

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Same yo

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Mines harder

severe mortar Sep 16, 2019, 10:15 PM

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yeah true

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📎 image0.jpg

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would rlly appreciate it 🥺

vagrant karma Sep 16, 2019, 10:15 PM

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Lemme see what I can do I’ll prob give you the wrong answer lmao

severe mortar Sep 16, 2019, 10:16 PM

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delightful

vagrant karma Sep 16, 2019, 10:16 PM

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📎 image0.jpg

primal lynx Sep 16, 2019, 10:17 PM

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y=x-3

severe mortar Sep 16, 2019, 10:19 PM

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🥺

silent plank Sep 16, 2019, 10:22 PM

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how big is the angle BAC?

severe mortar Sep 16, 2019, 10:23 PM

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uhh I don’t know all I knwo is that

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📎 image0.jpg

silent plank Sep 16, 2019, 10:24 PM

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given the bearings to B and C from A, you have enough info to find angleBAC

severe mortar Sep 16, 2019, 10:25 PM

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200-140

silent plank Sep 16, 2019, 10:26 PM

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are you able to find BC now?

severe mortar Sep 16, 2019, 10:26 PM

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Yes thanks

latent wave Sep 16, 2019, 10:35 PM

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Hello, I'm new to Trigonometry, and I have a few simple questions: 1) How do I differentiate the graphs of Sine and Cosine visually? 2) When writing the equation of a Sine or Cosine function when only a graph is given, how do I know if it has been reflected?

vagrant karma Sep 16, 2019, 10:54 PM

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I have more lol im dumb sorry

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📎 image1.jpg 📎 image0.jpg

silent plank Sep 16, 2019, 11:05 PM

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what have you tried?

vagrant karma Sep 16, 2019, 11:46 PM

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I don’t where to start💀💀💀

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know^

silent plank Sep 17, 2019, 12:14 AM

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distance formula?

stray spire Sep 17, 2019, 2:53 PM

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could anyone help me with geometry ?

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this is the task -the lengths of the edges of the square are in the ratio 1: 2: 3. Prove that the surface of its slope is 9 times larger than the surface of its base

upper karma Sep 17, 2019, 4:28 PM

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📎 image0.jpg

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Not sure what to do

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the only way I can think of is using the law of sines to find the height! good luck : )

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SOHCAHTOA

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oh right there is that too haha. yeah pretty much similar

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Would it be sine=72/12

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Since Sine = opp/hyp

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sin(angle) = opp/hyp

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Wdym by Angel?

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oops

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My bad

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How do I use it to find length

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Do I need to use inverse of sin

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yep!

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wait

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And how do I do that?

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actually no haha

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Ah shit

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sin(72) = x/12

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Why?

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I am so confused rn

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so for the triangle you have. the hyp=12 and the angle is 72. we are trying to find the height (opp)

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Right

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just direct plugging in sin(angle) = opp/hyp

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sin(72) = opp/12

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Ihhh

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Makes sense

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: )

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So then I need to rearrange

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Or not?

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yes, what do you think the answer is?

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11.41

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I did 12sin(72)=x

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yep!

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Calculator says 11.41 (1.d.p)

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Yes!!

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Thank you so much

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you're very welcome

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Then

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I do height

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Times base

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So 11.41x16?

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correct

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Yes

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Thank you

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My retarded Spanish maths teacher can’t even tell me this

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Honestly, people like you are revolutionising the art of teaching mathematics

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haha thanks! I'd love to bring you to say that to the people I tutor xD

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Lmao lol

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May I ask for assistance for another question please?

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Or would that be too much?

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go for it, if it's under geo-trig post it here, if not just mention me in the appropriate channel

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I did some work on this one

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📎 image0.jpg

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But I need help on figuring out perpendicular height

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you're almost done. you have a similar problem to the previous one: 7.2 is the adj, and you want the opposite. try drawing the triangle AVM in 2 dimensions, and you will definitely see it better. you have 2 known values and trying to find a third

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Not sure what the other value is

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so the angle is 58°, and the side length (adjacent) of the angle is 7.2cm, and we are trying to find VM (opposite), which trig uses that information?

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I can’t see 58

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I only see 68

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sorry 68. I'm blind

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Would we use Cos

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COH

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CAH*

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not quite, give me a moment

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SOG

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SOH*

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Sin

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📎 image0.jpg

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Try now

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Further hint

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📎 image0.jpg

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It’s not Sin?

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COS

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CAH

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Nope! You have Adj, and you are trying to find VM, Opp

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Adjacent/hyp

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Yes but, VM isn’t hyp

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AV is hyp

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Yes

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That’s what I thought

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Wait, how do you use SOHCAHTOA

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Is it by what values there are

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On specific positions

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On a triangle?

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Yes! So
Sin is Opp/Hyp
Cos is Adj/Hyp
Tan is Opp/Adj

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Which value do we have in our triangle? And which value do we want to find??

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Cos

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We only have hyp and adj

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So cos

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Which is adj/hyp

wet portal Sep 17, 2019, 5:16 PM

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V=1/3Ah

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h=?

upper karma Sep 17, 2019, 5:18 PM

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So yes we are trying to find h

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Would it be tab

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Tan

wet portal Sep 17, 2019, 5:19 PM

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yes

upper karma Sep 17, 2019, 5:19 PM

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Wait why?

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I am so confused

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I was told

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Haha it’s ok. So h is VM

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Right

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And that’s the opposite

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Yes

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Yep, and you have an angle and the adjacent

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Yes

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Which one uses those 3

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Wdym?

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Which trig uses an angle, the adjacent, and the opposite?

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Tan?

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Yesh

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But tan

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Is opp/adj

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Yep

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How is angle involved

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It belongs to the triangle. If you look at the given information. It says VAM measures 68°

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And you already found that AC is 14.4, half of it is AM, which is 7.2

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Makes sense now? : )

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But I could use 7.2

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As adj

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Why is angle important to me?

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So you can find the height. Which you need to find the volume

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So what would tan equation look like

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Tan(angle) = opp/adj

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Assuming

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We needed to work out the angle

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And not height

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What would be equation

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Tan

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though we do have the angle, it's 68°, given from the question

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our goal is to find VM, the opposite

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What would be the general equation

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To find height

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Not angle

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I mean angle

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Not height

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there is nothing you need to do to find the angle, since it's already given as 68°

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Okay

wet portal Sep 17, 2019, 5:37 PM

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if assuming u needed to find it

upper karma Sep 17, 2019, 5:37 PM

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So I got 7.2/tan(68)

wet portal Sep 17, 2019, 5:37 PM

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it would depend on things that you knew about the triangle

upper karma Sep 17, 2019, 5:37 PM

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Is that correct

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To find opp

wet portal Sep 17, 2019, 5:38 PM

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if the equation is Tan(angle) = opp/adj

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you put in Tan(68) = x/7.2

upper karma Sep 17, 2019, 5:39 PM

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Yes

wet portal Sep 17, 2019, 5:39 PM

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how would you solve that

upper karma Sep 17, 2019, 5:39 PM

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You need to rearrange

wet portal Sep 17, 2019, 5:39 PM

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how

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you want to find x so what do you do next?

upper karma Sep 17, 2019, 5:40 PM

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7.2tan(68)=x

wet portal Sep 17, 2019, 5:40 PM

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yes

upper karma Sep 17, 2019, 5:40 PM

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My bad

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I mixed up opp and hyp

wet portal Sep 17, 2019, 5:40 PM

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yeah, it happens

upper karma Sep 17, 2019, 5:40 PM

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Thanks though

wet portal Sep 17, 2019, 5:41 PM

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i remember the time when i was studying sin,cos,tan and i was super tilted

upper karma Sep 17, 2019, 5:41 PM

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I got 17.82 (1.d.p)

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📎 image0.jpg

wet portal Sep 17, 2019, 5:41 PM

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yep

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what next?

upper karma Sep 17, 2019, 5:42 PM

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Put it in equarikj

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Equation

wet portal Sep 17, 2019, 5:42 PM

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yes

upper karma Sep 17, 2019, 5:42 PM

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So 1/3x100

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X17.82

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100 being area of base

wet portal Sep 17, 2019, 5:43 PM

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yes

upper karma Sep 17, 2019, 5:43 PM

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👍

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Thank you guys

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For everything

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Damn

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anytime

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I still have a few questions to do

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Is that okay?

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Please tell me if it’s too much of your time

wet portal Sep 17, 2019, 5:44 PM

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yeah

upper karma Sep 17, 2019, 5:44 PM

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It is

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Okay

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My bad sorry I will leave then

wet portal Sep 17, 2019, 5:44 PM

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nono, you can ask

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anytime

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😄

upper karma Sep 17, 2019, 5:44 PM

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Oh sorry lol

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📎 image0.jpg

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I made one annotation to this question

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I'm fine with as many questions as you want to ask, as long as you attempt them first

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I’ve tried to see if there are a few triangle theorems

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They didn’t work

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I just made one statement

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Saying it could be perpendicular

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But I am probably wrong

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Because I do know

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This one triangle theory

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Where one side doubles

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Like this

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📎 image0.jpg

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So I did 45x2

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Yeah I am not so sure on the SAS, SSS, ASS system

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I know what they stand for

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But I don’t know how to implement them in the answers I make

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Basically, I don’t know how to use that system

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I just know they mean side,angle side etc

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that's fine. 2 triangles would be congruent under 3 conditions: SSS, SAS, ASA

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Only 3?

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yep

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No more no less?

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nope

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So how I do I know which one is which

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In other words, which one do I know is the correct one for this question?

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it depends really... I don't think there is a 100% way to tell, but most of the time if you are given an angle then you are using either SAS or ASA. Again, I never thought of which one to use, I just kept going until I satisfied one of them

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Hm okay

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for example, start by writing 10s all over the sides of the squares and you are basically almost done

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On the question?

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if you are allowed yes, or on a separate paper

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so now you have 2 identical pairs of sides of the triangles

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I am so confused

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Oh lord

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lol that's ok

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Why do I need to put 10s?

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because they are squares

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and squares have the same side lengths

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ABCD and AEFG

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so you have 8 side lengths of 10cm

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So where do I put 10s

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On sides of square?

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yeah, does that make sense though?

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Yes

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Because all sides are the same

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On a square

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yep haha, okay

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But I why you putting all over the squares?

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so that you can see now the triangles

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they both have 2 sides of 10s

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This okay?

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📎 image0.jpg

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yep

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So now what do I do,

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okay let's think... we are at the point of S_S, where we just need to prove either the last side is identical, or the angle in between the sides is identical, does this make sense?

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Where did you get the 2 Ss?

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From*

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look are the triangles again, each has 2 sides with 10 cm

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alright, so back to what I said, you either need SAS or SSS, but here, finding the last side to be the same can be quite challenging if at all possible. The question gives us an angle of 45°, and that's a good sign. Do you think you can finish it from here?

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Hmm

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Not sure

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Why is 45*a good sign?

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because it can be used to find the angles of the triangles that are in between the sides

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Okay

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And how would I do that?

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Would I use the triangle theorem

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Where the bottom part of triangle doubles

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nope... much easier than that

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try and see which angles you are trying to say are the same

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The 2 bottom ones

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📎 image0.jpg

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not quite. you want the angle between the 10s of each triangle

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which ones are those? can you name them using letters?

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ABC and ACF,

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?*

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AGF*

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@upper karma

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those are the angles of the squares though... the angles we need to prove are the same are BAE and GAD

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those are the angles which are in between the 2 pairs of sides that are identical, can you see it?

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Yes I can

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What do we do with them

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find those angles! : )

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then tell me what you find out

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📎 image0.jpg

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Red line represent 2 angles

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yep! those are 90°

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What is?

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EAG and DAB

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so draw a little square (90°) on each

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and now find BAE and GAD

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Why is it 90*

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Oh because they are on a square!!!

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😂 👍

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So wdym

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By find BAE and GAD

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the angle measures! what are they

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90*

harsh panther Sep 17, 2019, 6:35 PM

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Hey,try highlighting the triangles you are looking to prove for themselves in some colour it may help with the imagination

upper karma Sep 17, 2019, 6:37 PM

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Would BAR

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BAE

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Be 90+45?

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@upper karma

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YES!! and GAD?

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should also be 90+45

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So 135

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yep... so now you can tell they are congruent using SAS

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can you see it? does it make sense

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Oh because the 2 angles in the 2 other angles I am trying to prove are equal angles?

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nooo you proved 2 sides are the same

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Right

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that's what the 10cm everywhere are for

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Right

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and then you are proving the angle between the 10cm pairs is the same as the other triangle

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which turned out to be 135°

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In the exams

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Can I just draw on diagram

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To get marks

harsh panther Sep 17, 2019, 6:43 PM

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That would depend on your system

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School system*

upper karma Sep 17, 2019, 6:45 PM

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^

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Okay

vagrant karma Sep 17, 2019, 8:48 PM

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I have no idea what to do or where to start

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Please help

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I still need help on this and it’s due tomorrow

📎 image1.jpg 📎 image0.jpg

silent plank Sep 17, 2019, 8:57 PM

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how do you find the distance between two points?

vagrant karma Sep 17, 2019, 9:08 PM

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x1-x2/2

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NO PLUS

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and

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y1+y2/2

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NO WAIT THATS MIDPOINT

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LOL IM DUMB

weary drift Sep 17, 2019, 9:09 PM

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that'll get you the coords of the midpoint, plus you're missing parentheses

vagrant karma Sep 17, 2019, 9:09 PM

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but it’s the square root of (x2-x1)^2+(y2-y1)^2

#

GIRL IDEK HOW I GOT INTO HONORS GEOMETRY LOLLL

weary drift Sep 17, 2019, 9:10 PM

#

that looks better

somber coyoteBOT Sep 17, 2019, 9:11 PM

#

RokettoJanpu:

$d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

📎 150414892225134592.png

vagrant karma Sep 17, 2019, 9:11 PM

#

What

#

That is not the answer to my question but ok lol

weary drift Sep 17, 2019, 9:12 PM

#

you recalled the correct equation for the distance. i just typed it out in latex to confirm this is what you're talking about

vagrant karma Sep 17, 2019, 9:12 PM

#

Ohhh

#

Yes thank you

#

Also is this right or do you see mistakes ??

📎 image0.jpg

weary drift Sep 17, 2019, 9:17 PM

#

isn't that hw site supposed to tell you if your answers are right?

vagrant karma Sep 17, 2019, 9:18 PM

#

No

#

I wish lol

weary drift Sep 17, 2019, 9:18 PM

#

looks good

sly marlin Sep 17, 2019, 9:18 PM

#

Isn't TRL an acute angle?

weary drift Sep 17, 2019, 9:20 PM

#

@vagrant karma actually i don't know what "not an angle" means

sly marlin Sep 17, 2019, 9:21 PM

#

I think maybe, syntax is wrong

#

$\angle ABCD$ is totally not an angle

somber coyoteBOT Sep 17, 2019, 9:21 PM

#

Element118:

📎 104219409991626752.png

sly marlin Sep 17, 2019, 9:21 PM

#

$\angle 45$ is totally not an angle too

somber coyoteBOT Sep 17, 2019, 9:21 PM

#

Element118:

📎 104219409991626752.png

sly marlin Sep 17, 2019, 9:22 PM

#

And maybe involving nonexistent points, so for your diagram,
$\angle ABC$ is not an angle in the diagram

somber coyoteBOT Sep 17, 2019, 9:22 PM

#

Element118:

📎 104219409991626752.png

vagrant karma Sep 17, 2019, 9:43 PM

#

@sly marlin trl is not an angle, bc it goes to the vertex then back.

dusty sleet Sep 17, 2019, 10:52 PM

#

anyone here that could help me with differentials?

astral cedar Sep 17, 2019, 11:17 PM

#

Hello

#

m angle = 50 and m angle sgh = 120, prove angle fgh=170

#

how to do

crude pelican Sep 17, 2019, 11:23 PM

#

I checked the resources but i can't exactly find what i'm looking for in Geometry, Can anyone link me to a resource that may help me study? I'm doing Long-distance Formula with Shapes, finding Area & Perimeter and what not.

#

Sorry if it's not clear enough, I'll add more context if required.

astral cedar Sep 17, 2019, 11:32 PM

#

<@&286206848099549185> m angle = 50 and m angle sgh = 120, prove angle fgh=170
how to do

zenith ember Sep 17, 2019, 11:49 PM

#

Please read #❓how-to-get-help

scarlet halo Sep 18, 2019, 1:51 AM

#

can i get some help ?

📎 unknown.png

spark stag Sep 18, 2019, 1:53 AM

#

what are you asked to find

#

where are you stuck

scarlet halo Sep 18, 2019, 1:54 AM

#

find x

#

but no idea where to start

spark stag Sep 18, 2019, 2:00 AM

#

are you familiar with sine/cosine/tangent? maybe the acronym "SOH CAH TOA"?

scarlet halo Sep 18, 2019, 2:01 AM

#

like legnth/since(angle)

spark stag Sep 18, 2019, 2:02 AM

#

...what?

scarlet halo Sep 18, 2019, 2:02 AM

#

i guess not then

#

We did ratios today where y/x is the ratio and it should equal the side legnths ratio

spark stag Sep 18, 2019, 2:03 AM

#

uhh

#

do you have any more information?

#

from the question, that is

scarlet halo Sep 18, 2019, 2:04 AM

#

not really the question was

📎 unknown.png

#

this was another one i didnt get

📎 unknown.png

spark stag Sep 18, 2019, 2:06 AM

#

yeah, you cant do these without trigonometric functions

scarlet halo Sep 18, 2019, 2:10 AM

#

so u cant do it with the ratio based off an angle ?

#

it shows me the answer but i dont get where they got it from

📎 unknown.png

silent plank Sep 18, 2019, 2:19 AM

#

who wrote the approx 5/1?

scarlet halo Sep 18, 2019, 2:19 AM

#

the book

silent plank Sep 18, 2019, 2:20 AM

#

was the answer also from the book?

scarlet halo Sep 18, 2019, 2:20 AM

#

yes

silent plank Sep 18, 2019, 2:21 AM

#

nfi why they gave the approximation like that

#

if they didn't plan on approximating the answer

scarlet halo Sep 18, 2019, 2:23 AM

#

so u dont get it ?

silent plank Sep 18, 2019, 2:23 AM

#

but yeh to get the answer you'd need the trig ratio for tan

scarlet halo Sep 18, 2019, 2:24 AM

#

So u dont understand the way they got the answer

silent plank Sep 18, 2019, 2:24 AM

#

i do.
tan(theta) = opp/adj

scarlet halo Sep 18, 2019, 2:25 AM

#

can u explain how to do it ?

silent plank Sep 18, 2019, 2:25 AM

#

have you leaned any trig at all?

scarlet halo Sep 18, 2019, 2:26 AM

#

i dont belive so

silent plank Sep 18, 2019, 2:27 AM

#

i'll provide enough to get you through the question

#

in a right triangle tan(angle) will give you the ratio of the opposite/adjacent sides

upper karma Sep 18, 2019, 2:28 AM

#

anyone down to help me with reflections and translations

silent plank Sep 18, 2019, 2:29 AM

#

to make calculations less tedious we are going to use theta which was calculated from 90-79 = 11 degrees

#

what is the length of the side opposite theta?

scarlet halo Sep 18, 2019, 2:30 AM

#

is it just tan( 11 ) ?

silent plank Sep 18, 2019, 2:31 AM

#

tan(11) will be the ratio
but you need to know the sides involved

#

so which side would be opposite theta (the side that theta isn't touching)

scarlet halo Sep 18, 2019, 2:32 AM

#

x ?

silent plank Sep 18, 2019, 2:32 AM

#

yeh

#

and which side would be adjacent to theta?

#

(the side that theta is touching that isn't the hypotenuse)

scarlet halo Sep 18, 2019, 2:33 AM

#

O ok

#

so the 95 length side

silent plank Sep 18, 2019, 2:34 AM

#

yeh

#

so what would the ratio be for tan(11 deg)

scarlet halo Sep 18, 2019, 2:35 AM

#

x/95 ?

silent plank Sep 18, 2019, 2:35 AM

#

yep

scarlet halo Sep 18, 2019, 2:36 AM

#

so tan(11) = x/95 ?

silent plank Sep 18, 2019, 2:36 AM

#

yeh

#

and then solve for x and use a calculator

#

hmm, it seems they rounded incorrectly

quiet mason Sep 18, 2019, 2:36 AM

#

,w tan 11

somber coyoteBOT Sep 18, 2019, 2:36 AM

#

📎 wolf.png

Display results online and refine query

quiet mason Sep 18, 2019, 2:37 AM

#

,w cot 11

somber coyoteBOT Sep 18, 2019, 2:37 AM

#

📎 wolf.png

Display results online and refine query

quiet mason Sep 18, 2019, 2:37 AM

#

theyre right

scarlet halo Sep 18, 2019, 2:37 AM

#

for some reason tan(11) *95 is -21465 on my calcuator

quiet mason Sep 18, 2019, 2:37 AM

#

/95??

#

uhm convert to degrees

silent plank Sep 18, 2019, 2:37 AM

#

||18.466... rounds to 18.47 not 18.46||

quiet mason Sep 18, 2019, 2:38 AM

#

tan 11 radians is -225

scarlet halo Sep 18, 2019, 2:38 AM

#

um do u know how to change that is it in mode ?

#

O nevermind i got it

#

thanks

#

for this one is the answer 173.26 ?

📎 unknown.png

quiet mason Sep 18, 2019, 2:41 AM

#

,w cot22*70

somber coyoteBOT Sep 18, 2019, 2:41 AM

#

📎 wolf.png

Display results online and refine query

quiet mason Sep 18, 2019, 2:42 AM

#

ye

#

youre correct

scarlet halo Sep 18, 2019, 2:42 AM

#

thanks for all the help

upper karma Sep 18, 2019, 3:05 AM

#

can anyone help me with translations and reflections

#

Sure.

#

Like the entire concept of it?

#

like homework

#

lol

#

Oh.

#

Send problem.

#

📎 unknown.png

#

Okay let's look at from A to B first.

#

What x movements and what y movements does it take?

#

im not sure

#

what ya mean

#

like a rgid motion?

#

Yes.

#

It is a translation after all.

#

so i need

#

to find

#

bdc

#

?

#

You just need to find the horizontal movement and vertical movement from a to b...?

#

oo my b

#

i get it lol

#

👍

#

wait hold up

#

📎 unknown.png

#

was i suppoed to do that

upper karma Sep 18, 2019, 3:36 AM

#

@upper karma No.

#

From A to B.

#

It moves 2 horizontally and 3 vertically.

#

~~Wherever you got -5 and 1 there's no such thing.~~

#

So... if translations are marked by $T_{(x,y)}$

somber coyoteBOT Sep 18, 2019, 3:38 AM

#

leviosa:

📎 604465962053533709.png

upper karma Sep 18, 2019, 3:38 AM

#

Where x and y are the x and y movements.

#

What would the first translation be?

dusty sleet Sep 18, 2019, 9:51 AM

#

📎 unknown.png

#

📎 unknown.png

#

use the graph below to find wich interval changes the smallest on avg

#

is this just y2-y1/x2-x1?

dark sparrow Sep 18, 2019, 10:04 AM

#

PARENS

dusty sleet Sep 18, 2019, 10:06 AM

#

whats that

dark sparrow Sep 18, 2019, 10:08 AM

#

you forgot parentheses

dusty sleet Sep 18, 2019, 10:08 AM

#

oh sorry

dark sparrow Sep 18, 2019, 10:08 AM

#

you didn't mean $y_2 - \frac{y_1}{x_2} - x_1$, did you?

somber coyoteBOT Sep 18, 2019, 10:08 AM

#

Ann:

📎 183668144404037632.png

dusty sleet Sep 18, 2019, 10:08 AM

#

i ment this

#

📎 unknown.png

dark sparrow Sep 18, 2019, 10:09 AM

#

(y_B - y_A)/(x_B - x_A)

dusty sleet Sep 18, 2019, 10:10 AM

#

allright, well is that all? all there is to differentials

#

how ever it is called in english

dark sparrow Sep 18, 2019, 10:12 AM

#

difference quotient

#

yes

dusty sleet Sep 18, 2019, 10:13 AM

#

okay thanks !

worthy root Sep 18, 2019, 2:15 PM

#

Thats is called slope? @dark sparrow

dark sparrow Sep 18, 2019, 2:15 PM

#

it's the slope of a line going through two points sure

worthy root Sep 18, 2019, 2:15 PM

#

Yes.

rigid wharf Sep 18, 2019, 7:53 PM

#

Which statement accurately describes how to perform a 180° rotation of point A (−2, 3) around the origin?

A- Create a circle with the origin as its center and a radius of the origin and point A, then locate a point on the circle that is 180° from point A.

B- Create a circle with point A as its center and a radius of the origin and point A, then locate a point on the circle that is 180° from point A.

C- Create a line parallel to the x‒axis from point A, and locate a point on the perpendicular line that is equidistant to the distance between the y-axis and A.

D- Create a line perpendicular to the y-axis from point A, and locate a point on the perpendicular line that is equidistant to the distance between the y-axis and A.

zenith ember Sep 18, 2019, 7:55 PM

#

Which one do you think it is?

rigid wharf Sep 18, 2019, 8:04 PM

#

A?

#

It is A

novel flax Sep 18, 2019, 10:43 PM

#

A

silent plank Sep 18, 2019, 11:13 PM

#

when is cosec positive?

#

which quadrants

#

does option A represent that?

rancid mantle Sep 19, 2019, 12:58 PM

#

.

#

A car whose speed is 20 m/s passes a stationary motorcycle which immediately gives chase with a
constant acceleration of 2.4 m/s2

#

help plz

silent plank Sep 19, 2019, 12:59 PM

#

this is neither geo nor trig nor a question

rancid mantle Sep 19, 2019, 1:24 PM

#

a.How far will the motorcycle travel before catching the car?

#

oops LOL

sudden locust Sep 19, 2019, 6:56 PM

#

If anyone want a deeper understanding of The Vertical Angles Theorem this might help: https://youtu.be/YDbfSTHXzLE 🥰

YouTube

Simple Solutions Math Videos

Vertical Angles Theorem 🔺 | (Mini Series Part 1) | The Proof

Welcome to part 1 of my Vertical Angles Theorem Mini Series. In this video, we are going to take a look at the proof, which shows why vertical angles are equ...

▶ Play video

jagged lantern Sep 19, 2019, 8:02 PM

#

Can someone help me with something?
I have a question that asks "Prove the following trig identities: tanY cosY = sinY"

#

i don't know how to

#

if anyone can help please ping me 😄

harsh panther Sep 19, 2019, 9:06 PM

#

Tell me what is sinY/cosY
For a right angled triangle where h represents the height of the side opposite to angle Y and b is the adjacent side while H is hypotenuse
Then tell me what tan Y is.

#

@jagged lantern

coarse bronze Sep 19, 2019, 11:11 PM

#

stop advertising smh @sudden locust

upper karma Sep 20, 2019, 12:01 AM

#

@sudden locust Imagine having anime music as your intro. Cool video ig.

novel flax Sep 20, 2019, 2:04 AM

#

if you have a triangle with sides a b and c how do you find the area of the triangle

silent plank Sep 20, 2019, 2:09 AM

#

have you ever heard of heron's?

median crown Sep 20, 2019, 2:12 AM

#

@upper karma you can use heron's or law of cosines

#

you should probably use heron's

#

📎 unknown.png

#

📎 unknown.png

#

a, b, and c are the side lengths

silent plank Sep 20, 2019, 2:16 AM

#

(s is called the semi-perimeter)

hard lotus Sep 20, 2019, 2:54 AM

#

I have been stuck on this problem for a very long time. I have to prove they intersect and i just don't get what to do, I've asked friends who also try and fail and I even looked it up once after just having given up and I just don't get it.

Given a hexagon ABCDEF inscribed in a circle with AB = BC, CD = DE, EF = FA, show that AD, BE, and CF are concurrent.

📎 5617482c5a0ac352c6414a2869debcf13f2c721f.png

#

gotta use i think inscribed angles and such to prove it but I'm just not getting it

#

can someone help please

#

i just feel stupid...

scarlet halo Sep 20, 2019, 5:21 AM

#

is the answer to this 69.21 ?

📎 unknown.png

upper karma Sep 20, 2019, 5:30 AM

#

Yes.

#

But should be 69.22.

#

You end up with: 69.2204 which ~ 69.22.

upper karma Sep 20, 2019, 8:08 AM

#

Hello there

#

Anyone here ?

#

;-;

dark sparrow Sep 20, 2019, 8:11 AM

#

no obviously not

#

not until you post your question at least

#

@upper karma

tepid bone Sep 20, 2019, 11:05 AM

#

Need help with Q3 please, been stuck on it for half an hour now woke

📎 JPEG_20190920_190452.jpg

dark sparrow Sep 20, 2019, 11:17 AM

#

,rotate -90

somber coyoteBOT Sep 20, 2019, 11:17 AM

#

📎 624561870509965333.png

jagged sandal Sep 20, 2019, 11:20 AM

#

Being discussed in #help-2 for those that want to get updated on this.

dark sparrow Sep 20, 2019, 11:24 AM

#

oh what the fuck

#

what part of "don't post the same problem across multiple channels" do people not understand

sly marlin Sep 20, 2019, 1:41 PM

#

Update, finished being discussed in #help-2

sudden locust Sep 20, 2019, 3:07 PM

#

@upper karma Haha thanks!

#

@coarse bronze Sorry if it came of as shameless

upper karma Sep 20, 2019, 7:05 PM

#

📎 image0.png

#

Here is my question

#

Do ping me if anyone wants to discuss

dark sparrow Sep 20, 2019, 7:06 PM

#

@upper karma what have you tried so far

upper karma Sep 20, 2019, 7:07 PM

#

Yeah here is my approach

#

📎 image0.jpg

#

The thing is in the video using a different approach a solution has been shown with answer as around 32.6 (approx.) while by using my very simple reasoning i am getting 32.8 , is my approach to the problem correct ?

#

Here s the link to the video

#

https://youtu.be/YGyR4UiMRpc

YouTube

MindYourDecisions

What Is The Square's Area?

This is a tricky geometry problem with an elegant solution! Sources Adapted from past SEAMO problem (emailed to me) Cut The Knot similar problem https://www....

▶ Play video

dark sparrow Sep 20, 2019, 7:10 PM

#

what's this meant to be

📎 unknown.png

#

sqrt(3)?

upper karma Sep 20, 2019, 7:11 PM

#

Yes , but is it not clear ?

dark sparrow Sep 20, 2019, 7:11 PM

#

your three barely registers as such

#

anyway

#

this uh

#

this would make the vertical leg of that triangle sqrt(22)

upper karma Sep 20, 2019, 7:12 PM

#

Ha ha i wonder you remember or not you helped in a permutation and combination a month ago and gave the same remarks about my handwriting

dark sparrow Sep 20, 2019, 7:12 PM

#

and yet you've got it marked there as sqrt(3)

#

to the left

#

so what gives

upper karma Sep 20, 2019, 7:12 PM

#

What ? Let me check

#

Thats just rough work not a marking

dark sparrow Sep 20, 2019, 7:13 PM

#

your thing just falls apart

#

it all

#

falls apart

upper karma Sep 20, 2019, 7:14 PM

#

Yes i recognised my mistake

#

I assumed the values , got lucky and therefore got a right answer

#

There are other combinations which would result into sqrt 17 but my foolish self failed to understand that

#

Thanks for help tho.

dark sparrow Sep 20, 2019, 7:15 PM

#

📎 unknown.png

#

and this all goes against your figure being a square

#

bc 4 + sqrt(3) is not 1 + sqrt(22)

upper karma Sep 20, 2019, 7:17 PM

#

Yes , in excitement i kind of messed up the whole thing , but how did you do those marking so quick ? You have an apple pencil or stylus kind of thing ?

#

Oh wait those are typed

#

My bad

dark sparrow Sep 20, 2019, 7:23 PM

#

yes those are typed

#

i did that in microsoft paint lol

hard lotus Sep 20, 2019, 11:25 PM

#

i love how my question was just ignored

sly marlin Sep 20, 2019, 11:44 PM

#

@hard lotus shall we go through it?

#

can't seem to find it, mind posting again?

#

https://discordapp.com/channels/268882317391429632/326138757474680852/624438277482610718

#

okay this

hard lotus Sep 20, 2019, 11:46 PM

#

yeah

sly marlin Sep 20, 2019, 11:46 PM

#

Oh, the concurrency point is a centre of triangle ACE

#

Proof is literally about 5 lines at most

#

Focus on triangle ACE

#

What are lines AD, CF and BE?

hard lotus Sep 20, 2019, 11:47 PM

#

they all converge at the same point

#

i mean intersect

sly marlin Sep 20, 2019, 11:47 PM

#

no, what are they

hard lotus Sep 20, 2019, 11:47 PM

#

and i think they are same length?

sly marlin Sep 20, 2019, 11:47 PM

#

Not necessarily

hard lotus Sep 20, 2019, 11:47 PM

#

they are angle bisectors?

sly marlin Sep 20, 2019, 11:47 PM

#

How do you show that?

hard lotus Sep 20, 2019, 11:48 PM

#

show they are angle bisectors?

sly marlin Sep 20, 2019, 11:48 PM

#

yeah

#

of which angles?

hard lotus Sep 20, 2019, 11:48 PM

#

they bisect angle B angle F and angle D

#

but how do i prove they are bisecting them

#

i mean i think they bisect those angles

sly marlin Sep 20, 2019, 11:48 PM

#

focus on triangle ACE

#

well, you can find the angles based on the chord of the circle

#

@hard lotus

hard lotus Sep 20, 2019, 11:50 PM

#

ok

sly marlin Sep 20, 2019, 11:51 PM

#

try out

#

what angle relations do the equal chords give you?

night dew Sep 21, 2019, 9:14 AM

#

Hello, I'm trying to understand what my notes are lul , i wrote that to check if something is linearly dependant, we can use this theorem:

somber coyoteBOT Sep 21, 2019, 9:14 AM

#

Alaanor:

📎 213397906571395072.png

night dew Sep 21, 2019, 9:15 AM

#

however, I also noted that we can use his contraposed version:

#

but I can't understand what does the second one implies

somber coyoteBOT Sep 21, 2019, 9:19 AM

#

Alaanor:

📎 213397906571395072.png

spark stag Sep 21, 2019, 9:22 AM

#

if vectors $\vec{a}_1, \vec{a}_2, \vec{a}_3$ are linearly independent, then $\lambda_1 \vec{a}_1 + \lambda_2 \vec{a}_2 + \lambda_3 \vec{a}_3 = 0$ if and only if all the scalars are 0 (i.e. $\lambda_1 = \lambda_2 = \lambda_3 = 0$)

somber coyoteBOT Sep 21, 2019, 9:23 AM

#

Namington:

📎 105878844061286400.png

night dew Sep 21, 2019, 9:24 AM

#

omg yes, i'm dumb

spark stag Sep 21, 2019, 9:24 AM

#

if they're linearly dependent, then there exist some scalars not all equal to 0 such that $\lambda_1 \vec{a}_1 + \lambda_2 \vec{a}_2 + \lambda_3 \vec{a}_3 = 0$

somber coyoteBOT Sep 21, 2019, 9:24 AM

#

Namington:

📎 105878844061286400.png

spark stag Sep 21, 2019, 9:24 AM

#

so if you can show that those scalars exist and are not all zero

#

you've proven linear dependence

night dew Sep 21, 2019, 9:25 AM

#

awesome ! thanks

spark stag Sep 21, 2019, 9:25 AM

#

(this is usually done by just giving the scalars)

night dew Sep 21, 2019, 10:33 AM

#

Problem: Be ABC a triangle, let I, J, K being defined as $\vec{AI} = {1\over 3}\vec{AB}, \vec{CJ} = {1\over 3}\vec{CA}, \vec{CK} = {1\over 3}\vec{BC}$. Show that I, J, K are aligned.

somber coyoteBOT Sep 21, 2019, 10:33 AM

#

Alaanor:

📎 213397906571395072.png

night dew Sep 21, 2019, 10:34 AM

#

So I'd check if IJK are colinear by proving by example $\vec{IK} = \lambda\vec{IJ}$

somber coyoteBOT Sep 21, 2019, 10:34 AM

#

Alaanor:

📎 213397906571395072.png

night dew Sep 21, 2019, 10:35 AM

#

but idk how I should actually prove that, how can I find lambda ? I'm not asking for a complete solution, just which way I should take, how should I approach this.

sly marlin Sep 21, 2019, 10:36 AM

#

hmm

#

it looks like Menelaus

#

https://en.wikipedia.org/wiki/Menelaus's_theorem

night dew Sep 21, 2019, 10:38 AM

#

Haven't got that far in the theory, I'm at uni since 1 week only. Learned vector this week. LUL there must be an another way to prove it

#

but this indeed look like this case

sly marlin Sep 21, 2019, 10:39 AM

#

~~Project onto an arbitrary vector~~

#

no, not that

#

Well you can check that two vectors are parallel

#

so you probably can use that to prove that the 3 points are collinear

night dew Sep 21, 2019, 10:41 AM

#

which is $\vec{IK} = \lambda\vec{IJ}$

somber coyoteBOT Sep 21, 2019, 10:41 AM

#

Alaanor:

📎 213397906571395072.png

night dew Sep 21, 2019, 10:43 AM

#

mhhh in the corrected exercices they compare IJ to JK and prove that it is equals

#

WUT

#

I mean yes but I would never know that without drawing the situation

sly marlin Sep 21, 2019, 10:49 AM

#

yeah, this is probably the cleanest way to do it

#

without invoking menelaus

sudden locust Sep 21, 2019, 10:52 AM

#

Pythagorean Theorem 🕊️ But with semicircles... I put a lot of effort in to this one and think many of you guys will genuinely find it interesting... 🥰 Hope you have a wonderful day. 🐛https://youtu.be/Ir3MUW_VkgQ

YouTube

Simple Solutions Math Videos

Pythagorean Theorem 🕊️ - But With SEMICIRCLES?!

I think most of you guys are familiar with the Pythagorean Theorem. But im not quite sure you are familiar with the theorem but with semicircles? Join the co...

▶ Play video

dark sparrow Sep 21, 2019, 11:50 AM

#

familiar to the Pythagorean theorem

#

megathink

somber coyoteBOT Sep 21, 2019, 12:46 PM

#

euler/fermat:

📎 73695295149453312.png

#

euler/fermat:

📎 73695295149453312.png

dark sparrow Sep 21, 2019, 12:47 PM

#

yes, $(a \sin(C))^2 = a^2 \sin^2(C)$

somber coyoteBOT Sep 21, 2019, 12:47 PM

#

Ann:

📎 183668144404037632.png

dark sparrow Sep 21, 2019, 12:47 PM

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$(xy)^2 = x^2y^2$

somber coyoteBOT Sep 21, 2019, 12:47 PM

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Ann:

📎 183668144404037632.png

dark sparrow Sep 21, 2019, 12:53 PM

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I just know there are some rules about inverse trig functions and so on
:?

somber coyoteBOT Sep 21, 2019, 12:57 PM

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euler/fermat:

📎 73695295149453312.png

dark sparrow Sep 21, 2019, 12:58 PM

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cos^-1 is an awful notation that should never be used because cos doesn't have a true inverse

upper karma Sep 21, 2019, 3:27 PM

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<@&286206848099549185> what is section formulae?

upper karma Sep 21, 2019, 4:04 PM

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@dark sparrow what should be used? arccos?

dark sparrow Sep 21, 2019, 4:04 PM

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yes

upper karma Sep 21, 2019, 4:05 PM

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Why so? I think its common knowledge on the restriction of the domain

dark sparrow Sep 21, 2019, 4:05 PM

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i've seen it misused one too many times

quiet mason Sep 21, 2019, 4:08 PM

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@upper karma a+b=b+a

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and please read rules

upper karma Sep 21, 2019, 4:31 PM

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i said the section formula in coordinate geometry?

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how do you prove it?

dark sparrow Sep 21, 2019, 4:43 PM

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what is the "section formula"

dark sparrow Sep 21, 2019, 5:20 PM

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extended law of sines?

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what... is that meant to be

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oh the circumcircle radius thing

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...honestly there's no good answer whether that's "3rd grade" or not

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why does it matter

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like... idk, i wouldn't say trig itself is a thing that could be meaningfully taught as early as 3rd grade

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you're not seriously suggesting giving a problem like that to 9 y/os are you

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thonk

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i mean

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i guess a smart 9 y/o could have the necessary algebra background i guess

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but this all sounds a bit... out there to me

flint pelican Sep 21, 2019, 6:01 PM

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📎 image0.jpg

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Dude that problem is insane I can’t even do that and you want to give that to a 3rd grader

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Kids these days are so smart

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Guys for this I can see now that 12^2 144 so if we double the force since they are inversely related that means that we need to divide 144 by 2 I think so 72

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But the problem factors 72 into 6 square root of 2 and I’m not sure why

flint pelican Sep 21, 2019, 6:40 PM

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<@&286206848099549185>

lunar crow Sep 21, 2019, 6:40 PM

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is that

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the ACT

flint pelican Sep 21, 2019, 6:43 PM

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Yes this is a question from an act practice test

upper karma Sep 21, 2019, 6:53 PM

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📎 unknown.png

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how do i find ps? im retarded

mighty gull Sep 21, 2019, 7:05 PM

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so the just want the radial angle right

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have you heard of inscribed angle theorem

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@upper karma

upper karma Sep 21, 2019, 7:05 PM

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yes it htought that the angle was 95 or 85 but im not sure

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the arc*

mighty gull Sep 21, 2019, 7:06 PM

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can you apply inscribed angle theorem here

upper karma Sep 21, 2019, 7:06 PM

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i only had 2 opportunity to check if the answer was correct so idk if my answer was correct

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and i wanted to know

mighty gull Sep 21, 2019, 7:06 PM

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95 and 85 arent answers

upper karma Sep 21, 2019, 7:06 PM

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ok

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oh

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its fucking 80

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lol

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im retarded

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wtf

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i didn't even see it

mighty gull Sep 21, 2019, 7:07 PM

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yeah. dw its fine, we all miss things like this all the time

upper karma Sep 21, 2019, 7:07 PM

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haha

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i was trying to figure out the left and right angles from the center of the circle

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and i didn't even see that the 40

mighty gull Sep 21, 2019, 7:09 PM

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its fine

flint pelican Sep 21, 2019, 8:13 PM

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📎 image0.jpg

primal lynx Sep 21, 2019, 8:20 PM

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well if ur asking about the first one

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choice k is the choice that makes the least sense

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bc it says varies inversely with the square of the distance

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so farther away less powerful

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k is the only choice thats further

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but if you were to write an expression with proportionality constant say k

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you can write the inverse square as like 1/d^2 is equal to k 1/r^2

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but k is 2 right

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bc twice as strong

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and d is already 12

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then you get j

flint pelican Sep 21, 2019, 8:52 PM

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Yeah the answer is j but like

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This is what I said earlier

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Guys for this I can see now that 12^2 144 so if we double the force since they are inversely related that means that we need to divide 144 by 2 I think so 72

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But the problem factors 72 into 6 square root of 2 and I’m not sure why

spark stag Sep 21, 2019, 8:56 PM

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they're not inversely related

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there's an inverse relation with the square of the distance

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so suppose the distance is 2, then $F \propto \frac{1}{d^2} = \frac{1}{2^2}$

somber coyoteBOT Sep 21, 2019, 8:57 PM

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Namington:

📎 105878844061286400.png

spark stag Sep 21, 2019, 8:58 PM

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in this case, we have $F$ and we want $2F$

somber coyoteBOT Sep 21, 2019, 8:58 PM

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Namington:

📎 105878844061286400.png

spark stag Sep 21, 2019, 8:58 PM

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we know $F \propto \frac{1}{d^2}$, so how do we double F?

somber coyoteBOT Sep 21, 2019, 8:59 PM

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Namington:

📎 105878844061286400.png

flint pelican Sep 21, 2019, 9:00 PM

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By dividing 144 by 2

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Right?

spark stag Sep 21, 2019, 9:01 PM

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so you're saying that it needs to get farther away for the magnet to get stronger?

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what we're trying to do is change the distance

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144 represents the distance squared

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not the distance

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you're right that we want the distance squared to be 72

flint pelican Sep 21, 2019, 9:02 PM

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Ooooooooooooooooooooooooooooo

spark stag Sep 21, 2019, 9:02 PM

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so what does the distance have to be?

flint pelican Sep 21, 2019, 9:02 PM

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The square root of 72

spark stag Sep 21, 2019, 9:02 PM

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right

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now make that a mixed radical

flint pelican Sep 21, 2019, 9:02 PM

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Because 72 would be the distance squared

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Yeah then I see that factors to 6 square root of 2

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Thank you

spark stag Sep 21, 2019, 9:03 PM

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sounds good.

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and we can test this

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,calc 1/(12^2)

somber coyoteBOT Sep 21, 2019, 9:03 PM

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Result:

0.0069444444444444

spark stag Sep 21, 2019, 9:03 PM

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,calc 1/((6sqrt(2))^2)

somber coyoteBOT Sep 21, 2019, 9:03 PM

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Result:

0.013888888888889

spark stag Sep 21, 2019, 9:03 PM

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,calc 0.13888888 / 2

somber coyoteBOT Sep 21, 2019, 9:03 PM

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Result:

0.06944444

spark stag Sep 21, 2019, 9:04 PM

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er

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missed a 0

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,calc 0.01388888 / 2

somber coyoteBOT Sep 21, 2019, 9:04 PM

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Result:

0.00694444

spark stag Sep 21, 2019, 9:04 PM

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there we go

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yep, thats correct.

primal lynx Sep 21, 2019, 9:07 PM

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lol

median crown Sep 21, 2019, 11:11 PM

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Flip everything upside down

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nice dude

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spilled your gfuel

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ye u did it right

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except

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your scale labels are wrong

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you put pi/2<pi/4

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yeah

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yeah

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you have pi/3 on the left side of the y axis too

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wtf

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why

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you should fix your handwriting

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the handwriting litearlly makes it look like a 3

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sure

solid nexus Sep 22, 2019, 12:39 AM

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hi can someone help me?

sly marlin Sep 22, 2019, 12:44 AM

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JustAsk @solid nexus

delicate lake Sep 22, 2019, 3:30 AM

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Am I supposed to use sin, cos or tan?

📎 20190921_202830.jpg

wise pawn Sep 22, 2019, 3:31 AM

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do you know SOH CAH TOA?

silent plank Sep 22, 2019, 3:31 AM

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which one do you think it is and why?

delicate lake Sep 22, 2019, 3:33 AM

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This trade school is giving me trig questions and I've never even took the class.

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Trying to teach myself through online videos

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Trying to find hypotenuse.. so i use Tan? @silent plank

silent plank Sep 22, 2019, 3:36 AM

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no

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what is the ratio of tan(x)?

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do a quick search and/or use the mnemonic shown above

delicate lake Sep 22, 2019, 3:37 AM

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This is above my pay grade. I have TI-84 Plus calculator but not every experienced with it

silent plank Sep 22, 2019, 3:38 AM

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what i'm asking has nothing to do with a calculator

delicate lake Sep 22, 2019, 3:38 AM

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My notes to help me solve the problem. Idk the answer to your question

📎 20190921_203713.jpg

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I'm literally trying to teach myself

silent plank Sep 22, 2019, 3:39 AM

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and what does it say on your notes?

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tan(\theta) = ?

delicate lake Sep 22, 2019, 3:39 AM

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Opposite/Adjacent

silent plank Sep 22, 2019, 3:40 AM

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would that be appropriate here?

delicate lake Sep 22, 2019, 3:40 AM

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Oh no.. lol but there is no 2 sides given.

silent plank Sep 22, 2019, 3:41 AM

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which side is given relative to the angle?

umbral snow Sep 22, 2019, 3:42 AM

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c isn't given, but pretend it is for now

delicate lake Sep 22, 2019, 3:42 AM

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Adjacent and Hypotenuse so i use COS

silent plank Sep 22, 2019, 3:42 AM

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yes

delicate lake Sep 22, 2019, 3:42 AM

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?

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Oh snap

silent plank Sep 22, 2019, 3:43 AM

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well to be more specific, adj is given, hyp is unknown so you use cos

delicate lake Sep 22, 2019, 3:43 AM

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.883 is the answer

silent plank Sep 22, 2019, 3:43 AM

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no

delicate lake Sep 22, 2019, 3:43 AM

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Rip

silent plank Sep 22, 2019, 3:44 AM

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can the hypotenuse be shorter than the other sides?

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you must've done something wrong in your calculattions

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what equation did you start with?

delicate lake Sep 22, 2019, 3:45 AM

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📎 20190921_204328.jpg

umbral snow Sep 22, 2019, 3:45 AM

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You're getting ahead of yourself. What's your cos equation?

delicate lake Sep 22, 2019, 3:47 AM

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Adjacent divided by hypotenuse but hypotenuse isn't given

graceful hemlock Sep 22, 2019, 3:47 AM

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Is it in radians or degree mode LOL

silent plank Sep 22, 2019, 3:47 AM

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cos(28 deg) = adj/hyp = ?

umbral snow Sep 22, 2019, 3:47 AM

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That's okay! Write it out anyway

silent plank Sep 22, 2019, 3:50 AM

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what is the adjacent side?

delicate lake Sep 22, 2019, 3:50 AM

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1.5

silent plank Sep 22, 2019, 3:50 AM

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what is the hypotenuse represented by?

delicate lake Sep 22, 2019, 3:51 AM

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C

silent plank Sep 22, 2019, 3:51 AM

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what is adj/hyp

delicate lake Sep 22, 2019, 3:52 AM

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1.5/C

silent plank Sep 22, 2019, 3:52 AM

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and what will cos(28 deg) be?

delicate lake Sep 22, 2019, 3:52 AM

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Thank you.

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1.5/.8829

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=1.69

silent plank Sep 22, 2019, 3:54 AM

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that wasn't what i asked for but you found the expression for C
be careful with rounding

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technically the value they provided is wrong. but i think that's what they want you to choose.

delicate lake Sep 22, 2019, 3:57 AM

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I will re-read this convo for future simliar problems.

"which side is given relative to the angle?" Helps me decide which function to use.

But i do see where I was confused.

silent plank Sep 22, 2019, 3:57 AM

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,w 1.5/cos(28 deg) 4 decimal places

somber coyoteBOT Sep 22, 2019, 3:57 AM

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📎 wolf.png

Display results online and refine query

delicate lake Sep 22, 2019, 3:58 AM

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Ok

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Much appreciated.

silent plank Sep 22, 2019, 3:59 AM

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based on the options it seems that they rounded before dividing

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which you shouldn't do

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,w 1.5/0.883 4 decimal places

somber coyoteBOT Sep 22, 2019, 4:00 AM

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📎 wolf.png

Display results online and refine query

silent plank Sep 22, 2019, 4:01 AM

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as mentioned above, it seems they want you to pick 1.6988, but the "correct" value rounded to 4dp would be 1.6989

delicate lake Sep 22, 2019, 4:06 AM

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Hypotenuse & opposite given so I would use SIN.

SIN (39) =.62932

.62932 × 30 = 18.88

📎 20190921_210302.jpg

quiet mason Sep 22, 2019, 4:09 AM

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👌

delicate lake Sep 22, 2019, 4:11 AM

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This is 2.1849

📎 20190921_210756.jpg

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Sweet thanks. Appreciate the patience

novel flax Sep 22, 2019, 1:02 PM

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sin x = opp/hyp
sin 39 deg = 1.375 / c
c * sin 39 deg = 1.375
c = 1.375 / (sin 39 deg)
c = 2.1849

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you are correct @delicate lake

flint pelican Sep 22, 2019, 3:43 PM

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📎 image0.jpg

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Lost in 55 and 56

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For 56 isn’t the 5 the radius so 5 squared is 25 and then the height is 12 since this is a 51213 triangle so that’s 25pi*12=942.478 and divided by 3=314

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So that’s 100pi but that isn’t the answer the answer is j

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And for 55 what I guessed was that since 2 angles are 60 that must mean the other 2 are 120

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So I multiplied 4 by 2 to get 8 since 60*2=120