help-27

[eager cairn]

i just learnt that

they didnt say the name tho+

aight i got you

[chrome pilot]

yah

so just apply this theorem to the problem

all is solve

[eager cairn]

is it intercepts on parallel lines?

[eager cairn]

w

[chrome pilot]

parallel lines

GC and FB

[eager cairn]

i can just write that?

[chrome pilot]

yes

[eager cairn]

aight

that is it right for that question

if u dont mind can i send another one ive been struggling on?

@chrome pilot

[chrome pilot]

sort of quadrilateral?

[eager cairn]

ye

like a parallelogram or smth like that

[chrome pilot]

I means i don't understand the 'sort' means

[eager cairn]

like the type

e.g. square, rectangle, rhombus etc

[chrome pilot]

ahh ok

[eager cairn]

can u help me on this or nah?

[chrome pilot]

ok then

[eager cairn]

tysm

[chrome pilot]

do you know the midpoint polygon

[eager cairn]

yes

[chrome pilot]

ok so

as you see in the drawing, P, Q, R are midpoint of 3 edges of the triangle ABC

[eager cairn]

yes

[chrome pilot]

so that means: PQ, QR, RP is middle way (I don't know what it called in english)

[eager cairn]

ok sure

[chrome pilot]

and the middle way has the property that it will parallel with the line of the other edge

[eager cairn]

ye

is that the proof?

[chrome pilot]

so

yes

[eager cairn]

can u say the middle way in ur language so i can google translate it

i need to be precise with my answer so the tecaher doesnt get mad for saying improper terms

[chrome pilot]

Midline

[eager cairn]

the english version?

[chrome pilot]

yes

[eager cairn]

ok thank you

[chrome pilot]

midline of a triangle

[eager cairn]

pq is midline of triangle?

[chrome pilot]

yes

[eager cairn]

ok

can i say that about PR?

[chrome pilot]

yes too

[eager cairn]

ok i did first question now the second part of the question

how to do?

@chrome pilot

[chrome pilot]

let me see

[eager cairn]

ok

[chrome pilot]

you can prove that ARPQ is the parallelogram right? so that you can do it for CPRQ and QRBP too

[eager cairn]

and then after that?

[chrome pilot]

yah and

if you see in a parallelogram

divided it by the line

hmm

[eager cairn]

?

[chrome pilot]

2 parts that are divided will be the same right?

[eager cairn]

yea

[chrome pilot]

so turn back to the thing you have

3 parallelogram

[eager cairn]

yea

[chrome pilot]

so you can now prove that 4 triangles in the drawing

have the same area

[eager cairn]

oh

[chrome pilot]

remember to close the channel then

[eager cairn]

.close

[devout snow]

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[devout snow]

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[urban harbor]

having trouble with this one, how would you determine if this series is convergent or divergent?

[frozen aurora]

comparison with $\frac{1}{2\ln(n)}$ ?

[woven radish]

artemetra

[urban harbor]

well i was thinking that too and was trying to think of 1/ln(n) and tried doing integral test with that, turns out its a tough integral

[frozen aurora]

hm

what about comparison with 1/n

[urban harbor]

oh yeah 1/ln n seems to be larger than 1/n

[frozen aurora]

cuz just thinking in terms of asymptotics ln(n) grows slower than n

[frozen aurora]

exactly

[urban harbor]

yeah that would do it, thanks : )

[frozen aurora]

and 1/n is def divergent

[urban harbor]

.close

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[red bay]

Put exp(z) = e^z . Solve the equation

[red bay]

How do I solve this one?

[robust dust]

write i in exponential form on the right so you can solve for exp(z)

then repeat, making sure to keep track of any multiples of i*pi you can add at each step

[red bay]

Thanks!

so i=e^iPi/2+2nPi

[robust dust]

yeah, so you can take logs on both sides, and now we have exp(z) = ...

which you can solve similarly

(again, keep track of any multivalued stuff you get from the exps)

[red bay]

so we will have e^z*times e^z on LHS

right`?

[robust dust]

where is that from?

[red bay]

exp(exp(z))

isn't that equal to e * e^z

[robust dust]

it would be e^(e^z)

[red bay]

oh

[robust dust]

you want to use this property

[red bay]

yeah

Why do we get this

[robust dust]

that's the definition of the principle branch log (where Arg is the principle branch argument)

[red bay]

oh okayu

Thanks

do we always get two solutions

[robust dust]

you get infinitely many solutions because of the +2pi's

[red bay]

I thought about -pi/2

and pi/2

where did the negative pi/2 come from

[robust dust]

I'd need to see other working but it's probably from being able to negate something in one of the steps

and that'd reverse the argument

[red bay]

Okay thanks!

[devout snow]

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[normal forum]

When I want to solve for x (not x_0) with system of equations:

[normal forum]

$$
\left{
\begin{aligned}
ax + by &= -c \
bx - ay &= bx_0 - ax_0
\end{aligned}
\right.
$$

[woven radish]

Bennxy

[normal forum]

I got $x=\frac{b^{2}x_0-aby_0-ac}{2ab}$ instead of $x=\frac{b^{2}x_0-aby_0-ac}{a^{2}+b^{2}}$. I'm not sure what went wrong. But what I did was multiplying the first equation by 'a' and the second by 'b'. And then add the equation together.

I know that we could use substitution for this. But I need to know how we will can do this by elimination.

[rugged sparrow]

can you show your working

because this is a pretty simple one

you must've been doing some obvious mistake

[normal forum]

Ok so, after multiplying the first equation by 'a' I got: $abx+aby = -ac$. And multiplying the second equation by 'b' results $abx-aby = b^2x_0-aby_0$

[woven radish]

Bennxy

[normal forum]

Wait

I think I found the mistake

[rugged sparrow]

for the equation 1, you should get a^2*x smth

[normal forum]

It should be a^2

[normal forum]

Hahahah yea

That's really stupid to be honest

[rugged sparrow]

yeah, you should get it now

[normal forum]

I should take a break

[normal forum]

Thanks for the help

Have a nice day 😉

.close

[devout snow]

Channel closed

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[rugged sparrow]

[devout snow]

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[restive river]

Im not great wth numbers

mm

.close

[devout snow]

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[devout snow]

Please don't occupy multiple help channels.

[hard knot]

How do I find the range for x

[candid lance]

Idk if this works But I can give it a try

as u know the range of t

[twilit comet]

range = all possible outputs

[twilit comet]

wat

[candid lance]

u can write equaton in terms on x

and then use t to get the range of x

[twilit comet]

snow we don't need to do that???

[candid lance]

idk how to explan

[twilit comet]

actually ykw

u might b right

[candid lance]

I think I did that

[hard knot]

wada

[twilit comet]

BUT range = all possible outputs

[candid lance]

ik

[twilit comet]

so what are all possible outputs for $\frac 3t + 2$, basically

[woven radish]

ren

[crisp trench]

Range of 3/t are all numbers except 0. Finding range of 3/t + 2 will be....

[twilit comet]

!nosols

[devout snow]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[twilit comet]

wrong

[crisp trench]

Sorry if that was a spoiler i found it very trivial

[twilit comet]

$0 = \frac 3t + 2$ is possible

[woven radish]

ren

[candid lance]

thats a domain

what he said

[twilit comet]

yes

that's my point

[hard knot]

it is

Im just idk

[crisp trench]

I said range of 3/t

[twilit comet]

@hard knot range is outputs
2 is a possible output

[twilit comet]

well we're discussing 3/t + 2

not 3/t

[crisp trench]

And let him "figure out" range of 3/t + 2

[twilit comet]

i-
ok??

idm

[hard knot]

t> -3/2

[twilit comet]

go ahead lmao

[twilit comet]

??

[hard knot]

wait but that gives 0

so

[twilit comet]

what

[hard knot]

t> -3

i dont know

[twilit comet]

what are you talking about

i don't understand what ur tryna say

bc t > -3 isn't very helpful

[hard knot]

Ok i dont know how to find ranges or domains of jack

i forgot

[crisp trench]

Typically it is best to just write your function as composition of many functions and find the range of every one of them. But since this function is pretty simple you can just find an inverse and find the domain

[hard knot]

ok how would that look like?

[twilit comet]

okay ykw

case's approach was good

[twilit comet]

@hard knot what is the one value that 3/t can never take on

[hard knot]

-3/2

[twilit comet]

no

ofc it can be -3/2

[hard knot]

Why

it would make x = 0

[candid lance]

it would give 0

and thats valid

what did bro write

[woven radish]

ren

ren

ren

[twilit comet]

that's a valid output

[crisp trench]

Like for example finding a range of $\frac{1}{\sqrt{1 + (x - 3)^2}}$ the easiest thing to do is find the range of $(x - 3)^2$ then of $1 + (x - 3)^2$ and etc... going out as you go. I hope im not confusing you

[twilit comet]

so no, not -3/2

[woven radish]

casework

[twilit comet]

case ur confusing me lmao
(jk)

[candid lance]

the main problem is that

what u both talking bat

abt

is of no means to equity

[twilit comet]

no

it is

[candid lance]

Idk what is he confused about

[twilit comet]

wait for it snow

[candid lance]

but hes stuck somewhere

[twilit comet]

@hard knot

what is the one value that $\frac 3t$ can never be??

[woven radish]

ren

[hard knot]

0

Oh thefore t /= 0

so t /= 2

[crisp trench]

t isnt 0 either way

[twilit comet]

wh-?

[crisp trench]

I think he is confusing x with t

[twilit comet]

the hell?!?!!

@hard knot

look can 3/t ever be EXACTLY zero

[hard knot]

3/t = x t cannot be 0, but if its 3/t + 5? t can be 0

?

[crisp trench]

No

t cant be 0

[hard knot]

👽

[crisp trench]

Do you mean x cannot be 0 in the first case and x can be 0 in second?

[twilit comet]

okay i have no idea what equity is saying.

case, you're on ur own

[hard knot]

imma do a few questions and spot patterns

and then ill let ya know

Probably better

bc i did this stuff before

ty lol

:3

.close

[devout snow]

Channel closed

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[crisp trench]

I didnt even get to say all the cool stuff you can do with ranges. Like if $f(x) \in \mathbb{R} \setminus S$ then $ f(x) + c \in \mathbb{R} \setminus (S + c)$

[woven radish]

casework

[twilit comet]

that's... obvious?

[crisp trench]

Yeah it really is

But its kinda easy sol for this one

[hard knot]

yesnt

frl frl

.reopen

[devout snow]

✅

[hard knot]

ranges for y

I got y = -2

How do I know if its

? < t <-2
or ? < t <_ -2
or t< -2
or t> -2

[crisp trench]

You mean y?

[crisp trench]

You draw the parabola and see

I mean its really just looking at the leading coefficient

But just draw if unsure

[hard knot]

hmm

yeah im ganna have to redo the whole topic

I remember i did it via drawing

Ill have to do it again

ty

.close

[devout snow]

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[devout snow]

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[woven radish]

Anastasia_
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[devout snow]

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[devout snow]

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[safe knoll]

start from left hand side

by finding f(g(x)) first and then trying to show that its equal to x^2 -5x +4

[fresh elbow]

Do you know how to graph linear equations

Do you know how to graph y=x

yeah sure

so first can you convert the 2nd equation where the y variable is on one side

Can you show how you got there

it’s not correct

yes

but you did it incorrectly

The equation is not equal to x+y=3

[granite zealot]

Can anyone help me with math

[fresh elbow]

@granite zealot #❓how-to-get-help

How do you move the X to the other side

only -x

What is the equation now

correct

y=x+2 is the other equation

To graph it easier

Nono we want to graph it

Don’t do that method

they ask to graph

Do you know how to graph y=x

What happens

If you have y=x+2

No it will go left

2 units

uhhh

Draw the line 2 units to the left

Another way to think of it

Is that the y value is 2 higher than y=x

Because y=x+2 basically means: the y value is equal to the x value plus 2

So at each x value, add 2 to it to get your y value

What is the y value at the x value 5

Hmmm

That’s +1

At the x value 0, the y value is 1 when it should be 2

y=x+2

you added 1 tho

You need to “add 2 squares” to the top line as well

So the button should hit the graph at 3,5

nice

y=-x+3

Do you know what y=-x looks like

it should look like a line from the top left to bottom right

Nono

Remember the slope

what is y when x is 5

huh

I have to go now <@&286206848099549185> help please

[devout snow]

@restive river Has your question been resolved?

[devout snow]

@restive river Has your question been resolved?

[devout snow]

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• Be polite and have a nice day!

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[celest grove]

can someone help

[quaint citrus]

@celest grove stay here

what have u tried @celest grove

[celest grove]

one min

[quaint citrus]

,rotate

[quaint citrus]

ngl im not too sure what u did after that equals sign

[celest grove]

idk fr but icant solve that

[quaint citrus]

ok do u know log change of base formula

[celest grove]

can u show me

I started studying today

[celest grove]

what c equals?

[quaint citrus]

c can be anything u want

so now we have this formula

[quaint citrus]

can u use this to change each log in this

[celest grove]

how idk fr

[quaint citrus]

ok for example

$\log_3 {12} = \frac {\log 12}{\log 3}$

[woven radish]

Stephen

[celest grove]

oh understand

[quaint citrus]

you can assume we're using log base 10

[celest grove]

i use it to each log ?

[quaint citrus]

yes

use the property for each log and then show me what you get

[celest grove]

there or when i started

[quaint citrus]

where u starte

[celest grove]

like this yep?

@quaint citrus

[quaint citrus]

yes

thats the first fraction

[celest grove]

i already wrote second

[quaint citrus]

where?

[celest grove]

one min

now what to do?

divide

~

?

[quaint citrus]

yes do the individual fraction divisions first

[celest grove]

devide this or

them

[quaint citrus]

them

[celest grove]

like this?

but now what to do?

do same to the second division?

[quaint citrus]

simplify this first

[celest grove]

how ?

[quaint citrus]

just combine the terms

(log12 * log36)/(log3)^2

now do the same for the second set of fractions

[celest grove]

okey one min

oh rotate need

srry one min

[celest grove]

done

correct?

[quaint citrus]

yea looks good, just make sure in the denominator its (log3)^2 and not log(3^2)

after u do that, combine the fractions

[celest grove]

ah one min np

[celest grove]

one min

[quaint citrus]

good

now we want to simplify the numerator

lets start with log(12)

can u give me the prime factorization of 12

[celest grove]

223?

2 * 2 * 3 ?

[quaint citrus]

yes

so we have

actually just do this

log(12) * log(36)

what are some common factors

of 12 and 36

[celest grove]

2 * 2 * 3 * 3

now 2 2 3

?

[quaint citrus]

hold on a sec

[celest grove]

np

[quaint citrus]

lemme work thru this a bit before i bring u any further

[celest grove]

okey waiting

[quaint citrus]

ok it simplifies nicely

so heres what we wanna do

log 12 = log 4 + log 3 correct?

[celest grove]

yeah

one min

lemme think

[quaint citrus]

dont write it down yet

just follow along

[celest grove]

yeah

yeap

[quaint citrus]

write log36 as a sum of log3's and log4's

[celest grove]

log4+log3^2

?

[quaint citrus]

yes, which can also be written as log4 + 2log3 correct?

[celest grove]

yeap

[quaint citrus]

ok so now we have

(log 4 + log 3)(log4 + 2log3)

for that first part

can u mutliply that out

[celest grove]

log4 * log4+log4 * 2log 3?

like that?

[quaint citrus]

thats only part of it

(a+b)(c+d) = ac + ad + bc + bd

[celest grove]

okey one min

correct?

[quaint citrus]

yes

now combine terms

[celest grove]

lg4*lg4=lg16?

[quaint citrus]

no

lg4 * lg4 = (lg4)^2

[celest grove]

lg4^2

?

[quaint citrus]

u need the parentheses

dont forget them

[celest grove]

one min

but what to do in second

lg4*2lg3 ?

[quaint citrus]

its just 2lg3lg4

what does the entire expression simplify to?

[celest grove]

icant understand question

[quaint citrus]

what does the numerator become

[celest grove]

one min

correct?

2lg3lg4+lg3lg4

=3lg3lg4?

[quaint citrus]

yes

[celest grove]

What should I do with the denominator?

one min

[quaint citrus]

we're not done yet, we still have to simplify log4 * log108

[celest grove]

one min

in the numerator

i get : (lg4)^2+3lg3lg4+2(lg3)^2

correct?

[quaint citrus]

yes

(lg4)^2+3lg3lg4+2(lg3)^2 -lg4lg108

[celest grove]

one min

[celest grove]

we cant simplifcation -lg4lg108?

[quaint citrus]

we can

turn lg108 into a sum of log 3's and log 4's

[celest grove]

3^3*4?

[quaint citrus]

yes thats the factorization

but make log108 into a sum of log3's and log4's

[celest grove]

3lg3+lg4?

[quaint citrus]

yes

so now we have lg4(3lg3 + lg4)

multiply that

[celest grove]

3lg3lg4+(lg4)^2?

[quaint citrus]

yes

so now in numerator we have

(lg4)^2+3lg3lg4+2(lg3)^2 - (3lg3lg4+(lg4)^2)

simplify all of that

[celest grove]

one min

[quaint citrus]

@celest grove got it?

[celest grove]

help pls how

i cant

one min i will show

(lg4)^2 -(lg4)^2 yep?

and next 3lg3lg4-3lg3lg4

and we have left 2(lg3)^2

?

[quaint citrus]

yes

that is numerator

and what is denominator

[celest grove]

(lg3)^2

answer is 2?

[quaint citrus]

yes

[celest grove]

2(lg3)^2/(lg3)^2

and answer is 2?

[quaint citrus]

.

[celest grove]

omg thank u so much

[quaint citrus]

np

[celest grove]

can you check one example

?

one min i will show and say if it correct

[quaint citrus]

yea

[celest grove]

thank u so muchh

i appriciate it

have a good night ❤️

[quaint citrus]

u2

[devout snow]

@celest grove Has your question been resolved?

[devout snow]

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[flint hemlock]

I know how this works if one of the x's is a constant, but not when both parts of the interval are variable

[inland seal]

Let's say cos(t^5) has antiderivative G(t)

What would the integral be equal to in that case?

@flint hemlock

[flint hemlock]

sin(t^5)

[inland seal]

?

The integral

In terms of G

[flint hemlock]

um

integrating something is the same as finding the antiderivative, right?

I'm sorry- I'm not great with the terminology

[inland seal]

If our function was, 4x^3 for example

The antiderivative would be x^4

[flint hemlock]

yes

[inland seal]

In this case, oos(t^5) doesn't have an antiderivative that you can express easily

But, let's just say it's G(t)

Meaning G is a function such that G'(t) = cos(t^5)

Now, use the fundamental theorem of calculus to evaluate the integral.

Don't take any derivatives yet

[flint hemlock]

oh okay

so it would be F(G(b)- F(G(a))?

[inland seal]

F isn't related here

[flint hemlock]

oh

oh yeah

it would just be G(b) - G(a)

[inland seal]

What are a and b?

[flint hemlock]

a is x and b is 6x

um

G(6x) - G(x)

[inland seal]

Yes

So F(x) = G(6x) - G(x)

Now take the derivative

[flint hemlock]

G(6) - G(1)?

or... no would it be 6G(6x) - G(x)?

[inland seal]

No

[flint hemlock]

So the first one? or neither?

wait no

G'(x) = cos(t^5)

so derivative would be 6cos((6x)^5) - cos(x^5)

Maybe

[inland seal]

Yup

Well done

[flint hemlock]

Thank you!!!

[inland seal]

Np

[flint hemlock]

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[lament basin]

Hello, I was wondering if the equation posted is linear

I am saying it is not linear because the e^y is exponential

[restive river]

yeah non linear

your intuition is correct

[lament basin]

Ty!

.close

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[lament basin]

.reopen

[devout snow]

✅

[lament basin]

Sorry, what if it was e^x(dy/dx) would that equation still be linear?

[devout snow]

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[lament basin]

<@&286206848099549185>

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[nova whale]

Im not sure where to go from now

[jovial stump]

youre trying to show the top line correct?

[nova whale]

yes

[jovial stump]

the 1- gives me the idea and solve for sinx by cos^2 x +sin^2 x =1 if that cosx is a given

imma tell you tho i got no clue what the o is doing

[nova whale]

xd

its fine

[jovial stump]

i got the whiteboard out gimme a minute

[devout snow]

@nova whale Has your question been resolved?

[nova whale]

ok

[jovial stump]

alr good news is i checked it

bad news is it doesnt work

sry i couldnt help u out

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[snow merlin]

taking a proof class rn, what would be the best way to write out a justification for going between 7 and 8? using more steps is fine

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[inland kiln]

hello, I've been stuck on this question for a good 30 minutes

[inland kiln]

I've written out the first few values for each sequence but can't find a clear pattern

x0 = 1
x1 = 0
x2 = -6
x3 = -30
x4 = -114
x5 = -414

y0 = 1
y1 = 6
y2 = 24
y3 = 84
y4 = 300
y5 = 1020

[devout snow]

@inland kiln Has your question been resolved?

[devout snow]

@inland kiln Has your question been resolved?

[severe nest]

@inland kiln this is a pretty tough problem but I did manage to find that
[x_n = 1 - \sum_{k = 0}^{n-1}y_k]
idk if this is helpful at all but if you can find a closed form for ${y_n}$ then this relationship gives you the closed form for ${x_n}$ automatically

[woven radish]

Awesam

[devout snow]

@inland kiln Has your question been resolved?

[inland kiln]

No

[devout snow]

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[pure socket]

im stuck on this integral and not sure how to proceed

i tried doing trig sub (not in the screenshot above) but i think it just made it worst

[covert root]

Is u a constant?

[pure socket]

thats supposed to be 4

sry for bad hand writing

[elfin pewter]

xd

[lunar owl]

substitute x = 2tanΘ

dx = 2sec^2ΘdΘ

then solve

[pure socket]

this was my trig sub attempt

[lunar owl]

i dont think u can factor the x out either btw but what you have in the beginning is ok

so

we have

1/ sqrt(4tan^2Θ+4)

i just finished it lmk if u need anything else

[pure socket]

then u = 4tan^2 + 4 right?

[lunar owl]

i solved it a different way but show me what you did

well

are you doing a u substitution?

[pure socket]

yes after the trig sub

[lunar owl]

theres no need yet

how can you simplify the denominator algebraically?

[pure socket]

factor out the 4 in the sqrt or rationalize it

[lunar owl]

yes factor the 4 out

[pure socket]

ohhh then you can use the indentity right

[lunar owl]

mm

not just yet

[pure socket]

ok

[lunar owl]

factor the 4 out and show me

[lunar owl]

right

we can do one more thing to the denominator

whats the sqrt of 4

[pure socket]

2

[lunar owl]

yes

so we can move that out

of the sqrt

[lunar owl]

so when we have two terms multiplied within a square root

[pure socket]

oh shoot

yeah

should be 2 not 1

[lunar owl]

wdym

for which

[pure socket]

2tan^2 + 2

[lunar owl]

no not quite

[pure socket]

or should the 2 be factored out still

[lunar owl]

remember your algebra rules

sry its sideways lol

[urban harbor]

,rotate

[lunar owl]

w soosh

[pure socket]

im so dumb rn

lmao

[lunar owl]

you good

i had trouble too

[pure socket]

the 2 should be out

yeah

[lunar owl]

yes

[pure socket]

my b

[lunar owl]

dw

i was 100x worse

whne i took calc 2

[pure socket]

lmaoo its ruff rn

do you use the indentity now?

[lunar owl]

mm

well

we have a constant in our fraction

so

we can pull that to the left of the integral

we need to simplify the fraction as much as possible we're not nearly done

[pure socket]

yup

[lunar owl]

but to answer your question after we factor that out then yes

its the trig identity

[lunar owl]

yes

hey

bro spoiled it

[uneven notch]

oh doing step by step?

sorry

[lunar owl]

nw

but yes thats what u get

[lunar owl]

yes

do u know the integral of secant

[pure socket]

is there a way to derrive this anti derrivitive or should i just memorize it

[lunar owl]

its just a memorization thing

[pure socket]

ah ok

[lunar owl]

ln |secΘ + tanΘ| + c

i believe

[pure socket]

ok cool

so now sub theta for x?

[lunar owl]

yeah convert back to x

remember x = 2tanΘ

so tanΘ = ?

and then Θ = ?

[pure socket]

arctan x/2

[lunar owl]

well

u gotta set up the triangle too

dyk how to do that

[pure socket]

yeah one sec

[lunar owl]

yea

[pure socket]

first fraction should be flipped

[lunar owl]

uhh i got ln| sqrt(x^2+4) + x| + c

[pure socket]

is this part right?

[lunar owl]

wait

give me a second bro my bad lol

can someone else pick this up

[pure socket]

lol all g could i jus see your final subsitution work?

nvm figured it out

was silly mistake

.close

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[lunar owl]

my bad my girlfriend called

good that u figured it out

[pure socket]

Mine differred by a constant of ln2 I think but since it’s just a constant I think it’s still good thx for the help

[devout snow]

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[tranquil lark]

how do i do this

[outer kestrel]

have you learned function transformations

[loud field]

Do you know what the graph of 1/x looks like?

[tranquil lark]

i have but i cant remember much

[tranquil lark]

yes

[outer kestrel]

so first describe with words how the function changed

[tranquil lark]

its moved to the right from the origin by 1 unit ?

[outer kestrel]

how many units?

[tranquil lark]

oh 2

mb

[outer kestrel]

yeah and also its shifted up by 2 right

[tranquil lark]

oh yes

[outer kestrel]

theres one more thing

see if you can find it

[tranquil lark]

is it reflected ?

[outer kestrel]

right

so 3 things

how do we describe each one in the function?

lets start by the one that shifts up

this should be the easiest

[tranquil lark]

+2

[outer kestrel]

right

and how do we right shift?

[tranquil lark]

-2 ?

[outer kestrel]

right, (x-2) right

ok finally how do we reflect

[tranquil lark]

i dont know. shouldve payed attention in the lectre

[outer kestrel]

thats alright weve all been there

to reflect horizontally you just put a negative in front of the x term

[tranquil lark]

oh okay

[outer kestrel]

cs now all the originally negative values will come to the other side

so it flips

[tranquil lark]

yes, makes sense

[outer kestrel]

so what's g(x)

[tranquil lark]

(1/-x) +2 -2

-1/x ?

[outer kestrel]

oh no

[tranquil lark]

oh lol

[outer kestrel]

that horizontal shift

you put the -2 in the x term

so it would be

test some values for this and it wil make sense

[tranquil lark]

yes

thank you

[outer kestrel]

np

[tranquil lark]

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[echo kraken]

can someone quickly explain please why when i have the equation:

8sin^3 x = cos^3 x it becomes: (8sin^3 x) / (cos 3^x) = 1

[echo kraken]

like why does it equal 1 and not 0?

[limber atlas]

because we divided by cos^3x

[limber atlas]

typo here, should be cos^3x in in the last part

[echo kraken]

next to the 1?

[limber atlas]

it would be zero if we subtracted cos^3x from both sides

[limber atlas]

yes

[echo kraken]

basically i was doing this trig question and this is part of the working out for the answer

so i was confused why it becomes 1

[limber atlas]

this is what we did

[echo kraken]

ah ok thanks

wait is it 1 because its a divide then

if it was a minus it'd be 0 yea

[limber atlas]

yes

[limber atlas]

yep

x/x = 1

x-x = 0

[echo kraken]

thanks yea haha

.close

[limber atlas]

ofc

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[iron geyser]

how to do (viii)?

you need to show the coefficients of the FFFs are equal but idt they are

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[iron geyser]

<@&286206848099549185>

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[steep cipher]

Im confused at number 10

[steep cipher]

29=23+11x+2
=25
x=4/11

Ohh mb

I forgot 1/2

I got 3 now

Help for 3

[haughty apex]

what does | this mean

[steep cipher]

Idk

[haughty apex]

...

[steep cipher]

Probably line segment

[rustic jetty]

length

[haughty apex]

oh

ok

[steep cipher]

Oh alr

[rustic jetty]

Its an isosceles trapezoid, so AM = WR

[haughty apex]

yeah

so that means

[rustic jetty]

AM = AE + EM

So AE + EM = WR

[steep cipher]

Ohhh

Damn

I see it now

Ty

Sqrt of 10???

Does it go like

5√10
5√(2.5)(4)
(5)(2)(.5)
=5

Is it though???

EK is 13 but IT is 5

Wtf?

Help plz

<@&286206848099549185>

Yall

Plz

Help me

Help

Pls

Its almost time

Imma have to pass this tomorrow

Pls

I dont get it at all

[storm relic]

hello

I tried to explain it in paper as much as I could, hope it helps

it's the 7th question's answer though.

[steep cipher]

Oh alr

[steep cipher]

Could you explain it to me?

[steep cipher]

Kites area*

[storm relic]

in the note part I added how we can get the root out

[devout snow]

@steep cipher Has your question been resolved?

[storm relic]

which part did you not understand, let me help

[steep cipher]

Huh

Oh

Actually

I got it

Lol

[storm relic]

oh ok

[steep cipher]

So uh

Thx alot

[storm relic]

did you understood the root part?

[steep cipher]

Yeah

[storm relic]

Alright then :)

you can use .close so the room can be avaible to others

[steep cipher]

Alr

.close

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[restive river]

It is known that the cube ABCDEFGH and T is the midpoint of DG as in the figure to the side. The distance between point F and line DG is

A. FD

B. FT

C. FG

D. FH

E. FC

[restive river]

,rotate

[woven radish]

Couldn't find an attached image in the last 10 messages.

Mar 5, 2024, 2:02 PM

[frozen aurora]

it would help to make a diagram

oh

,rccw

[restive river]

FG or FT?

[frozen aurora]

the shorter one

FG

[restive river]

It's always the shortest ?

[frozen aurora]

yes

[restive river]

Alright then

Tysm

.solved

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[broken fiber]

can someone pls help me asap😭 🙏🏼

[upbeat pawn]

A= 4x
B=4x+12
a/b=2/5

solve x (this is one side length of A) not final

[broken fiber]

ty!

[loud field]

!done

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[broken fiber]

.close

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[restive river]

H

[twilit comet]

@restive river what is your doubt

.close

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[torn coyote]

Any hints ?

Using the iteration method btw

Or do I solve directly ?

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[thorn sage]

I have to show that the inequality holds and it states that I can use induction.

Below is a picture of what I’ve tried myself, but I can’t really figure out how to do it

[devout snow]

@thorn sage Has your question been resolved?

[thorn sage]

No… 😦

[spare crypt]

if you plug in the bounds of summation into the 2's exponents, you see that the ones on the left side go from 1 to (j-i) and (j-i) to 1 (backwards), so actually they're going to be the same thing when you add them

that'll turn the left side into 2*[sum of 2^k from 1 to (j-i)] - 2

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[atomic drum]

i need help solving this problem

[crisp trench]

Multiply by lcm

[atomic drum]

what is lcm

[crisp trench]

Least common multiple

Of the denominators

[atomic drum]

x?

[crisp trench]

No

[valid iron]

basically try to make all the denominators the same by multiplying/dividing