geometry-and-trigonometry

[upper karma]

ah thanks

[worn echo]

the first equation makes a sphere

and the second is a plane

to find their intersection do i have to substitute z =1 in the first equation?

if i do so i get $x^2 + y^2 = 3$

[somber coyote]

Abdul

[worn echo]

a circle centered at origin with $\sqrt{3}$ radius

[somber coyote]

Abdul

[worn echo]

but it's not centered at origin here

in the diagram

[upper karma]

more like at (0,0,1)

[worn echo]

ah

oh alright

that makes sense then

thanks !

[timid plaza]

@upper karma

[urban oyster]

first thing

@upper karma

get all of the coordinates

write them down

for A B E D

[olive cove]

Because the triangle is reflected in y=x, you can say (a,b) becomes (b,a). So (-2,6) becomes (6,-2)

Wait that's a test?

Im sorry but its against the rules

Its bannable

1. Requesting help during an exam is a bannable offense.

Its in #rules

[upper karma]

@olive cove next time feel free to tag @ Moderators

[olive cove]

Ok

[supple onyx]

<@&268886789983436800> in case unseen

[forest stump]

uh oh

[tepid finch]

so what im seeing is that he cheated so he went from 4/13 to 10/13

[copper valve]

sigh

shouldve just banned the first time

[sacred ore]

Can anybody help me with this? Like how do i find the missing side.

[amber star]

you mean the missing angle?

[sacred ore]

Ye

[amber star]

alright so

ill walk you through it

[sacred ore]

In struggling im studying right now for my test on thursday

[amber star]

how many degrees make up a triangle, for the inside i gues

idk how to word it

[sacred ore]

180?

[amber star]

360

wait

[sacred ore]

Oh

[amber star]

is it 180?

ah it is 180

[sacred ore]

I thought triangle was 180

[amber star]

yeah youire right

its been years but ik how to do it

anyway

you know the angles of 2 of the angles on the bottom triangle, so how do you find the third angle?

[sacred ore]

I dont know

[amber star]

ok so

180 = angle A + angle B + angle C

angle A = 25 degrees
angle B = 69 degrees
angle C is unknown

[sacred ore]

So this is a supplementary angle?

[amber star]

180 = 25 + 69 + X

well yes but youre solving for the internal angles

not the outside ones

[sacred ore]

Wait how do you figure out which is angle a b and c?

[amber star]

you can just assign them as you please

[sacred ore]

Oh so it does not have to be in order.

[amber star]

yep

[sacred ore]

If its 180=25+69+x what happen to 53

[amber star]

ignore it for now

we're solving for the bottom triangle

we'll get to it in a sec

[sacred ore]

Ok im writing it on my notebook

So what do i do next?

[amber star]

solve for X

[sacred ore]

Is it -25 both sides of 180?

Or -69?

[amber star]

the total i guess, angle is 180 degrees for a triangle

you have 2 of them

to solve for the unknown, or the third one

this one right here

you do, 180 - 25 - 69 = X

X is the angle of that quesiton mark

[sacred ore]

86?

Can you tell me what lesson is this?

[amber star]

i have no clue

and yes that angle is 86 degrees

[sacred ore]

Wait if its 86 degreees

Its mirror right 86 other side too?

Or no?

[amber star]

yes

[sacred ore]

Oh i get it nowee

[amber star]

i guess mirror works as a word for it

[sacred ore]

86+53=139

And then 139-180=41

Right?

[amber star]

yes

that's your answer

[sacred ore]

Thank you.

[amber star]

no problem

[sacred ore]

Can anybody help me here.

[silent plank]

where are you stuck?

[mellow zephyr]

simple algebra, QIH + JIQ = 179, solve for x and substitute back in the equation for QIH. Hard to say which side of JIH is 179 though.

[silent plank]

the measure of the minor angle is implied

[sacred ore]

How do i start the problem

Im writing it on my reviewer

[mellow zephyr]

I can see that, but at a glance it’s hard to tell wether the left or right side.

[sacred ore]

Left or right is what?

[silent plank]

this is an application of angle addition postulate,
set up your equation appropriately

[glacial ice]

Can someone help me with this, please? sec(a) - csc(a) / sec(a) + csc(a)

i've simplified it down to sin(a) - cos(a) / sina(a) + cos(a) but don't know where to go from there

[silent plank]

have you used sufficient parentheses to clearly indicate your numerators and denominators?

also what are you being asked to do with that?

[glacial ice]

i got it, thanks

[coral badge]

Can someone help me for this

[upper karma]

@coral badge what have you tried so far?

[coral badge]

m1=47.5°
mBD=100°

[mossy valve]

how would one find the perimeter of a shape inscribed inside of a circle? and example would be a 8 sided polygon inscribed inside of a circle with a radius of 8

is there a formula i can use for this?

[coral badge]

They jus said that basing on theorems

[heady sapphire]

right i feel dumb not being able to solve this

anyone?

[plain wyvern]

similar triangles

[heady sapphire]

right right but what are the conditions?

for using similar triangles?

[plain wyvern]

https://brilliant.org/wiki/congruent-and-similar-triangles/#:~:text=AAA%3A-,Compare Similar Triangles,by translation%2C rotation or reflection.

[heady sapphire]

YOOOOOOOOOO MAKES SENSE POG

but what are the conditons?

like a right angler or somewhere or a certain measurement???

[waxen quest]

I just need to know is this possible with the data given

[slate blade]

make compound shapes

[upper karma]

Oh I hate crap like that, youhave to make some assumsions, assume all the curves are circels

[slate blade]

The "ends" combine to make a circle (as they are semicircles)

And the "inside dents" (there are 4 of them) combine to make another circle as well

[waxen quest]

the inside dents dont really make a circle since the heights are longer so there ovals

[slate blade]

wait what is the 4.6 supposed to be lol

[waxen quest]

ya i dont know what thats supposed to be its just kinda a random length its not the radius or diameter so i dont know

[slate blade]

I thought that was the diameter of the big end circles

and 3.2 is the "side of the dotted square"

[waxen quest]

an 10in is total length

[slate blade]

yeah I Think so

[thorny bough]

wait is the frequency of acosbx

b

or 2pi/b

monkas

[slate blade]

Frequency = 1/period

period = 2pi/b

so therefore the frequency = b/2pi

@thorny bough

[carmine kernel]

hey guys, how do i find the horizontal asymptote of "r(x) = 2 + x+1/x-3"

I know the horizontal asymptote is 3 because I graphed it, I just dont know how I can tell from just the equation. Help would be great :))

[thorny bough]

wait rip

[versed river]

you can write the function as r(x)=3+4/(x-3)

[thorny bough]

lmao imagine losing a full letter grade on an ode assignment cuz of that

cant b me oh wait it is me

[shrewd grotto]

is this the place for help with angles

[lilac pine]

if you havent solved yet, it is 6 from 3*12 = x^2

@heady sapphire

[slate blade]

@shrewd grotto Yes

[heady sapphire]

yes after figuring that out i questioned my existence and whether or not i still deserve to be in this server

[lilac pine]

pfft

[heady sapphire]

my geometry skills lack

how do we know its similar

[lilac pine]

you can study

[heady sapphire]

like conditons

https://tenor.com/view/oh-what-detective-charles-boyle-joe-lo-truglio-brooklyn-nine-nine-shocked-gif-16300873

[lilac pine]

you use the "öklid"

[heady sapphire]

DA WHAT

[lilac pine]

idk whta it is called in english

eucledean?

[heady sapphire]

google shall translate that dark energy

[lilac pine]

euclid,

[heady sapphire]

icelandic?

[heady sapphire]

huh???

[heady sapphire]

U LIVE IN ICELAND???

[lilac pine]

You can use these if the is a right angle coming from another right angle

[heady sapphire]

i never knew of this things existence

[lilac pine]

God I don’t know how to explain questions in English

Nah

[heady sapphire]

what

[lilac pine]

I live in Turkey

[heady sapphire]

right angle coming from another right angle

[lilac pine]

Yes

[heady sapphire]

their on top opposite sides

[lilac pine]

Dikten dik indirirsen

[heady sapphire]

u mean a bisector splitting a right angle?

MAN WHAT

[lilac pine]

No

[heady sapphire]

When lowering from a vertical

i now speak fluent turkish

[lilac pine]

In that case, the median would be equal half of the side it divided

If only

If it’s a right triangle

Would you like me to write them down for you to understand?

[heady sapphire]

yea but i just used a proportion lol: 12/x = x/3

[heady sapphire]

write what down

[lilac pine]

The

Triangles

I feel like you are conf

Used

[heady sapphire]

what triangles???

[lilac pine]

Sorry if I sound condescending, my intentions are good

The euclid triangle

And the right triangle with a median

[heady sapphire]

isnt this the elucid triangle???

[lilac pine]

Yeah

But it’s in turkish

[heady sapphire]

but the variables arent

[lilac pine]

Yeah

[heady sapphire]

just need the varibles

right?

unless there is something else?

[lilac pine]

Only this one

[heady sapphire]

what does the dot represent

[lilac pine]

The right angle

[heady sapphire]

okay okay

so whats this triangle for?

[heady sapphire]

like this is the only triangle we cant use the formula on?

[lilac pine]

So

I said something wrong

It’s not a bisector, a median

[heady sapphire]

im still following keep going

[lilac pine]

A median coming from a right angle in a triangle divides he side in two, and the median is equal to half of the side it divided creating 3 equal sides

That’s as far as my English goes

I just wanted to share, thought it may be helpful

In the future

[heady sapphire]

yes that makes sense but what does it have to with the formula???

[lilac pine]

Nothing

[heady sapphire]

but i can already see this just by looking at the picture

[lilac pine]

Yeah

[heady sapphire]

except the 2nd part

[lilac pine]

Uh

[heady sapphire]

wait half the side it divided?

it divided an entire shape not one side?

its a median

[lilac pine]

If the side is 2n, the median will divide the side to n and n and the median will be n also

If it’s a right triangle

[heady sapphire]

ohhh makes sense

[lilac pine]

So the median will be equal to half of the side it divided

[heady sapphire]

what side???

can u draw it out?

like just the 2n and n thing?

OHHHHHHHHHHHHHHHHHHH

[heady sapphire]

see that makes a whole bunch of sense now

thank you

actually no

[lilac pine]

:(

[heady sapphire]

teşekkür ederim

i think idk

[lilac pine]

Hahahah

You’re welcome

Here’s an image I found on the internet

Good luck with geometry

[heady sapphire]

lmao i left geometry in like 8th grade. im in precalc now

[lilac pine]

I am not sure whether i have seen precalc before

[hard warren]

The longest side of a triangle is 11 meters longer than the shortest side. The medium side is 15 meters long. The perimeter of the triangle is 46 meters. Find the length of the shortest side of the triangle. Solve it on linear equation

Help pleasee

[silent plank]

introduce a variable for shortest side

[solemn axle]

How would I go about factoring a polynomial like this?

[hard warren]

x

[hollow raven]

@solemn axle Rational Root Theorem + Factor Theorem could be a start

[solemn axle]

What’s that?

[hollow raven]

oh wait no

since 6x^2 * y = 2xy *3x

you can factor out GCF

for first two and last two terms

6x^2 + 2xy + 3x + y
= 2x(3x + y) + 1(3x+y)
...

[hard warren]

Can anybody help me with this?

[hollow raven]

let x equal shortest side value

what would the longest side value be?

[hard warren]

Idk bro

[silent plank]

The longest side of a triangle is 11 meters longer than the shortest side

if the shorter side has a value of x,
how would you denote something 11 units larger than that?

[viscid tulip]

how do i find the coordinates

[sacred ore]

Does anybody know this problem.

[earnest echo]

Are you familiar with Properties of parallel lines

In this question you need to know about corresponding angles

[limpid phoenix]

um help please?

[patent plume]

Uh

Are you allowed to use cosine rule?

Cause you then directly get the answer for x, let alone an inequality

but I suppose what you could do is find the range for the common segment via triangle inequality and you should get a range for x.

Wait

This is weird

<@&286206848099549185> I don't want to ping, but I'm honestly confused about that question xD. There should be only 1 value for x right?

[upper karma]

The equation is y=2x-4. Need help with number 3

Hi let A be a point outside the circle K, how can I prove that the maximum distance from K to A is 2r + m, where m is minimal distance from K to A and r radius

I've searched for this online and a guy told that this can be proven using triangle inequality but I wonder how

[earnest echo]

Are you sure that r is perimeter and not radius??

[upper karma]

Radius sorry

[earnest echo]

Max Distance=m+(length of the longest cord)

[upper karma]

Why do we even consider m here?

Like, let AT = m, so AT + longest cord = longest distance

Now, let AT' = m' > m

What guarantees that AT + longest cord > AT' + new longest cord

[earnest echo]

I thought m is shortest distance as a fact is given to us

[upper karma]

It is

[earnest echo]

Here first we'll need to prove that m is the shortest distance

Ig this is where you'll need the triangle inequality

[upper karma]

We don't lets take that as granted

So we def know AT < AT'

[earnest echo]

Then I don't see a problem in that

Because the longest cord will be the diameter

[upper karma]

But wait let c be diameter and c' < c

How do we know that m + c > m' + c'

Huh?

Oh nvm

[earnest echo]

Ahh, I see what you are asking now

Okay, let's look at it this way

From the point A

Make a triangle with point of tangency and end point of line segment 2r+m as vertices of triangle

Let's name the points

A,B and C, where B is point of tangency

Now clearly this is a right triangle and line segment AC is hypotenuse which longest

The maximum value the line segment AB can take if we try to move it around the circle will always be less than AC

Hence AC is the longest distance possible

@upper karma

Do it more formally when actually writing the proof

[upper karma]

that's satisfiable

[earnest echo]

What can you say about angles PDC and angle APD

[warm night]

equal?

[earnest echo]

Okay, why are they equal?

[warm night]

corresponding angles i think

[earnest echo]

Alternate angles

Moving on

In triangle ADP, Angle ADP=angle APD

What can you say about the sides

[warm night]

isosceles have same base so ad=ap cos of that reason

[earnest echo]

Correct reason is

Sides opposite to equal angles are equal

[upper karma]

damn you use so much

space

[earnest echo]

---

**
**

---

[loud steeple]

@upper karma you can't use the pythagorean on the "big" traingle since it's not a right-angled triangle. But you know two smaller traingles which is right-angled

[upper karma]

Yeah, forgot to say i tried using it on the smaller triangles.

Was trying it on the bigger triangle than the smaller one but I couldn't figure it out.

Can you give me a idea of how I would approach it?

[loud steeple]

what do you think the length of x is?

[sacred ore]

Anybody know the answer for this?

[upper karma]

can i get help

https://gyazo.com/430e9db42fc78e83c2210684371a658b

[upper karma]

Ur dumb

[tender prawn]

thats the question, that's what i have

oh the labels are a bit messed up sry

i can easily prove A A' M N is a cyclic quadrilateral

im having trouble proving that O lies on the circle

D here is the center of rotation

ahhhh i got it nvm

no i dont got it

wait yes i got it

[upper karma]

https://gyazo.com/430e9db42fc78e83c2210684371a658b

https://gyazo.com/ee7a68e1d042a85af076e52a03daae18

can anyone help me please

[formal peak]

Can anyone help me with this

[thorny maple]

help please

[whole carbon]

https://cdn.discordapp.com/attachments/473918103424729099/788193478512214057/image0.jpg

[whole carbon]

<@&286206848099549185>

[humble pulsar]

@whole carbon what are you stuck on?

[whole carbon]

I dont know how to do any of this

1-5

[humble pulsar]

do you know how to calculate means/averages?

[whole carbon]

Yeah

the mean

[humble pulsar]

Oh whoops I was looking at 2 not 1

Ok so 1 is asking 66 and 2/3 % of $(something) = $50

[whole carbon]

yup

[humble pulsar]

so in math, "of" means multiply, and for simplicity im gonna use a variable for the something, what variable would you like?

[whole carbon]

x

[humble pulsar]

Ok so x * 66(2/3)% = 50 right?

[whole carbon]

yeah

[humble pulsar]

Ok so percents are weird in math, so let's convert it into a decimal

Divide the percent by 100 to get a decimal/fraction (2/3)

[whole carbon]

what percent

the 66

[humble pulsar]

66(2/3)%

yeah

[whole carbon]

0.0044

[humble pulsar]

[66*(2/3)]/100 = 2/3

where did you get .0044?

[whole carbon]

divided the 66

by 100

66(2/3)

[humble pulsar]

$\frac{66\frac{2}{3}}{100} = \frac{(\frac{200}{3})}{100}=\frac{2}{3}$

[whole carbon]

0.44

[somber coyote]

moshill1

[humble pulsar]

you need to deal with the 66(2/3) together, not after

[whole carbon]

Oh ok

so its just 2/3

[humble pulsar]

Or, just move the decimal place 2 places to the left to divide by 100

and you get .66666666 etc. which is 2/3

ok so (2/3) * x = 50 right?

[whole carbon]

yeah

50

[humble pulsar]

So how you isolate/solve for x?

[whole carbon]

would u times the 50

by 2/3

[humble pulsar]

not times

what's the opposite of times?

[whole carbon]

divide

[humble pulsar]

yeah, so divide both sides by 2/3

[whole carbon]

so

75

[humble pulsar]

yep

[whole carbon]

Okay

Thank you for that

[humble pulsar]

np

[whole carbon]

On to 2nd one

if you dont mind

[humble pulsar]

2 is just find the mean/average of the data

(add up all the scores) / (number of scores)

[whole carbon]

ok

so 35

on the left

do I just add them both up

https://cdn.discordapp.com/attachments/473918103424729099/788193478512214057/image0.jpg

[humble pulsar]

35?

[whole carbon]

yeah

[humble pulsar]

where are you getting 35?

[whole carbon]

I added up the left side

Got 35

[humble pulsar]

Oh no

you add up the grades

[whole carbon]

oh

[humble pulsar]

the left column is just whatever test it is

[whole carbon]

then divided by 5?

[humble pulsar]

(assuming the 23 is a typo)

yes

[whole carbon]

77.4

[humble pulsar]

Yep

[whole carbon]

so

now 3rd one

[humble pulsar]

Ok so you want to add a 6th score to the mean such that you get the mean = 80

$\frac{S_1 + S_2 + S_3 + S_4 + S_5 + S_6}{6} = 80$

[somber coyote]

moshill1

[whole carbon]

ok

what is it

[humble pulsar]

what's the sum of the 1st 5 scores?

[whole carbon]

387

[humble pulsar]

$\frac{387 + S_6}{6} = 80$

[somber coyote]

moshill1

[humble pulsar]

can you solve for S_6?

[whole carbon]

im not sure

[humble pulsar]

ok so first you can multiply both sides by 6 to get rid of the fraction

[whole carbon]

so 480

[humble pulsar]

then what do you do to get rid of the 387 from the left side?

[whole carbon]

I divide that too?

[humble pulsar]

no, it's being added

so what's the opposite of addition?

[whole carbon]

subtract

[humble pulsar]

yah

so S_6 = 480 - 387

[whole carbon]

ok so 93

[humble pulsar]

yes

[whole carbon]

Alright

[upper karma]

Can someone please help me with this?

[whole carbon]

https://cdn.discordapp.com/attachments/473918103424729099/788193478512214057/image0.jpg

Need help with 4 and 5

[night basin]

how do i do the first one

[humble pulsar]

very 1st or 1st magnitude question?

[night basin]

very first lmao

i’m like 10 minutes late

@humble pulsar

[humble pulsar]

Ok so if you take 1 u and add 1 v tip to tail, where does the tip end?

[whole carbon]

@humble pulsar hey

[vestal pendant]

hello

[whole carbon]

@vestal pendant hey you mind helping me with a few questions

[vestal pendant]

I have a question here too

can you use corresponding sides / angles as a reason of proof

[night basin]

yeah say

CPCTC

[upper karma]

I Need Help Here's The First 2 Questions

[quartz imp]

do you know what the sum of angles in a triangle is @upper karma

and what is angle TRP

[upper karma]

I Figured It Out But Thanks @quartz imp

[quartz imp]

nice

[fair marten]

https://cdn.discordapp.com/attachments/708389565076865024/788216849212768256/unknown.png

i need some help on this

[polar summit]

distance formula

find coordinates of A, B and C

[formal goblet]

An equal rectangle has a square area 120cm^2 if the diagonal is 10cm what is the length of the side

how do we find the side

[south cargo]

can someone help me with my homework? its really easy, I can dm u the picture of it

its like the basics of geometry

[silent plank]

post here

[unborn bison]

How would you find the internal side lengths?

[silent plank]

since you clearly understand stuff like that discussing that publicly is wrong...
fk

wrong ping sry

just clarifying is the test live?

sry modwalker

[last verge]

ahaha wtf :D

[silent plank]

was trying to spell moderator

[last verge]

np

[silent plank]

i clearly failed

[last verge]

:)

[mighty wharf]

ok

[upper karma]

I Could Use Help PLEASE

[dusky surge]

Hello! @upper karma
They look good!

Except some places

Like x=8

And question 10, you have to set up an equation, so 7 x 2 =14 is not appropriate

[upper karma]

Those Are The Problems Not My Answers

Except For The First 2

X=9 My Fault For 2

[silent plank]

the above are comments on your disregard for case sensitivity

as well

and don't use x for multiplication in plain text in the presence of the variable x

use * instead

[upper karma]

I Still Have No Clue What To Do For The Rest Of The Problems

[silent plank]

and don't chain equations like that

54/2 = 24/3 = 8
is VERY BAD
(even if you wrote the 27 instead of 24)

[upper karma]

why do you capitalize every starting letter?

[upper karma]

Habit I Have

[silent plank]

phone issues maybe

it is recommended that you start with a clear equation telling you the relation between yoursides

[upper karma]

Got It But How Do I Even Start The Next Few Questions

[silent plank]

eg for Q10,
start with 2x = 2 * 7

and it is clear that you'd get x by dividing both sides by 2

[upper karma]

No I Mean Like 12-15

[silent plank]

same idea for the rest of them

it is recommended that you start with a clear equation telling you the relation between yoursides

all the similar triangles in these questions have the same ratio.
the long marked edge is double the shorter marked edge

and that's what should be represented in your initial equation

eg for Q11,
54 is double 3x, i.e.
54 = 2 * 3x
and solve for x from there

and apply the same idea for the rest of them

[upper karma]

Alright I'll Try

[silent plank]

also note that in your work:
2 * 7 = 14 (that's a true statement)
and that tells me nothing about how you concluded that
x = 7 other than magic

[stable gull]

can someon help with this/

[woeful summit]

https://gyazo.com/9ae71059fdf11ece073ec2f7fb073cf3.

[stable gull]

6x+3=33

@woeful summit

[woeful summit]

alright thanks

https://gyazo.com/a705e6f82bf680459824a501f675748c

[stable gull]

12x+4+80=180

[woeful summit]

https://gyazo.com/ddcf71e71020d15f460a4bd37aa5ece2

[stable gull]

all angles are congruent so all sides are congruent

[woeful summit]

https://gyazo.com/9bd93fff60ae53aaff761bb6f77a7987

[teal mica]

Can anyone help me with this

I use law of sines and get 1.157

[woeful summit]

https://gyazo.com/bb8c789da3356dc68be131c950396b7f

[high geode]

Anybody know how to solve these type of problems? I'm stuck on thess

[soft zodiac]

I’m not sure if I set this problem up correctly

[humble pulsar]

@zenith garnet i suggest labeling the sides firstly

[zenith garnet]

nvm sorry realized how easy it is

[humble pulsar]

Nice

[zenith garnet]

lol

[upper karma]

For A’ I got (-2,-4) but I’m confused if the enlarge at of 2 correct

[upper karma]

Can someone tell me how my teacher got the 6 outside the square in the 3rd step

[upper karma]

@upper karma wdym?

The 6 is still there on the 4th step

@upper karma like where does she get the 6 from in 9+6 and so on were does that 6 come from in the thing

Oh like

The 2nd line to the 3rd one i'm assuming

yes @upper karma

Ok hold up

Remember $(a+b)²=a²+2ab+b²$, in our case, $(\overbrace{{\color{green}{3}}+{\color{blue}{\sqrt{2x-7}}}}^{({\color{green}{a}}+{\color{blue}{b}})²})²$ applying it we get \ $\overbrace{{\color{green}{3²}}+2\cdot {\color{green}{3}}{\color{blue}{\sqrt{2x-7}}}+{\color{blue}{\sqrt{2x-7}}}²}^{{\color{green}{a}}²+2{\color{green}{a}}{\color{blue}{b}}+{\color{blue}{b}}²}=9+6\sqrt{2x-7}+2x-7\ =2+6\sqrt{2x-7}+2x$

[somber coyote]

Al𝟛dium

[upper karma]

@upper karma does this massively maybe unnecessary detailed tex help you understand

@upper karma

y

e

s

thank you

[upper karma]

@upper karma can you help me with one more thing

I have this /3=-9+/k

how do i get my answer from here

What's /3 and /k supposed to mean

square root

${\sqrt{3} - ${\sqrt{k} = -9$

[somber coyote]

SynxGG
Compile Error! Click the reaction for more information.
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[upper karma]

ayy that @upper karma

ik you move the k over so

you get

Solving for k right?

yes

${\sqrt{3} = -9 + {\sqrt{k}$

[somber coyote]

SynxGG
Compile Error! Click the reaction for more information.
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[upper karma]

so thats what i have so far

Try to isolate sqrt(k) first

@upper karma by sqauring everything?

That's the move yes, after having sqrt(k) on one side alone

so it leaves 9 = 81 +k?

[upper karma]

Because if you square root both sides right now with the -9 with the sqrt(k) on that side, you'll get another term with sqrt, and it's not something you want specifically

[upper karma]

Did you.. just said 9=81?

[upper karma]

but when squaring everything

[upper karma]

Ok no, 2 things to note, reread what i said here

And $(a+b)²\scalebox{4}{≠}a²+b²$

[somber coyote]

Al𝟛dium

[upper karma]

This is very important, this mistake is very common, the earlier you fix it the better

ok

still dont get it tho

[upper karma]

.

Try to have ONLY sqrt(k) on one side

so move the 9 to the left?

Yes, the -9

ok

so i have /3 +9=/k

${\sqrt{3}+9 = {\sqrt{k}$

[somber coyote]

SynxGG
Compile Error! Click the reaction for more information.
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[upper karma]

yep

there are 2 unnecessary { 's on your tex

$\sqrt{3}+9 = \sqrt{k}$

[somber coyote]

Al𝟛dium

[upper karma]

ok

then you would square root im guessing

square both sides

not sqrt both sides

so the 3 and k

wait

no

wait

idk anymore

remember that $(a+b)^2=a^2+2ab+b^2$

[somber coyote]

Al𝟛dium

[upper karma]

what's your problem

so far it looks like ${\sqrt3} +9 = {\sqrtk}$

/k

there's no need to put {} around \sqrt

ok

$\sqrt{3} = -9 + \sqrt{k}$

it's \sqrt{} not {\sqrt}

there we go

thats what it is

but you went back an step

but with the 9 on the left lol

$\sqrt{3} +9 = \sqrt{k}$

[somber coyote]

SynxGG

[upper karma]

ok

from here

by squaring both sides i mean $(\sqrt 3 +9)^2=\sqrt{k}^2$

[somber coyote]

Al𝟛dium

[upper karma]

ok

[somber coyote]

Al𝟛dium

[upper karma]

try to apply it here

so

${3+9} = 3\sqr2 + 2 \sqrt{k} + k$

[somber coyote]

SynxGG
Compile Error! Click the reaction for more information.
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[upper karma]

not 32

just 3^2

wait.. how did you

can you translate that into normal text

as in sqrt(3) or whatever

[pure garnet]

@upper karma Could you help me right now?

[upper karma]

wym

ok

wait

[upper karma]

this channel is occupied, proceed with a free one.

3+9=3^2 + 2/k + k

like that?

use sqrt() to indicate $\sqrt $

what do i use to indicate squared

to indicate square root of something on plain text, use sqrt(whatever)

${3+9} = 2\sqrt{k}+k}$

uhh?

if you don't know latex don't use it

there's no need to communicate by latex

there

thats what i think it would be

i've no clue how we have 3 terms on the right side

ok forget that then

no but like

[somber coyote]

SynxGG
Compile Error! Click the reaction for more information.
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[upper karma]

$(\sqrt 3 +9)^2=\sqrt{k}^2$

[somber coyote]

Al𝟛dium

[upper karma]

from here, can you explain your thought process to get there?

sure

so you said to square root which would make the /3 on the left just 3 and 9 into 81 then on the right side would the /k just be k

don't twist my words

i did not say that.

i don't know where you are getting these things from

do you actually read my messages or do you like only read the shorter ones

sorry you did not say that

yes i am reading your messages its just im just slow im sorry

don't be sorry, i can fully understand someone to be slow, what i can't stand is someone ignoring my messages aiming to help you

i didnt ingnore any of your messages its just i just dont understand to much about this and i was trying to take everything in account at once.

if you don't understand something, ask

i'm not in your head to know what's confusing you

ok i will from now on i just wanted to try to do it on my own and that probably just scuffed me up

ok so lets start off fresh what do i do after your step you provided

try things on your own is probably what my method of helping is based, now you know somewhat what your mistakes are and how to avoid them. If i were to give you the solution or "do this" or "do that", you'd not know what mistakes you actually have when doing something on your own, which is why i'm here to help you to prepare for your future test

so that's good, making mistakes is a great way of learning.

kk

ok so back to the problem

$(\sqrt 3 +9)^2=\sqrt{k}^2$

[somber coyote]

Al𝟛dium

[upper karma]

alr what do i do from here

let's think about the right side first of all

ok

how can the right side be simplified

get rid of the square root?

yes, and what do you get

just k

notice that we have $\sqrt{k}^2$ and NOT $\sqrt{k^2}$ on the right side, both are different

yes

[somber coyote]

Al𝟛dium

[upper karma]

so the right side is just k

o got it

now for the left side

$(\sqrt 3 +9)^2=k$

[somber coyote]

Al𝟛dium

[upper karma]

this is what we have so far

since its on the outside it means its going to canncel it out and if its on the insdie it means that vaule is squared

kind of yes

k

so now left side

ok

do you notice a similarity with $(a+b)^2$

[somber coyote]

Al𝟛dium

[upper karma]

yes

which i remember is NOT a^2+b^2, but instead (a+b)^2=a^2+2ab+b^2

let me color this

k

$({\color{green}{a}}+{\color{blue}{b}})^2={\color{green}{a}}^2+2{\color{green}{a}}{\color{blue}{b}}+{\color{blue}{b}}^2$ and we have\ $({\color{green}{\sqrt 3 }}+{\color{blue}{9}})^2=$ ?

[somber coyote]

Al𝟛dium

[upper karma]

yup

try to do it on your own

if stuck look at the colors

take your time

so you want me to expand it

yes

k

so

$\sqrt3^2+2{3}{9}+9^$

like that?

if you want to go with latex, remember that commands work with the argument {} and not ()

ok

not exactly, but close

ok

[somber coyote]

SynxGG
Compile Error! Click the reaction for more information.
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[upper karma]

the middle term is $2\cdot 9\mathbf{\sqrt{3}}$

$({\color{green}{a}}+{\color{blue}{b}})^2={\color{green}{a}}^2+2{\color{green}{a}}{\color{blue}{b}}+{\color{blue}{b}}^2$ and we have\ $({\color{green}{\sqrt 3 }}+{\color{blue}{9}})^2=$?

look at the colors

[somber coyote]

Al𝟛dium

[upper karma]

i dont get how you got 9

for the a

[upper karma]

\sqrt{3}^2 and if you want to express multiplication put a * in middle, {3} does nothing, {} are used for commands like \sqrt or \frac or whatever

wound in not be 2 time 2 time 9

to be sure..

we are talking about the middle term right?

yes

ok

why would you think it's 2*2*9

don't be afraid

sorry 2 times 3 times 9

not 2

ok

why times 3

i think because the a value is 3 and b is 9

the term a is sqrt(3) and NOT just 3

oh

ie we are expanding (sqrt(3)+9)^2 not (3+9)^2

ok

that makes sense

[somber coyote]

Al𝟛dium

[upper karma]

for reminder

k

notice the term a is the whole, not just 3

ok so

we'd get $\sqrt{3}^2+2\cdot 9\sqrt{3}+9^2=k$ do you agree?

[somber coyote]

Al𝟛dium

[upper karma]

this is just what we just said

yes

be sure you don't actually have any doubts

i encourage asking if any

i get everything i and still just a bit just a tiny bit confused on the 9 in the middle

ok what's your confusion about it

i get where your coming from just i stilll am confused about the 9

wdym

ok remember in the first thing i asked when you explained $(a+b)²=a²+2ab+b²$, in our case, $(\overbrace{{\color{green}{3}}+{\color{blue}{\sqrt{2x-7}}}}^{({\color{green}{a}}+{\color{blue}{b}})²})²$ applying it we get \ $\overbrace{{\color{green}{3²}}+2\cdot {\color{green}{3}}{\color{blue}{\sqrt{2x-7}}}+{\color{blue}{\sqrt{2x-7}}}²}^{{\color{green}{a}}²+2{\color{green}{a}}{\color{blue}{b}}+{\color{blue}{b}}²}=9+6\sqrt{2x-7}+2x-7\ =2+6\sqrt{2x-7}+2x$

[somber coyote]

SynxGG

[upper karma]

when you explained the 6

yes

thats what im tring to put back in my head for the 9

okay well

waiitttt

I got it

I got it

basically we've just applied the formula $(a+b)^2=a^2+2a\mathbf{b}+b^2$

[somber coyote]

Al𝟛dium

[upper karma]

ok good

kk we can move on

great, the hardest part is done

now just some little algebra

yay i still have like 5 more problems of these for hw to lol

k

$\sqrt{3}^2+2\cdot 9\sqrt{3}+9^2=k$

[somber coyote]

Al𝟛dium

[upper karma]

what's $\sqrt{3}^2$

[somber coyote]

Al𝟛dium

[upper karma]

just 3

yep

they cancel out

and what's 9^2

18

you may want to recheck that again

no

91

81

right

now the middle term

thought it was 9 times 2 for a sec lul

what can we simplify about $2\cdot 9\sqrt{3}$

[somber coyote]

Al𝟛dium

[upper karma]

something very simple

umm the square?

nope, the sqrt(3) can't be done anything with it

ok the 2 x 9

18

yes

so we finally get

$3+18\sqrt{3}+81=k$

[somber coyote]

Al𝟛dium

[upper karma]

yes

now what can you do as your final step

combine like terms

remember you can't add 3 and 18sqrt(3) because of the sqrt(3)

[upper karma]

well, that's not the term but i know what you mean

ya

i get it

what do you get

the 3 and 18/3 cant be multiplied

no, they are not even multiplying theirselves

ik ik

they are terms

so we have 18sqrt(3) + 84 = k

did you meant to put "+" instead of "="

yes lol

and please to express a square root on plain text, use sqrt(3), not /3

k

yes

and that's your answer.

$k=18\sqrt 3 +84$

[somber coyote]

Al𝟛dium

[upper karma]

the answer key says something way different

...

i only took your problem where you told me

ik

the answer says k=3

but im not going to worry about that

the solution of what you gave me $\sqrt 3 -\sqrt{k}=-9$ is $k=18\sqrt{3}+84$

[somber coyote]

Al𝟛dium

[upper karma]

maybe you formulated the eqn wrong

no that was it

its o

k

thank you so much for the help imna go try tacking the other problems alone

the answer of the equation you gave me, is what we got. there are no other solutions, i'm 100% you formulated the eqn wrongly, or the book has a mistake.

ya its weird

the equation i gave you was the right thing

everything we did was correct

then if you are fully sure about that, the book is wrong.

yup that sucks

my next problem is $\sqrt{x+4}+\sqrt{x-3}=7$

[somber coyote]

SynxGG

[upper karma]

imna go do this

ok i'm leaving to sleep, gl and post away any doubts you have, and you might get answered by another one.

thank you have a good night sleep

good morning

[sleek axle]

Is anyone here familiar with the intercept theorem and able to spare a minute?

[silent plank]

what intercept theorem

[sleek axle]

I have a question which needs answering to obtain a set of coordinates, here is the question in full

A man was out walking through the woods and he discovered an interesting tree. As he stood there admiring it, he began to contemplate how tall it was. He tried to work it out and started off by noting down some measurements based on what he could see. It then started to rain so he ran for cover to the nearby café where he could sit down with a nice brew and a slice of lemon drizzle cake to finish off the calculations. Unfortunately the rain had spoilt the piece of paper the measurements were written down on and all he could make out from his scribblings were 2 things:

The distance from his eye to the base of the tree was 51.32m, and the angle from his eye level to the top of the tree was 42.44°.

He then measured his eye level to be 1.8m from the ground and realised that he had enough information to be able to work out how tall the tree was.

If the distance from his eye to the top of the tree was AB.Cm, and the total height of the tree is DE.Fm, then the cache can be found at N51 48.ABC W000 46.DEF

Can you help me with the answer so I can work out what the values of A A, B, C, D, E & F are please?

[humble pulsar]

did you draw a diagram?

[sleek axle]

I'm thinking Thales' Intercept Theorem might solve it but I'm lost when it comes to maths

[humble pulsar]

did you draw a diagram?

[sleek axle]

A diagram was not provided

[humble pulsar]

Read what I wrote

did YOU draw a diagram?

[sleek axle]

Sorry, no I haven't

[humble pulsar]

then start there

[silent plank]

AB.Cm, and the total height of the tree is DE.Fm, then the cache can be found at N51 48.ABC W000 46.DEF
also is that what's literally written in your question?

[sleek axle]

@silent plank yes

[humble pulsar]

Geocaching pog

[sleek axle]

@humble pulsar yes haha

Any chance you would be able to share the answer?

[silent plank]

did YOU draw a diagram?
then start there

[sleek axle]

Yes, i've drawn a diagram

[humble pulsar]

^

share the diagram at all?

[sleek axle]

I've written it down, is there a tool to draw one up on here?

[silent plank]

phone image or some shit

jesus...are A,B,C,D,E,F supposed to denote single digits in their respective values?
was that mentioned in the question or do you need to be psychic?

[sleek axle]

Just drawing up a diagram now

https://www.geocaching.com/geocache/GC7X9H3_treegonometry?guid=82cde4d1-d328-4e32-a7fb-8734495cb11a

Here's a link to the page

[humble pulsar]

bruh it's an actual geocache wtf

[humble pulsar]

42.44 isnt a side

and you dont have the height of the person included

[silent plank]

the diagram shows complete disregard for the human and the ground

[sleek axle]

The humans eye level is 1.8m so that would be the height

[silent plank]

start by drawing a horizontal line for the ground

a stick figure person

and a (long) vertical line a reasonable gap away for the tree

[sleek axle]

Excuse the drawing

That lollipop is the tree

[silent plank]

the thing you labelled isn't 51.32m

read the text carefefully

[sleek axle]

the distance from the eye to the base is 51.32

[silent plank]

metres, yes.

and does your horizontal line connect the eye to the base of the tree?

[upper karma]

[sleek axle]

no

[silent plank]

also are the human and tree magic or something?

hovering over thin air like that

[upper karma]

😂

[silent plank]

ok that's better

now

and the angle from his eye level to the top of the tree was 42.44°.

also you can ditch that circle for treetop, not needed

[sleek axle]

ok

[silent plank]

and maybe make the tree a bit taller in your diagram

[silent plank]

and the angle from his eye level to the top of the tree was 42.44°.
start by drawing a horizontal line to indicate the eye level

and draw a line from the eye to the top of the tree

and label the relevant angle

[silent plank]

no

the side isn't 42.44°

[sleek axle]

Where would you like me to depict the angle?

[silent plank]

angle

at the angle

[silent plank]

ok, continuing on:

He then measured his eye level to be 1.8m from the ground

[sleek axle]

yes

shall I note this on the far right?

[silent plank]

if you want

now that you've successfully drawn a diagram, that's pretty much 75% of the problem done.
the rest is applications of phythagoras and basic trig

[sleek axle]

ok

[silent plank]

why vertical...ugly af

just write it as 1.8m normally

[sleek axle]

How would I use the Pythagorus theory?

[silent plank]

you have right triangles here

[sleek axle]

yes, three

[silent plank]

apply it to a triangle where you know 2 of the sides

[sleek axle]

54.32m and 1.8m

Length of side c = 54.34m

[silent plank]

what's c?

[sleek axle]

the third side of the triangle

the straight line from the face

[silent plank]

in your triangle which side is your hypotenuse?

[sleek axle]

I don't know what that means, sorry

[silent plank]

you're blindly applying a^2 + b^2 = c^2
with complete disregard to what a,b,c represent

[sleek axle]

ok, where do I start?

[silent plank]

read up on pythag so you know what the variables represent so you can apply it properly

[sleek axle]

Pythagorean Theorem calculator calculates the length of the third side of a right triangle based on the lengths of the other two sides

I only have two confirmed distances on the diagram 54.32m and 1.8m

[silent plank]

yes.

there should be more

i.e. a formula and labelled diagram indicating what the variables represent

pythagoras does NOT state that adding the square of any 2 random sides will give you the square of the third

[sleek axle]

Can you draw on my diagram to explain where i should be labelling?

[silent plank]

...

label anything you want

just be clear about it

[sleek axle]

I understand you're trying to help which I'm thankful for but am struggle with the vague answers

I'm so confused at the moment

[silent plank]

ok. assign a random variable (not currently used in the question) to denote the horizzontal distance from the eye to the tree

since that's what you intend to find atm

also use those corner squares (or w/e you call them) to denote any right angles you can identify

and THEN refer to a guide to applying pythagoras and apply it properly to the triangle you're dealing with

[silent plank]

wheres 60m coming from?

[sleek axle]

You said to assign a random variable

[silent plank]

60m is not a random variable

[bronze jewel]

does anyone understand this?

[silent plank]

a variable is a symbol or letter to denote values

common used to denote unknowns you intend to solve / determine

[silent plank]

(not currently used in the question)

[sleek axle]

Sorry

[silent plank]

ok.

also use those corner squares (or w/e you call them) to denote any right angles you can identify
and THEN refer to a guide to applying pythagoras and apply it properly to the triangle you're dealing with

[thorny maple]

please help on 2,3,4

[sleek axle]

These are the correct degrees given it's a 90 degree angle

[silent plank]

WHERE THE FK IS 47.56° comeing from

how in the world is that a 90° angle

[sleek axle]

😆

[silent plank]

the horizontals are perpendicular to the verticals

the human stick figure is perpendicular to the ground

SURPRISE the tree is also perpendicular to the ground