#geometry-and-trigonometry

1 messages · Page 279 of 1

winged pagoda
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wl=a no?

dark sparrow
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this is not a rectangle.

winged pagoda
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uhh, how do i know that?

dark sparrow
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what do you mean

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do you not know what a rectangle looks like

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for one, a rectangle has four corners, while this shape has six.

winged pagoda
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oh, im thinking about overall shape, sorry.

dark sparrow
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you can view this shape as a 10 by (6+x) rectangle with a 6 by (10-x) rectangle cut out, if you want.

winged pagoda
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yep, thank you 😄

upper karma
upper karma
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@upper karma google "lateral area of a cylinder"

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the lateral area is the rectangular part

tepid marlin
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Circumference of circle

upper karma
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@upper karma post a picture of the problem

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along with any diagram they offer

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because I think you're missing something

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a diameter maybe?

silent plank
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consider the formulae for Volume and the lateral surface area

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the lateral surface area is just h*circumference (or 2 pi *r*h )

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$\begin{cases}
\frac{100}{360} \cdot \pi r^2h = 20\pi cm^3 \
2\pi rh = 48 \pi cm^2
\end{cases}$

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those would be the respective equations you can get from volume and lateral surface area

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solving that system will get you the values of r and h

upper karma
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pi r^2 h = 20 pi is for the 100 degree wedge

silent plank
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wait 1 sec

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glossed over that part

upper karma
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maybe if you turn 100 degrees into radians?

somber coyoteBOT
upper karma
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I thought about 100/360, but wouldn't it be better if it was in radians?

silent plank
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you'd end up getting the same thing

upper karma
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@silent plank how do you make a system on LaTex?

silent plank
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\begin{cases}
\end{cases}

upper karma
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Woah cool

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Thx

rich wolf
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,calc 100 deg to radian

somber coyoteBOT
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Result:

1.7453292519943 radian
silent plank
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additional conversions aren't really needed

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consider dividing eq1 by eq2

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yes

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you did understand why i put that 100/360 there right?

fringe canyon
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hello, i'need help with this exercise

silent plank
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in use

fringe canyon
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should i post in questions channel?

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i am new to discord

silent plank
upper karma
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why have I been put on slow mode for 1 hour?
EDIT: apparently there's an AP (advanced placement) test going on in America so we all have to suffer for 1h in order to prevent widespread cheating
EDIT 2: yep
EDIT 3: also, I solved the cylinder wedge problem

upper karma
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ok, did you solve the cylinder wedge problem?

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PS: if you can't post, close Discord and reopen

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@upper karma

pallid lichen
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@silent plank

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Can u help me with something quick?

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it will take a few min im just confused

silent plank
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probably

upper karma
pallid lichen
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@silent plank

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so

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whenever I do this

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then I used sin=0/h

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but i get different heights

silent plank
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oh discussions already going on in the calc channel

pallid lichen
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ye lol

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my parents made me do

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my brothers work..

upper karma
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law of sines

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find angle A

pallid lichen
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i dont remember this stuff

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ye i did

upper karma
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actually, wait

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it's easy

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all angles from a triangle add up to 180

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20 + 18 + ? = 180

pallid lichen
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ye

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142

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i did it

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i used sine law

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but then im getting 2 different heights

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i got 50.19=c
55.55=b

silent plank
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b and c aren't the altitudes of the triangle

pallid lichen
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i found the side lengths

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so then i can use

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sin=o/h

upper karma
pallid lichen
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ye but

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isn't what i did alright?

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i used the entire angle

silent plank
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,w 50.19sin(20deg)

pallid lichen
somber coyoteBOT
pallid lichen
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huh

silent plank
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,w 55.55sin(18deg)

somber coyoteBOT
pallid lichen
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lol i mixsed the angles

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did i

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wait

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no i did it right i thinl

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50.18sin18

silent plank
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you didn't label the diagram properly

pallid lichen
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where

silent plank
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from what i can make from the blur

pallid lichen
silent plank
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the work you did

pallid lichen
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yes

silent plank
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c is the side opposite angle C

pallid lichen
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oh right lol

silent plank
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i.e. AB

pallid lichen
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ohh OOPS

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omg

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i was more focused on the

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variable

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or idk what i was thinking lol

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yay thanks

silent plank
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there might be a shortcut

pallid lichen
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thanks

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i did it tho so

silent plank
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$h = \frac{base \cdot tan(\alpha)tan(\beta)}{tan(\alpha)+\tan(\beta)}$

pallid lichen
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what about that

somber coyoteBOT
pallid lichen
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wow when did u learn that

silent plank
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i forgot, i just derived it

pallid lichen
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is that uni level

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or u came up with it urself

silent plank
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from tan

pallid lichen
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wow cool

silent plank
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it should be documented

pallid lichen
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lol

silent plank
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,w 100*tan(18deg)tan(20deg)/ ((tan(18deg) + (tan(20deg))

pallid lichen
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uh

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lol

somber coyoteBOT
upper karma
pallid lichen
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lol ye

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i did it

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i just mixed the

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lengths up

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i labelled wrong

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thanks tho

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🙂

upper karma
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sure thing

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Given the cardboard inside the paper towel roll has a radius of 2cm and the distance from the center of the cardboard cylinder to the edge of the roll is 13 cm. If each paper towel had a thickness of 1 mm and the height of the paper towel roll is 19 cm, determine how long the paper towels would be when unrolled.
@upper karma

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the answer is probably something at the power of 110

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2 + 1.1^110 ?

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no, wait...

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2 pi r is the length of one ring

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so it needs to be multiplied by 110, but also taken into account that the radius increases by 0.1 each time

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sorry

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fixing it now

silent plank
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consider the formulae for the volume of paper when rolled and unrolled

upper karma
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consider the formulae for the volume of paper when rolled and unrolled
@silent plank
oh, man... that's pretty clever

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why didn't I think of that?

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sounds doable if you think of the unrolled paper as a giant rectangular prism with a very small height

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volume 1 = volume 2

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then you get the height lenght from there

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the width being the radius, of course

acoustic jungle
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ok

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👍

rotund vault
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Hello, may I have help with c? know the length is r = 41, and the center of the circle is (-1,3).

📎 unknown.png
upper karma
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Yep

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@rotund vault notice that O is the midpoint, use midpoint formula between both points

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Find midpoint

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Find radius

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Oh you got already the midpoint

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And then make it a equation of a circle

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Find radius

gilded parcel
upper karma
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Using distance formula

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@gilded parcel sine's law

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H,k Is Center of circle

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@gilded parcel understood what to do?

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(X-h) + (y-k) = r^2

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So just plug it in

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@rotund vault

gilded parcel
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Alr

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And yea

upper karma
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Lmk if you still need help

gilded parcel
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Thanks

upper karma
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Np catthumbsup

gilded parcel
upper karma
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what the...

gilded parcel
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Ikr

acoustic jungle
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there is nothing wrong with it

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this is just proof of cosine law.

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use pythagorean theorem to find b^2=DC^2+h^2

rotund vault
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Haha, yeah just got it. @upper karma

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x+1^2 + y-3^2 = 41^2

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Thanks though!

upper karma
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@upper karma I forgot about the surfaces on the inside 😐

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marked in green

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there's another one on the other side

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but you have the radius, you have the height...

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I'm gonna go to bed

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have fun 😉

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did you solve the paper towel question?

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One room is 5 m long, 4 m wide, and 3 m high. Up in one corner is a spider. It sees a dead fly in the lower opposite corner. The spider heads for a walk down to the fly. How long is the shortest route the spider can take to the fly?

📎 unknown.png
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spindel = spider
fluga = fly

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🙂

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the spider got drunk along the way

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Pythagoras' theorem

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well apparently not

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lol

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does it have glider wings?

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no xDDD

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this way might be shorter, idk

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how many metres is that?

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pythagora knows

vestal mauve
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Hello

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Can someone help me? I need help to pass this class lol

upper karma
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ok so

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5+5 = 10

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@vestal mauve as long as it's not an exam or a test, sure

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the answer is not 10 sadly :/

vestal mauve
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No it’s homework

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I’ma send a pic if that’s okay?

acoustic jungle
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@upper karma not shortest, not right

vestal mauve
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It like that like my@english not very good and I don’t know what it telling me

acoustic jungle
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@upper karma should be sqrt 41+3

upper karma
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@acoustic jungle nope :/

acoustic jungle
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it is.

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unless the spider can fly.

upper karma
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spider can't fly

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and there's another way

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which i personally don't understand

acoustic jungle
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ok if it's not sqrt(41)+3 what is it

upper karma
#

first of all

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you flatten the "room"

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so it's not three dimensional

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then you calculate the shortest

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it was something like that, give me one check i gotta check my notes

acoustic jungle
#

that doesn't make sense, what is the answer?

upper karma
#

8,6

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m

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so basically the spider opens a worm hole to another dimension that doesn't have a Y (up) axis?

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what do you mean flatten the "room"?

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so apparently you're suppose to think that the room is a box

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and then you flatten it out

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and from thereon you can calculate the shortest way

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Think of the room as a box. Unwrap the box so you'll see all the sides. Now point out the spider's position and the position of the opposite corner.
All you'll have to use the pythagora's theorem.

acoustic jungle
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ohhh

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nice

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looks like I'm dumb

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never thought of that

upper karma
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,w x^2 = (3+5)^2 + 4^2

somber coyoteBOT
upper karma
#

so

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4 * sqrt5

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,w sqrt(80)

somber coyoteBOT
upper karma
#

8.9

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8.944

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wrong answer :////

acoustic jungle
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depends on which side the fly is on

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the 4 meters away or 5 meters away

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it might be root of 7^2+5^2

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which is indeed 8.6

upper karma
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yeah? what if the spider falls down during the unfolding?

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well this is a problem where it's legit a robot

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it takes the shortest way

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without fail

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only success

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btw the pic you made

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pythagoras

acoustic jungle
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unfolding doesn't mean you unfold it

upper karma
#

64+36 = 10

acoustic jungle
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it means he's traveling on an angle to the ground

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then travaling on an angle to the fly

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very nice.

upper karma
#

the answer is 8,6

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i forgot my notes at school lol

acoustic jungle
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or sqrt(74)

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I learned something new today though.

upper karma
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yeah

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sqrt74 = 5,6

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8,6*

acoustic jungle
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yep.

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approx.

upper karma
#

how is it 74?

acoustic jungle
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sqrt((3+4)^2+5^2)

upper karma
#

why 3+4?

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hmmm...

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maybe the "room" is rotated 90 degrees?

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this is a 9th grade question

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lol

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math checks out

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wait what

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where it says 4, it's suppoe to be 5

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and where it says 5, it's suppose to be 4

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yes, because it's rotated 90 degrees

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or rather, you view it from a different angle

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the spider is still "up in one corner"

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doesn't specify which

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kind of a weird room, tbh

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if it's trig related, sure

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@vestal mauve which one did you have trouble with?

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I forgot about you 🙂

night summit
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oof

vestal mauve
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Honestly any I’ll get it quick

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It just different

acoustic jungle
night summit
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its fine disregard my question ill just do it tom

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wdym fish

acoustic jungle
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use the identity

night summit
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ill figure it out tom head isnt working anymore

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ty anyway

acoustic jungle
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you aren't supposed to derive the tan identity

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you can if you want.

upper karma
#

if that entire segment is 10

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and GF is 1

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how long do you think segment HG is?

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this is just addition and subtraction

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do you think it might be 10 - 1 ?

vestal mauve
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9

upper karma
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yes

vestal mauve
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Sorry I’m not good in math

upper karma
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can you figure out question 2?

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based on what you learned just now

vestal mauve
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Yes thak you

upper karma
#

what about this?

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the entire line segment is 26

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and this one?

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first you find the little "?"

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then you can get the bigger one

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@upper karma search the volume of a section of a circle

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@vestal mauve did you understand these so far?

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up until question 6

vestal mauve
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I just needded a like remindir

upper karma
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what did you get for 5) and 6)?

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@vestal mauve for question 5) it's 9+11 = 20, then you subtract that from the entire thing (which is 26)

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so 26 - 20 = 6

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question 6) is similar. First you get segment HI and then you add that 12 to get segment HJ

upper karma
#

@upper karma I wonder what happens to that 120 degree angle when it gets folded like this:

📎 unknown.png
candid terrace
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Hey everyone

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I’m having some trouble in geometry if anyone can spare some time to help me?

remote heart
upper karma
#

post your question

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yeah, don't ask if you can ask... just ask

candid terrace
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Ok

ionic bluff
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don't ask, don't tell

candid terrace
remote heart
#

do you know about inscribed angles?

candid terrace
#

Yes

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Are u asking about the theorem

remote heart
#

yes

candid terrace
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Yeah I know it

remote heart
#

okay

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so can you tell me what arc MNP is?

candid terrace
#

Yes

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Wouldn’t it be 146

remote heart
#

yes

candid terrace
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But that isn’t an answer

remote heart
#

and you want NP, which is part of that arc

candid terrace
#

Ohhhhh

remote heart
#

got it?

candid terrace
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Yeah thanks

remote heart
#

sure

candid terrace
#

Why did help me?

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What’s your motive?

remote heart
#

uhm

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I don't have one?

candid terrace
#

Just being nice

remote heart
#

to help you understand the problem, which is what I hope I did

candid terrace
#

Anyone that I’m stuck on

upper karma
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Exam @candid terrace

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?

candid terrace
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No

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It’s just homework

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A review

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Which is why it looks like a test

quartz pulsar
#

I also got a review

remote heart
#

try posting in a questions channel?

upper karma
#

@quartz pulsar plot it

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and maybe google what an "orthocenter" is

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@candid terrace search youtube for "circle theorems"

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the first 2-3 videos should be enough

quartz pulsar
#

Ummm, I need help from someone that actually knows the material.

candid terrace
#

Ok

remote heart
#

the people making the vids know the material

upper karma
candid terrace
#

I looked it

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I studied the theorems

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Let me know if anything looks wrong

upper karma
#

I should've been going to bed 3 hours ago

candid terrace
upper karma
#

g'night

candid terrace
#

Good night😴

barren tide
upper karma
#

Idk.

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isnt it 1/sqrt(2) ?

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I started by dividing by cos(x/3) first to get 1-tanx=0 then went from there.

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But ig that's wrong.

acoustic jungle
#

There is nothing wrong with the answer

upper karma
#

Yeah, ik.

acoustic jungle
#

they did cos(x/3)=sin(x/3) then squared both sides, converting sin^2(x/3) to 1-cos^2(x/3)

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and 1/sqrt2 =sqrt2/2

barren tide
#

thnx

prime barn
barren tide
#

why is it positive if it's in the third quadrant?

upper karma
#

It is what it is

barren tide
#

You sound like my mom

kindred girder
#

I have a coodinate geometry problem. I need to find the missing coodinate (-45, unknown) this coodinate is part of a parallel line which shares the same line with another coodinate (-90, 180). The line parallel to that one has the coodinate of (0,0) and (38, -535). They both have a gradient of -0.07. I need to find the missing coodinate.

#

<@&286206848099549185> (Sorry for the notify)

silent plank
#

the line connecting the points (0,0) and (38, -535) doesn't have a gradient of -0.07

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@kindred girder

visual mist
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im not sure how to work it out

vague pagoda
#

what have u tried?

kindred girder
#

What's their gradient then???

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@silent plank

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Sorry for the ping

visual mist
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well i found out that the cone volume is 18 pi?

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and so the ice cream volume should be 9pi since its half?

silent plank
#

what's a definition of the gradient?

upper karma
#

yes

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sushi

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wait

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yes

kindred girder
#

Like in the formula
Y=mx+c

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It's the m value

silent plank
#

and how do you find that value of m

kindred girder
#

I think I'm getting myself confused

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You use the formula (y2-y1)÷(x2-x1)

silent plank
#
  • m =
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(an expression by itself is not a formula)

visual mist
#

ok

kindred girder
#

That's how you find m

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Yes ik

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This whole thing is confusing

silent plank
#

and applying that, what do you get?

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so leave it as a fraction

kindred girder
#

That's how I got the negative number

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-0.07

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By using the 2 coordinates

silent plank
#

@visual mist finding the volumes isn't really needed, apply ratios of similar solids or something

visual mist
#

how

silent plank
#

tell me specifically how you are getting the value of -0.07

kindred girder
#

Ok

upper karma
#

@visual mist yes i got u

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i can explain

visual mist
#

ok

kindred girder
#

(38-0/-535-0) = -0.07 aka the "m" expression

silent plank
#

you're mixing your x and y coordinates

upper karma
#

nah i cant explain good

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ramanox u take it

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ramanov

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its just 1/2 cubed times the area of the cone

visual mist
#

lol

upper karma
#

and then thats the area of the ice cream

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volume i mean

kindred girder
#

Omg I am

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Woah thanks

upper karma
#

but ramonox will sure explain better

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@silent plank u got this

visual mist
#

ram is so busy XD

upper karma
#

ik lol

barren tide
#

How do you determine if there are one or two solutions that can be taken by a number?

📎 Problem_0001.PNG
silent plank
#

if the ratio of dimenstions (of similar objects) is:
1: a
the ratio of volume is
1 : a^3

visual mist
#

here comes another one lol

#

o

barren tide
#

oh shit ill wait

visual mist
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ohhh

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ok

upper karma
#

nah im wrong

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whatever i said awhile ago

visual mist
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lol

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wait so still how would i work out the depth

upper karma
#

whats depth

silent plank
#

should've said dimensions instead of sides

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anyway, you are told that for V,
the ratio is 1 : (1/2)

visual mist
#

yes

upper karma
#

cube root 1/2 times height?

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@silent plank

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for depth?

silent plank
#

yeh

upper karma
#

noice

visual mist
#

why cube root it

upper karma
#

ratio

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idk

silent plank
#

if the ratio of dimenstions (of similar objects) is:
1: a
the ratio of volume is
1 : a^3

visual mist
#

..

silent plank
#

cube root of a^3 is a

visual mist
#

yeah

silent plank
#

and 1: (cbrt(1/2) will be the ratio of the dimensions of the big to smaller cone

visual mist
#

wha

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im confused

upper karma
#

watch a video

#

i learned it better visually

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go look on youtube

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or khan academy

visual mist
#

gimme video link XD

upper karma
#

k wait

visual mist
#

ok

upper karma
#

lemme find one

silent plank
#

did you understand:

#

if the ratio of dimenstions (of similar objects) is:
1: a
the ratio of their volume is
1 : a^3

upper karma
silent plank
#

that doesn't seem related to volume

visual mist
#

yeah

upper karma
#

scale factor

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same idea here

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using scale factor stuff

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i guess

visual mist
#

but then the ice cream volume is half of the cone volume, why do u cube root it

upper karma
#

there u go

silent plank
#

if the ratio of dimenstions (of similar objects) is:
1: a
the ratio of their volume is
1 : a^3

#

in your question, 1/2 would be your a^3

visual mist
#

ok let me watch vid first

barren tide
#

Is it my turn now?

upper karma
#

yes son

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go for it

barren tide
#

How do you determine if there are one or two solutions that can be taken by a number?
@barren tide This

silent plank
#

properties of sin and the unit circle

barren tide
#

but what if its a number that isn't on the unit circle, like the example?

silent plank
#

there are 2 solutions to:
sin(x) = c
for: (0<c<1 and 0<x<180°)

#

wdym by isn't on the unit circle?

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(and then apply angle sum of a triangle or otherwise to determine whether the obtuse angle is a legitimate solution to the problem)

barren tide
#

Oh so I just have to test it? fuck me

dark sparrow
#

wdym by isn't on the unit circle?
they probably mean "the angle isn't an integer multiple of 30 or 45 degrees"

barren tide
#

ya that

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thnx

#

oh one more question

#

What about this? Why is it positive when it is considered in the third quadrant

upper karma
#

It is what it is 🤷‍♂️

silent plank
#

can you repost the image so i don't have to scroll

#

preferably one thats also clearer

upper karma
silent plank
#

like there's discord on the comp. and comps have screen cap tools

barren tide
#

kk gimme a sec

silent plank
#

pi < 3.805 < 3pi/2
is in quadrant 3

#

it doesn't matter if the angle is expressed as a negative value

#

what matters is the position / location

#

here it was shifted to quadrant 3 by doing
pi - (-0.663) = pi + 0.663 ~ 3.805

barren tide
#

Is it because it is between

pi < 3.805 < 3pi/2
is in quadrant 3
@silent plank

#

and negative isn't?

silent plank
#

,w sin( -999pi - arcsin(-8/13) )

somber coyoteBOT
silent plank
#

that angle is in Q3 and the sine of that will be negative

wicked raft
prime barn
graceful narwhal
#

@wicked raft what needs clarification?

#

@wicked raft A right cone has a slant height (m) twice as long as the radius of its base(r). Calculate

#

a) the ratio of the surface area to the base (M:A)

#

b) the central angle alpha of the lateral surface area

cosmic sun
#

acos(dot(normalize(r), vec2(0, 1)))

#

Should return the angle between the two vectors in radians, right?

#

Yet, the angle returned seemed to neither be in degrees or in radians, I've cheched

#

Could anyone help me with this?

#

Nvm, just used wrong vectors

proper citrus
#

need help with this lol

rich wolf
#

<@&286206848099549185>

proper citrus
upper karma
#

@rich wolf imagine tagging yourself xD

#

Im kidding

#

-------(free channel)-------

wicked raft
#

@graceful narwhal I got it now, thanks

dense sky
upper karma
#

Lol seems like a test @dense sky

dense sky
#

it isn’t

#

@upper karma

upper karma
#

Okok

dense sky
#

it’s like one of are final classroom assignments and the teacher decided to put a bunch of questions

#

cuz.. my teacher is annoying

#

so instead of like a 5 question assignment

#

hurr durr last week so it’s a bunch of questions

upper karma
#

State the circle's equation here on the chat.

dense sky
#

hk+yk= r2?

#

oh lol

#

x-h + y-k = r2

upper karma
#

$(x-h)² + (y-k)² = r²$

somber coyoteBOT
upper karma
#

@dense sky bruh you got answered already smh

dense sky
#

what?

upper karma
#

Yeah on another channel

#

You cant repeat questions on different channels

vague berry
#

hey guys

#

could anyone help me out

upper karma
#

@vague berry just post it JustAsk

vague berry
# 
sec x + tan^2 x = 0```
#

?

#

$tan(\pi +arcsin(-(2)/(3)))$

somber coyoteBOT
vague berry
#

how do you do this ^?

#

$cos(arcsc(tan(arcos 1/3)))$

upper karma
#

$cos(csc^{-1}(tan(cos^-1 1/3)))$

somber coyoteBOT
vague berry
#

please help

#

its not like that

#

it looks like cos(csc^-1 (tan (cos^-1 (-1/3))

#

like arccos

#

arccsc

upper karma
#

it looks like cos(csc^-1 (tan (cos^-1 (-1/3))
@vague berry so this is the question

vague berry
#

yeah

#

let's tackle this one 😄

upper karma
#

Are calc allowed

vague berry
#

like calculus?

#

no im in 10th grade

upper karma
#

Cos you can just plug those in

vague berry
#

no

upper karma
#

Calculator lol

vague berry
#

i cant use that

#

no

upper karma
#

Hm

vague berry
#

its gotta be like drawing those triangles

#

and like doing that

upper karma
#

Unit circle

vague berry
#

umm

#

idk

#

something like that i guess

#

@upper karma

#

please dont bail on me

#

<@&286206848099549185>

#

please help

#

im dying out here

#

$tan(π+arcsin(-2/3))$

somber coyoteBOT
vague berry
#

Test

versed aspen
#

what's troubling you?

vague berry
#

that ^^

#

monster above

versed aspen
#

Well like do you don't know how to start at all?

vague berry
#

im not allowed to plug in into a calculator or anything

#

i dont know how to at all

versed aspen
#

You wouldn't need to

vague berry
#

please help me

#

then there is also things like

#

$cos(arccsc(tan(arccos -1/3)))$

versed aspen
#

Alright so use the fact that $\tan(a + b) = \frac{\tan(a) + \tan(b)}{1 - \tan(a)\tan(b)}$

somber coyoteBOT
vague berry
#

I can't do that either

#

I have to use like triangles to draw it

somber coyoteBOT
vague berry
#

I'm a baby

versed aspen
#

I can't do that either
Wdym you can't do that, are you not allowed to?

vague berry
#

yeah

#

its like one of those draw a triangle use the angle thing and go from there

versed aspen
#

Alright well anyways $\tan(\pi + x) = \tan(x)$

somber coyoteBOT
vague berry
#

ok

#

i follow

versed aspen
#

That's because tan is pi periodic

#

After doing that you need to do all those triangle thingies you were talking about

#

To evaluate $\tan(\arcsin(-\frac23))$

somber coyoteBOT
versed aspen
#

Do you know how to do that, or no?

vague berry
#

no

#

well

#

i can

#

😄

#

actually

#

i can do it

#

wow that explained things easily

versed aspen
#

Oh well noice

vague berry
#

one more please

#

$cos(csc^-1(tan(cos^-1(-1/3)))$

somber coyoteBOT
vague berry
#

the -1 thing for csc and cos mean arccsc and arccos

versed aspen
#

you use cosec^{-1} or arcosec for that but yeah

#

Uhm well I would start by evaluating $\tan(\arccos(-\frac13))$ first using the triangle thingo

somber coyoteBOT
versed aspen
#

Then again do the same triangle thing for $\cos(cosec^{-1}(\text{The thing I found}))$

somber coyoteBOT
vague berry
#

ok

versed aspen
#

ok

vague berry
#

Tysm!

#

I really appreciate your help

versed aspen
#

np

valid wave
#

Can anyone help me with this i don't understand it Calculate the sizes of adjacent angles if one is an adjacent angle is 110 times smaller

fringe canyon
#

i am unsure how to evaluate with x=1, series i still didnt do as a chapter

#

so i am not sure how to just plop in x and do this

solid sequoia
#

Hey can someone explain the purpose of Log 10 and Natural Log?

upper karma
#

Can anyone help me with a problem

#

Yes

#

I’m here for you

#

Unless it’s super hard

#

Send it

#

Ok I’ll take a picture

#

Yes

#

Cause it’s kinda big

#

Ok

#

We can handle it

#

It’s number 4

#

Ok

#

Just change the equation to slope intercept form

#

Then plug them in

#

Plug x intercept and see if y intercept matches

#

I’m sure there are other ways too

#

You can ask other people

#

But that’s my way

#

Probably slower than other ways

#

I changed it to y= 7x -8 is that the right slope intercept form

#

Hold on

#

Nah It’s y= 7/2 x -4

#

Try your arithmetic again

#

Ok

#

And this applies to algebra

#

Not geometry

#

I think

#

It’s algebra since it has to do with the slope intercept formula

#

I learned that when I was in algebra

#

Ok

silent plank
#

it comes under coordinate geo which is fine here

upper karma
#

@silent plank Go for it

#

Yes

#

@silent plank It’s time for your lecture

#

You can explain it to him

#

If you need to

#

Explain in more depth

#

I guess

#

Ok I have another one it looks actually hard

#

Yes

#

Very hard

#

We can stand it

#

We are brave men

#

Send it!

#

Very brave

night ridge
#

Can someone please help me with this I’ve tried for a while but can’t come up with the answer? 🙂

#

“Write a expression of the area of figure N”

#

It’s a pattern

upper karma
#

Well do you see the pattern?

#

Try to interpret it and see the pattern

night ridge
#

I’m sorry but my English when it comes to math is not that good but i see that the increase is increasing with 2 every time

#

But I don’t know what to do with that info

upper karma
#

If you know the pattern you can determine the next figure

night ridge
#

Between figure 1 and 2 it’s increasing with 5 and between 2 and 3 it’s increasing with 7

#

Yes but what about if I want to figure out figure 135

#

Like I need an expression to explain the pattern

upper karma
#

You said you need to find the area of figure N

night ridge
#

Yes

upper karma
#

Which is after figure 4?

night ridge
#

Doesn’t matter any

#

So I could use the same expression for every figure

upper karma
#

@silent plank Help

night ridge
#

For an example it’s usually something like “Area=2n+1” but this one is much harder

upper karma
#

@paper oar Help

night ridge
#

Thank you ☺️

acoustic jungle
#

n(n+2)

#

one side is 2 larger than the other

upper karma
#

Yes bread man

#

You smart boi

night ridge
#

You’re right, how did you come up with that

high hemlock
#

got a math test tmr wish me luck

upper karma
#

I wish you luck son

#

Do well

#

Ace the test

paper oar
#

Why did I get tagged

acoustic jungle
#

One side of the rectangle is n, the other side is n+2. I multiply them together to get n(n+2)

upper karma
#

@paper oar Son it’s me

#

Weird tag

#

Do you remember me

acoustic jungle
upper karma
#

@paper oar

paper oar
#

Oh

high hemlock
#

i lowkey dont want to cheat on the test

#

but since its online

#

might have to

upper karma
#

Son it’s online

acoustic jungle
#

Don't do it peepee.

upper karma
#

Yes

#

Don’t do it

#

It’s not worth it

#

In the future you’ll have to do it by paper again

#

And you wouldn’t know how to do anything

#

Don’t cheat

#

Try ur best son

night ridge
#

Thanks for the help guys!!

acoustic jungle
#

||how do I write 10 sentences to get 1 point across?||

upper karma
#

@paper oar I love you too

#

10 sentences

#

Yes

high hemlock
#

you milk it

#

definitely

#

definitely ramble

upper karma
#

Milk it

high hemlock
#

wait what

#

if you want to get 1 point accross

#

wouldnt it be better to be concise

upper karma
#

Does anyone know how to do product of slopes

high hemlock
#

parallel or perp

#

arent there two

upper karma
#

I’m not sure the problem just says determine the product of the slopes of line y=10 and y=-2x-5

high hemlock
#

well i suck at math

#

but you can ask someoine else lol

acoustic jungle
#

y=10 is equal to y=0x+10

upper karma
#

Yessir

umbral snow
#

Find their slopes, multiply them together @upper karma

upper karma
#

@umbral snow yessir

#

I got u

#

It’s okay I solved it already

deep trail
#

halp

#

i dont understand what its asking

dreamy badge
#

So the two sides of the right-angled triangle are each 10 and 14 yards long. From that you should be able to use trigonometry to find the angle the sun is coming from

#

English isnt the language I learn math in, so I'm not completely sure whether the answer is supposed to be the top right angle or the bottom left angle '^-^

upper karma
#

Bottom left^

shell acorn
#

@upper karma can u help me pls

upper karma
#

Nice.

plush burrow
#

what is an example of an ambigous question?

#

can someone help me pls

dark sparrow
upper karma
#

Anyone know how I can solve this?

silent plank
#

draw a triangle

wanton prawn
#

hey guys i got following problem:
Prove: If ASB is a right angle, S lies on the circle with the center M of AB as the center
and | AM | as a radius.
Anyone an idea how to start such problems?
hope this is the right channel.

dark sparrow
#

mb write arctan(-2) as -arctan(2) first just to not have to deal with negative lengths

#

@wanton prawn this channel is occupied right now, please move

upper karma
#

@dark sparrow What is the step after -arctan(2)?

dark sparrow
#

"the" step

upper karma
#

From the calc this comes to decimal number

dark sparrow
#

no, you're kinda missing the point here, you don't need a calculator at all

#

anyway, you COULD simplify cos(-arctan(2)) to cos(arctan(2))

#

since you used the fact that arctan is odd, why not also use the fact that cos is even

#

and now you're dealing with positive numbers only, which is a slight step down in complexity

#

and it is now that you can follow ramonov's suggestion of drawing a triangle

#

label two of its sides 1 and 2 in a way that makes one of the angles have a tan of 2

upper karma
#

is the answer root5/5

silent plank
#

,w cos(arctan(-2)) = sqrt(5)/5

somber coyoteBOT
upper karma
#

Thank you

keen aspen
#

good point to always make

#

change everything in terms of cos and sin

rich wolf
#

Recognize that secx = 1/cosx

#

Etc etc

upper karma
#

^

#

And csc x=1/sin x

#

And cot x=cos x / sin x

#

@upper karma

#

like that @upper karma

#

Would this be the final solution?

upper karma
#

You can simplify further by doing common denominator @upper karma

#

Can you explain more? Would I multiply left by sinx and right by cosx, what about denominator?

#

By doing common deno on the upper part of the skycraper and on the bottom. You'd be doing common deno on $$\frac{1}{\cos x}+\frac{1}{\sin x}$$
And the same on the lower part
$$1+\frac{\cos x}{\sin x}$$

somber coyoteBOT
upper karma
#

On the first thing, multiply $\frac{1}{\cos x}$ by $\sin x$

somber coyoteBOT
upper karma
#

And $\frac{1}{\sin x}$ by $\cos x$

somber coyoteBOT
upper karma
#

While on the lower part of the skycraper, multiply $1$ times $\sin x$

somber coyoteBOT
upper karma
#

@upper karma

#

This is how common deno is done

#

You shouldn't have problems with this

#

Got it thanks

#

Np catthumbsup

upper karma
#

One calculator I got gives me 0.976018 for atan(1.4782) while another gives me 55.92. What am I doing wrong with the first one?

#

Under what situation can u get 0.976 for atan(1.4782)?

silent plank
#

calculator settings,
radians , degrees

upper karma
#

nope, I've tried converting 1.47 using angle to radian formula

#

and ive also tried converting 1.47 from radian to angle formula

#

why do you even need radians for tan anyways?

silent plank
#

1.47...isn't your angle. it is your ratio

#

,w convert 0.976018 radians to degrees

somber coyoteBOT
silent plank
#

$\arctan(1.4782) \approx 0.976 \text{(radians)} \
\arctan(1.4782) \approx 55.92\deg$

somber coyoteBOT
upper karma
#

uhhh

#

@silent plank I did that and it gives me 1.5592

#

I converted it using x * 180 / pi.

#

oh

#

lol turns out I converted the wrong thing

upper karma
#

@upper karma there's a button on your calculator that makes it work in either radians, degrees, or grads

#

a grad is 1/400 of the circumference of a circle

#

Ok thanks.

#

I couldn't tell because i was using a default atan() in a programming language

pastel anchor
upper karma
#

plot it

pastel anchor
#

let me try

#

actually i plotted all of them

#

and they all look wrong

#

@upper karma

upper karma
#

how do they look

#

post a picture

kind temple
#

Just take the options one by one and calculate the radius

#

The one which gives equal radius in every case is correct

upper karma
#

I just realized that I plotted the answers 😐

#

Jesus

deep trail
#

its not a right triangle so im confused

upper karma
#

(particularly the "More General Case" section)

#

@deep trail law of cosines

deep trail
#

i never learned that

wooden current
#

@deep trail Finding the width of the house is the same as finding the base length of the isosceles triangle. Let's draw a height from the top of the roof to the base of the roof. How do you find the length of the side next to the base angle in the formed right triangle?

deep trail
#

?????????????????????????????????????????????????????????????????????????????????????????????????

#

artist

winged hill
#

can’t we use the law of sines 😳

wooden current
deep trail
# 
.blue
wooden current
#

How do you find B

deep trail
#

oh

upper karma
wooden current
#

It's really simple at this point

upper karma
#

how about now?

#

do you see the right angle now?

deep trail
#

my drawing was the best

wooden current
#

Bruh

#

Ngl

deep trail
#

i got it ty

winged hill
#

u cant use the law of sines on that?

kind temple
#

law of cosines makes it very easy

winged hill
#

cosines can work aswell

kind temple
#

Cause of the common stuff

winged hill
#

saying we have the top angle between 2 known sides

deep trail
#

morinkashi why would you use it

winged hill
#

i’m used to using the law of sines and cosines

deep trail
#

you dont have the that side so you cant do sines

kind temple
#

Law of sines how would u find the thing for 35

upper karma
#

soh cah toa

#

you need cosine

#

cah

#

and the width of the house would be twice that

wooden current
#

I think they said they haven't learned about law of cosines/sines

kind temple
#

Ohhh

winged hill
#

u can use law of cosines by finding the top angle and and then you’d have an S.A.S case

#

(included angle thingy)

kind temple
#

Just take the options one by one and calculate the radius
Hey why doesnt this work

winged hill
#

cause the top angle can be found using 180 = 70 + x and solve for x

#

then plug in the values in the law of cosines

kind temple
#

Why are u making it so complex

winged hill
#

that’s easier for me 😔

kind temple
#

Top angle is 110

#

That can be seen at 1st glance

wooden current
#

law of sines easier doe

kind temple
#

Leave this

#

Tell me that

#

Why doesnt that work?

wooden current
#

what

kind temple
#

For this

#

We're supposed to find the center

#

I told we can go option wise and calculate by distance formula

wooden current
#

you can use the distance formula for every pair of points ( (2,1), (x,y) -> center ; (5,2) , (x,y) ; (3,4), (x,y) ) and get a system of equations

#

i think it reduces to linear equations

kind temple
#

I told we can just see from the options

#

And apply distance formula to every point

#

The option with same distance is correct

#

But I calculated and there's no option which satisfies this

upper karma
#

@kind temple post your work.

kind temple
#

Okay wait

#

Here

silent plank
#

what have you tried?

#

nope

#

segment is also the wrong word

#

set those two equal is a poor choice of words

#

their sum will be 360°

upper karma
#

looks like progress being made.

#

Flo would be happy.

silent plank
#

yeh. there is only one possible value of x

dim wren
#

Hey can someone help me with a math question

#

Im trying to find the measure of the arc

#

BC

#

D;

upper karma
#

because I'm feeling generous today 🙂

deep trail
#

srry

#

but im not ready that

upper karma
#

did you learn soh cah toa yet?

deep trail
#

yes