help-13

[crimson sedge]

I assume here

you convert sin2 into cos2

or no

you multiply out

[inland ocean]

None of that will be needed

Use $(a^2 - b^2) = (a+b)(a-b)$for numerator

[wraith dagger]

ColdTee

[inland ocean]

Don't you see the cos and sine in a similiar form

[crimson sedge]

ohy ep that works

I got

it

[crimson sedge]

yep I thought you were referring to using that

when multiplying out

because if you multiply out the denominator

next time both sides are the same

so both ways work

thank you bro

[inland ocean]

Yw

[crimson sedge]

.close

[cedar kiln]

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[cedar kiln]

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[novel kiln]

can i get some help

[cedar kiln]

Please don't occupy multiple help channels.

[obsidian coral]

Don't open multiple channels

.close

[cedar kiln]

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[trim helm]

just to make sure of my answer, is the answer C?

[trim helm]

I don't even know what I did to come to the conclusion that it's c

[frosty vapor]

yeah youre right

think about y=mx+b

b=0 because 0 hours means 0 folded

[cedar kiln]

@trim helm Has your question been resolved?

[trim helm]

Ah thanks!

[cedar kiln]

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[final glen]

How would I get x

[cedar hull]

You either have to be given a piece of information to calculate x, or the answers are to be written in terms of x. don't see it any other way.

[final glen]

Ok

[final glen]

Can you help with 2 more

[cedar hull]

2 more questions ?

[final glen]

Yes

[final glen]

^

[cedar hull]

well ask them. I will help if I can

[final glen]

^

[cedar hull]

I would like you to do it with me

the first question

how would you approach it ?

[final glen]

I would multiply it

[cedar hull]

and solve two equations in two unknowns ?

[final glen]

2(1,0)+3(0,1)

^ thats what id do

[cedar hull]

how would that help

[final glen]

Would I have to create 2 equations

[cedar hull]

yes

that's good

but judging by the questions you sent me so far

this is a matrix exercise

I want you to do it with matricies

[final glen]

Idk how to do this one with matrices

But

Here are my 2 eq

3a+b=2

2a+b=3

[cedar hull]

okay look

[final glen]

Yes

[cedar hull]

if you have something of the form

x * v1 + y * v2

this can be grouped like this (v1 v2)*(x y)

matrix multiplication holds on matrix blocks not just numbers

[final glen]

Alr

I got -1 and 5

[cedar hull]

$\begin{pmatrix}
a & b \
c & d
\end{pmatrix} \begin{pmatrix}
x \y

\end{pmatrix} = x \begin{pmatrix}
a \ c

\end{pmatrix}
+
y \begin{pmatrix}
b \ d

\end{pmatrix}$

[wraith dagger]

Mohamed Mohsen

[cedar hull]

that is what i was trying to say

[final glen]

Yes I understood that

[cedar hull]

that's good

so you can make it into a matrix formula and find inverse

you did that ?

[final glen]

I didn’t do this prt

[cedar hull]

If you can do it as exercise after the questions are done

i would like you to

I know it's easier to solve two equations the old way

but matrices will come in handy later on

[final glen]

Alr

[cedar hull]

now for the second question

[final glen]

X=y so I needa satisfy this condition

Which is why B should be correct

[cedar hull]

why does x = y

[final glen]

X-y=0

X=y

[cedar hull]

okay

[final glen]

And z should = 0

[cedar hull]

what about z though

[final glen]

^^^

Its given right

[cedar hull]

how does that help xD

okay

back a step

what is a null space

[final glen]

Based off of (x,y,z)

Since x = y

[cedar hull]

I understand your reasoning and it's correct

just tell me

what is the definition of a null space

[final glen]

Oh

Set of all vectors

[cedar hull]

I feel like you got what i'm hinting at

[final glen]

I don’t lol

[cedar hull]

xD

okay set of all vectors

such that the outcome is zero

you said if x = y

then

you get zero

that's correct

but you went to z and said the answer is already zero

but that is already a piece of info

z automatically going to zero means you can plug in anything in z and still get zero

this means all vectors of the form (0 0 z) go to zero

this adds to the basis doesn't it ?

[final glen]

Yes

Doesn’t it have to be linearly independent

[cedar hull]

yes

[final glen]

Then yes rhe answer is B

Lol

[cedar hull]

hold on

aren't they linearly independent ?

[final glen]

Yea

[cedar hull]

then why b xD

[final glen]

^

[final glen]

That is why

The other answer choices

Aren’t

[cedar hull]

first answer isn't ?

actually this question is confusing me xD

[final glen]

No it isn’t

[cedar hull]

reasoning ?

[final glen]

Vector in a set shouldn’t be scalar

For A

[cedar hull]

don't understand what that means

are the two vectors in the first answer

independent

yes or no and explain why using definition of linear independence

[final glen]

No

Bcz

No vector in the set should be a scalar multiple

[cedar hull]

is anyone a scalar multiple of the other ?

[final glen]

D is i think not sure tho

So Im between B or D

[cedar hull]

please answer my questions

D also works which is why i'm confusing

but apart from that

tell me the first answer

explain why it's wrong

and not in your head

I nead an example

show that one is multiple of the other

[final glen]

(0,0,1)=0*(1,1,0)

[cedar hull]

seriously xD

0 doesn't count

if it does that all vectors in existence and dependent

vectors $w,v$ are linearly independent if and only if $av+bw =0$, has one trivial solution, and that is $a=0,b=0$.

[wraith dagger]

Mohamed Mohsen

[cedar hull]

that is the definition of linear independence

so a multiple of 0 doesn't count

it should make sense

(0,0,1) = 0 * (1,0,0)

even though the point at completely different directions

the first points in z direction and the second points in x direction

and if so they are linearly independent.

[final glen]

Then idk then ig 😂

[cedar hull]

I'm just as confused as you xD

the first answer is linearly independent

but so is the last

and both are basis

so it's both A and D

who wrote this question 😂

[final glen]

My teacher

[cedar hull]

I think you may wanna ask him that

my thinking is he probably intended A as an answer and wanted to make other random vectors, but he made one of the random answers correct by mistake.

[final glen]

Idk whats goin on in this class

[cedar hull]

your teacher is confusing you, we need answers

[final glen]

She ain’t responding rn

She old

So prolly in bed

Ig

[cedar hull]

alright then wait for a response

but we know

the first two vectors are linearly independent

both individually give zero when plugged in

therefore this null space is two dimensional

so B can't be the answer

it's either A or C

[final glen]

Prolly C

Cuz thats the weird one

[cedar hull]

D I mean

xD

[final glen]

😂😂😂

[cedar hull]

no not C xD

[final glen]

Bruh u making me trip

[cedar hull]

yea the answer is horse

[final glen]

Ru cs major

[cedar hull]

No i'm way crazier

I'm an engineering bachelor who happens to be studying abstract math for 7 years

so i'm kinda doubly crazy

[final glen]

Im cs major

Math for 7 yrs thats crazy

[cedar hull]

could be more than 7 actually

started at 2015

[final glen]

What uni

[cedar hull]

you probably wouldn't know it

since i'm in egypt

[final glen]

Oh

[cedar hull]

it's called ain shams university

ASU for short

[final glen]

Yea don’t know it

[cosmic steppe]

Arizona state university

[cedar hull]

xD

[final glen]

Thats the only one Ik

[cedar hull]

yea mine is a crazy university

where exams are made for 3 hours and you are only given 2 to solve it

stuff like that xD

[final glen]

Thats happens

Which sucks

[cedar hull]

it happens in your college

in my college that's the only thing that happens

xD

[final glen]

I hate school in general

[cedar hull]

me too

they are bad at teaching

which is their only purpose

very ironic

[final glen]

Exactly

[cedar hull]

you had one job meme

xD

[final glen]

Fax

[cedar hull]

Okay so let us know whether or not the answer was horse

or maybe the answer was mount Everest

[final glen]

I will letchu know

[cedar hull]

thanks

[final glen]

Np

.close

[cedar kiln]

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[pale citrus]

Can someone explain to me why the hypotenuse/ the rope is equal to 150-(50-h)

[gritty viper]

the rope is 150 feet long

that length is divided between the hypotenuse segment and the vertical segment lifting the box

the box is h+2 feet off the ground, and the top of the pulley is at 52 feet

so the length of the rope not on the hypotenuse is 52-(h+2) = 50-h

[pale citrus]

oh ok. thank you!

.close

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[woven tartan]

Does anyone know how to solve the first order linear diffeq: x'=x{{1,1},{4,1}} for a matrix? I just need to general solutoin

[cedar kiln]

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[woven tartan]

Hey, what's the phase portrait look like for two equal eigenvalues of -3?

[woven tartan]

just one large omnidirectional sink?"

[slow thicket]

#linear-algebra

[woven tartan]

This is diffeq

[cedar kiln]

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[tropic oxide]

... do we have enough info to conclude that?

in this pair of congruent triangles, the angle which corresponds to S isn't A, it's M

you can't know any degree measures here

well ok no

only the degree measure of S is given

so you know that angle M = 38°

but you can't say anything about the measures of any other angles

[cedar kiln]

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[hexed blaze]

hi

[hexed blaze]

i have no clue how they got the bounds for part C

this is what I did on desmos

<@&286206848099549185>

[gentle flower]

send a clearer ss

[hexed blaze]

of what

[hexed blaze]

i sent it

[cedar kiln]

@hexed blaze Has your question been resolved?

[cedar kiln]

@hexed blaze Has your question been resolved?

[hexed blaze]

.done

.close

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[obtuse coral]

I’m confused here bc it talks abt central angles, but where even are they?

If that makes sense

Nvm, srry

Wrong pic

[cedar kiln]

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[foggy merlin]

gl bro

[tropic oxide]

WHY ARE YOU SHOUTING

[foggy merlin]

no

[tropic oxide]

are you in a test RIGHT NOW?

[crimson sedge]

Bro's trying to do a side quest

[tropic oxide]

are you in a test right now and do you want us to help you cheat?

then what do you want

[foggy merlin]

[crimson sedge]

Matrices?

:))))))

You're telling us your first question is 5×5

........the product?

A multiplication?

Literally.

[foggy merlin]

[crimson sedge]

Brother

I'm speechless.

I am sure this is some troll

Ah

[foggy merlin]

ah

we saw what u wrote

[tropic oxide]

@sonic loom how old are you

too young

come back in seven years

[crimson sedge]

I refuse to believe that

[foggy merlin]

[crimson sedge]

Bro's doing multiplication at SIX

[tropic oxide]

discord terms of service says you must be at least 13 years old.

this isn't about our server but about discord itself

[crimson sedge]

When I was 6 I was choosing my favourite ninjago

[tropic oxide]

<@&268886789983436800>

@sonic loom we do not allow underage people here

[crimson sedge]

That was a good laugh

[radiant topaz]

You are 6?

Cap

[crimson sedge]

You don't get the point

We're not giving you answers during a test

We're here to guide you and make you understand for other future problems.

[radiant topaz]

Who does multiplication in grade 1

[soft cloak]

in any case, you are too young for discord

[rapid halo]

Hang on

You're actually six years old?

[radiant topaz]

No

[rapid halo]

Okay

[crimson sedge]

This guy is asian
There's always an asian better than you, haha

[rapid halo]

Do not make racist jokes

[radiant topaz]

I'm Asian and there's always a six year old Asian better than me

[tropic oxide]

cut the racist talk

[crimson sedge]

It wasn't meant to be racist, my bad

[rapid halo]

That doesn't make it not racist

Anyway this help channel is now vacant

Move along

.close

[cedar kiln]

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[crimson sedge]

Thanks for not banning me.

I'll play safe

[radiant topaz]

Yes

[cedar kiln]

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[buoyant latch]

what can you glean about a matrix knowing the basis of its null space?

[cedar kiln]

@buoyant latch Has your question been resolved?

[cedar kiln]

@buoyant latch Has your question been resolved?

[crystal raptor]

The rank

[cedar kiln]

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[crimson sedge]

this is very easy

[crimson sedge]

where am i going wrong

apparently it should be 2.916666666

[cedar kiln]

@crimson sedge Has your question been resolved?

[buoyant latch]

,w 1/6 * sum i=1 to 6 i² - 3.5^2

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Frosst#1111

[buoyant latch]

,w 1/6 * sum i=1 to 6 i²

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Frosst#1111

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[crimson sedge]

i need help

[warped sierra]

I need help with this.

[gritty galleon]

#❓how-to-get-help

[crimson sedge]

Wow, 2 in 1

[crimson sedge]

This sticker is really ugly why do I keep using it

[warped sierra]

Do you mind helping me with mine?

[crimson sedge]

#❓how-to-get-help, again

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

.close

[cedar kiln]

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[restive flint]

so I work with an double integral and I drew the borders, I want to see if I am below the line or above but is there any good trick

[restive flint]

I want the know where x2 <= x1 - 10

so I drew the line x2 = x1 - 10

[restive flint]

I know a trick i choose an x1 and x2 below the line and an x1 and x2 above the line

the one that satisfies it means the part where I should look at

[cedar kiln]

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[drifting gulch]

you need to multiply the exponent

in the numberator sectin

numerator section*

distribute the ^3

so it would be 3.6^3 x (10^-2)^3

[cedar kiln]

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[zealous vale]

What is the length of the segment cut by the coordinate axes on the line AB, A(- 2, 4), B(4, 1)?

[zealous vale]

The picture above is what I've done so far. I think I have to use pythagoras theorem formula for this one. But I don't know how to apply it?

[tropic oxide]

find the points where the line AB intersects each axis.

[zealous vale]

(5.5, 3)

[tropic oxide]

i don't think you said this correctly.

you gave one point, not two. and this point doesn't lie on the x axis nor does it lie on the y axis, so it can't be either of the two points i asked for.

[zealous vale]

Oops. Let me fix that

(5.5, 0)
(0, 3)

[tropic oxide]

there we go.

now find the distance between these!

[zealous vale]

6.3

cm

I used a ruler to measure the distance for all of these. Idk if I'm supposed to do that

[tropic oxide]

no you're not

[zealous vale]

Yeah, definitely no. The answer sheet says 3√5. Must be a formula

[tropic oxide]

actually... hold on.

are you sure your axis intercepts are correct

bc i took those on faith but now they look sus to me

are you sure the intersection with the x axis happens at (5.5, 0)?

[zealous vale]

No, I measured that with a ruler. How else am I supposed to get that point?

[tropic oxide]

you're supposed to find the equation of line AB first

then in that equation, set y to zero and solve for x

[zealous vale]

Like a function equation?

[tropic oxide]

... yes

you have done those before, right?

perhaps the form y=mx+b might ring some bells?

[zealous vale]

Yes, of course

What do I input in the place of m and b though?

[tropic oxide]

you want the line which passes through your points A and B, whose coordinates you know.

m is the slope of your line.

b is the y intercept, which you can find by inputting the slope and one of your points into the formula.

[zealous vale]

So I know m is negative

[tropic oxide]

you do, but you can also calculate its exact value

do you know how to get the slope of a line going through two points?

[zealous vale]

I forgot, haven't done those in a long time. I watched a youtube video just now, I got

y=-2x+3

Idk if that is correct though

[tropic oxide]

it is not.

[tropic oxide]

perhaps you should review that part of algebra in general tbh

[zealous vale]

.close

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[wild belfry]

https://i.gyazo.com/22fe54a2fbcac0c15a746eaaf459e2aa.png

[wild belfry]

I know theres a way to identify the answer straight away when its in this form but I forget it

[sage forge]

What have you tried?

[wild belfry]

I know x intercept is 5, y intercept is 8, but those are the only ones I can identify

I forget what the 6 and 7 represent

I could guess but then I wouldnt understand how to get it haha

[sage forge]

x intercept is 5? So 6•7^(5-5)-8 = 6-8 = 0??

For the y intercept you plug in x = 0 and get that this is wrong, too

[wild belfry]

oh ok, I guess I have no idea what to do then haha

[sage forge]

No problem 🙂

[wild belfry]

Wait I just remembered

y=-7

Cause it usually goes through (0,1)

so need to add 1

[sage forge]

What about it?

[wild belfry]

Idk

it just helps a little

[sage forge]

Nah, that isn't correct, too

[wild belfry]

what hahaha

goddam

[sage forge]

I think you have some difficulties to unserstand what the 6,5,8,7 do

It is not really necessary to unserstand it for this particular question, but do you want to go through it anyways?

[wild belfry]

Yes please!!

That would really help

It will make it waaaay easier in the future if i understand it all

[sage forge]

So you know how the normal exponential function looks like?

[wild belfry]

yeye

[sage forge]

Like e^x?

[wild belfry]

hard to describe

ye

theres an invisible line that it doesnt cross

and it keeps going in a certain direction

getting closer to zero

also it usually goes through (0,1)

[sage forge]

Let^s say we have 7^x now. This function would grow faster for x > 0 than e^x as 7 > e. Would it still go through (0,1)?

[wild belfry]

Yes

[sage forge]

Very good

Now the 6 in 6*7^x takes the graph "stretches" everything in y direction

Does 6*7^x go through (0,1)?

[wild belfry]

Hmm I dont really know what you mean by this

[sage forge]

It multiplies everything by 6

[wild belfry]

so like the vertical asmptote changes, right?

[sage forge]

It does only have a horizontal asymtote

Sorry, got to eat for a second

[wild belfry]

yeye np

I will take a break aswell, gonna come back to this question in a while 😄

Thanks for your help!!!

❤️

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[random salmon]

help

[random salmon]

Is this true?

[livid hound]

no

without parentheses the tail takes priority

[random salmon]

The tails?

[livid hound]

$$2^{\red{x^2}} = 2^{(x^2)}$$
$$2^{\blue{{2^x}} = 2^{(2^x)}$$

the back ends

[wraith dagger]

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

[random salmon]

so the order matters

[livid hound]

so power of power doesn't apply there

[random salmon]

what does that mean ?

so is this true

$$2^{2x}=\left(2^2\right)^x$$

[wraith dagger]

forin

[livid hound]

yes

[random salmon]

what about this ?
$$2^{2x}=\left(2^x\right)^2$$

[wraith dagger]

forin

[livid hound]

yes

[random salmon]

really ?

[livid hound]

yes means yes

[random salmon]

so why is this NOT true

[rare glacier]

he is right

[livid hound]

middle line is wrong because there is no 2^x there to replace t with

last line is valid

[rare glacier]

yep

[random salmon]

so you are saying is this expression there is no 2^X ? $$\left(2^2\right)^x$$

[livid hound]

yes

[random salmon]

$$\left(2^2\right)^x$$

[wraith dagger]

forin

[rare glacier]

,help

[wraith dagger]

A brief description and guide on how to use me was sent to your DMs!
Please use ,list to see a list of all my commands, and ,help cmd to get detailed help on a command!

[livid hound]

$2^{\gray{(}{\red{2^x}}\gray{)}}$ has $2^x$ \
${(2^2)}^x$ does NOT. \
${(2^2)}^x$ is NOT the same as $2^{\gray{(}{\red{2^x}}\gray{)}}$ \

[wraith dagger]

ℝamonov

[random salmon]

hmm yeah that makes sense thank you , this makes it very clear thanks 👍

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[bronze briar]

does this proof require mathematical induction?

[bronze briar]

or would there be a different way

[sand pivot]

induction is a valid way to prove it

[bronze briar]

for induction, would my base case be k = 1? So i would have n, and obviously 1 divides n

[sand pivot]

well

remember how a proof by induction works

you assume the n case

[bronze briar]

yeah but there needs to be a base case first

right?

[sand pivot]

and then try to use the assumption to show the n+1 case must also be true

your base case of 1 works

[bronze briar]

oh okay. I just wanted to make sure the base case was correct. Ill see what i can do now. Thank you!

[sand pivot]

👍

[cedar kiln]

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[barren bridge]

How would I go about solving this?? I’m very confused I have no where to start or anything.

[worthy vale]

For the first question, make a pythagorean equation and make b the unknown

you know $a^2 + b^2 = 1$ and the radius is 1 because it is a unit circle

[wraith dagger]

Aorliei

[worthy vale]

Does that make sense?

[barren bridge]

Ohhhh

Yes

[worthy vale]

so it would be $-0.7^2 +b^2 = 1$

and once you have b you can figure out the angle using inverse of tan

[wraith dagger]

Aorliei

[barren bridge]

How would I do inverse of tan?

[worthy vale]

Uh how do you normally find out the angle

I'm not sure how you do it but the way I do it is by rearranging the equation tan angle = opposite/adjacent

opposite would be your y coordinate and adjacent would be your x coordinate

[barren bridge]

Ohh okay right

[worthy vale]

to get your angle you would divide the entire equation by tan

so it would be angle = (opposite/adjacent) x by inverse of tan

[barren bridge]

Ohh ok thank you

[carmine bronze]

Have you learned about inverse trigonometric functions?

[barren bridge]

I think I don’t know I’m super far behind in my class because I haven’t been listening so I’m trying to teach myself to catch up

It would probably help to explain how to do inverse

[carmine bronze]

Are you familiar with SOH-CAH-TOA?

[barren bridge]

Vaguely I learned about it like 2-3 years ago I took trigonometry in 10th or 11th grade so it’s been a couple years

[carmine bronze]

SOH-CAH-TOA is a mnemonic trick to recall the association between the sides of a Right Triangle and the the angles of the triangle.

SOH => sin(theta) = Opposite/Hypotenuse
CAH => cos(theta) = Adjacent/Hypotenuse
TOA => tan(theta) = Opposite/Adjacent

[barren bridge]

OH RIGHT

[carmine bronze]

An inverse trigonometric function does the reverse of the above. It takes the sides of a Right Triangle and gives you the angle.

We indicate an inverse trigonometric function by either ^-1, arctrig_function, or atrig_function.

$\sin^{-1}{(\theta)} = \arcsin(\theta) = \asin(\theta)$

[wraith dagger]

Kookiemon

[barren bridge]

So is sin^-1 equal to hypotenuse/opposite?

[carmine bronze]

No, an inverse trigonometric function takes the sides as input and outputs an angle.

$\sin^{-1}{\left(\frac{O}{H} \right)} = \theta$

[wraith dagger]

Kookiemon

[barren bridge]

Ohhh

Okay that makes sense

[carmine bronze]

There is a caveat though.

When using a inverse trigonometric function, there are always two possible results.

[barren bridge]

Ok

[carmine bronze]

The angle made by the x-axis and the points A and B both have the same angle.

[barren bridge]

But they are flipped along the X axis

[carmine bronze]

The angle in the 4th Quadrant is equal to 360 - 47.25 = 312.75.

[barren bridge]

Ohh

Okay wait question

I got this for the first answer is this right?

[carmine bronze]

That gives you the sides of triangle. You will need to use the inverse tangent function to find an angle. Because there are two possible angles, you need to determine which angle based on the quadrant it lies in.

[barren bridge]

It goes from top left to bottom right
2 1
3 4 right?

[carmine bronze]

Yes.

[barren bridge]

For quadrants?

Okay cool

So it’s in the second quadrant

[carmine bronze]

Yes.

[barren bridge]

So it’s 46 rounded to the nearest degree?

Or it’s the opposite

And its 360 minus 46

[carmine bronze]

The question is looking for the angle theta.

[barren bridge]

Oh right 46 doesn’t work there

[barren bridge]

Yeah that’s what I got

[carmine bronze]

Correct. When you use a trigonometric function, it calculates the angle inside of the triangle formed.

To calculate the correct angle, you will need to subtract the angle you found from 180 degrees.

[barren bridge]

OH RIGHT BECAUSE ITS A LINE

[carmine bronze]

Correct.

[barren bridge]

SO IT WOULD BE 134

[carmine bronze]

👍

[barren bridge]

Omg that makes so much sense

Okay I’m gonna try the second one on my own

So the second one should be 150

[carmine bronze]

Yes.

[barren bridge]

Omg thank you

[carmine bronze]

yw

[barren bridge]

One more problem I have a question on, i just looked through all of them and I know how to do law of sine and cosine but this one I don’t understand

Wait

Actually

It’s similar to what I just did isn’t it?

[carmine bronze]

Yes.

[barren bridge]

Okay then I think I’m good

You are a life saver thank you so much

[carmine bronze]

yw 🙂

[barren bridge]

.close

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[exotic zenith]

How do you do 7-8 and 9-11

[obsidian coral]

,rotate

[hushed birch]

@exotic zenith Well what is tan

like

what sides of a triangle is tan

rather ill just list u the steps

1. Tan is opp/adj, put that in terms of 3/1 since tan = 3.
2. Find the remaining side using pythag theorom a^2 + b^2 = c^2
3. Find sin by its identity of opp/hyp

[cedar kiln]

@exotic zenith Has your question been resolved?

[exotic zenith]

Oh

Ok thx

[cedar kiln]

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[trail bear]

I need help with the Volume of a Rectangular Prism

[gentle flower]

do you know what the formula is?

for volume

[dull oxide]

,rccw

[trail bear]

V=(bh2)L

[cedar kiln]

@trail bear Has your question been resolved?

[upper jacinth]

Area of triangle * prism height

Are of triangle is height of triangle * size of a side with height / 2

So 8 * 12 / 2 * 20

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[rotund escarp]

How would I go about showing that $\ell^\infty$ is a clopen subspace of $R^\infty$ with box topology?

[wraith dagger]

dream was legit

[cedar kiln]

@rotund escarp Has your question been resolved?

[rotund escarp]

idk how to start, like I've been asked to prove properties of spaces but never asked explicitly to show that something is clopen

Can I even use the metric here? induced by sup norm

or no

cause if I have a metric, it's easy if not, idk how to show that it's open

[rotund escarp]

@lusty birch i beg

<@&286206848099549185>

[cedar kiln]

@rotund escarp Has your question been resolved?

[cedar kiln]

@rotund escarp Has your question been resolved?

[rotund escarp]

nvm got it

.clos

.close

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[bronze briar]

so im trying to prove this by induction
Im not going to write the base case to save time
(but i wrote it down on my paper)
so for hte assumption i have assume for some k > 1, that the product of these k consecutive integers is divisible by k!
this means that k! | (n)(n+1)...(n+k-1)
then i said let Lk! = (n)(n+1)...(n+k-1) for some integer L.
then consider k + 1. We have (n)(n+1)...(n+k-1)(n+k)
I can rewrite this according the assumption to the following:
Lk!(n+k)
Then since n >= 1, I can say that (n+k) = [(k+1) + (n - 1)]. Thus Lk![(k+1) + (n - 1)]
now im confused on where to go and i can't really make any progress

[cedar kiln]

@bronze briar Has your question been resolved?

[bronze briar]

<@&286206848099549185>

[cedar kiln]

@bronze briar Has your question been resolved?

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[main oak]

|4-x^2|=2-x

[cedar kiln]

Please don't occupy multiple help channels.

[main oak]

.close

[cedar kiln]

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[main oak]

.close

[cedar kiln]

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[misty tinsel]

Help

[misty tinsel]

The midline is 0 I think

And the rest idek

[patent flame]

if f(x) = cos x then you're drawing 7f(4x) if that makes sense

maybe that makes it easier to draw idk

[misty tinsel]

Not drawing we gotta plot

If it was drawing it would’ve been kinda easier

[plain breach]

amplitude is 7

[patent flame]

still usable i think if you know where the maximum and minimum are on the normal cos graph

[plain breach]

y=Acos(Bx)

one cycle is 2pi/B

2pi/4=pi/2

[misty tinsel]

Oh

[misty tinsel]

We subtract 7-4 right

[plain breach]

no

[plain breach]

Using this the amplitude is A

[misty tinsel]

And 4-7

Ok

[patent flame]

no what i was getting at is what @plain breach described

[plain breach]

the graph hasn't been shifted vertically

so the range is from -7 to 7

[misty tinsel]

Oh

So the min is -7?

Max is 7?

[plain breach]

yes

[cedar kiln]

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[spark wraith]

I found 36

[vagrant elbow]

Yes

[spark wraith]

thank u

also..

is it c

[fair geyser]

yea

[spark wraith]

is it d

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[thin galleon]

i am dividing 1 by 243 with long division. after 0.004115226337 I get the remainder 9, and I add a 0 to this. 243 doesn't divide into 90, should I add a 0 to the quotient?

[gray blade]

why add a 0?

[thin galleon]

because 243 is greater than 90, that's my intuition

[livid hound]

yeh

with division this long, ideally you'd use grid paper

to see the alignment

[thin galleon]

thanks

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[zinc pagoda]

Can someone help me with this question please?

[zinc pagoda]

<@&286206848099549185>

[urban ermine]

well very quickly we can see that we can make the numerator have the same base as the denominator

[zinc pagoda]

okay what is the second step?

[urban ermine]

what can 9 be rewritten as?

[zinc pagoda]

3^2

[urban ermine]

perfect

now recall this:

what is x^3/x^2

[zinc pagoda]

x

[urban ermine]

or it can be written as this right:

[zinc pagoda]

yeah

[urban ermine]

so then what would we have?

3^(2x-2-(y+2))=81

now, how can 81 be re-written as in terms of base 3?

[zinc pagoda]

I don't get it

[urban ermine]

hm

[zinc pagoda]

how do you get this step

[urban ermine]

from here

[zinc pagoda]

an then

can I divide both side by 3

@urban ermine

[zinc pagoda]

okay

and what is the next step?

[urban ermine]

now can we make these the same base again?

[urban ermine]

uh

3^27 is not 81

[zinc pagoda]

how to solve for x

[urban ermine]

we need to make 81 have the same base

so 3^ what = 81

[zinc pagoda]

4

[cedar kiln]

@zinc pagoda Has your question been resolved?

[zinc pagoda]

nvm

[cedar kiln]

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[hollow geyser]

I have to find every real value of k so that the projection of x=(2, k, -2) over y=(k-1, 1, -2) gives me a vector of norm=2

[hollow geyser]

how would I go about solving this? I tried doing the projection with k and then making the norm of that equal 2 but I ended up with a really large polynomial that I had no way of solving

[tiny wing]

what polynomial did you get?

[cedar kiln]

@hollow geyser Has your question been resolved?

[hollow geyser]

k^4 + 10^3 - 24k^2 - 32k + 152 = 0

[cedar kiln]

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[hollow geyser]

:(

.reopen

[cedar kiln]

✅

[cedar kiln]

@hollow geyser Has your question been resolved?

[hollow geyser]

<@&286206848099549185>

[fallen moat]

Greetings
what do you mean by projection with k?
and how did you get such big polynomial?

[hollow geyser]

I mean that I tried to go through the steps of projecting x onto y to get a new vector with the unknown "k" as to solve it algebraically

[fallen moat]

i see

and i feel weird cuz i got 2nd deg poly rather than a 4th deg poly

can you show your work (pic)?

[hollow geyser]

Yeah, sure

one second

[fallen moat]

reading

[hollow geyser]

Did I already screw something up?

[fallen moat]

you might wanna use the scalar projection one instead

[hollow geyser]

oh

[fallen moat]

then you would have a 2nd deg one instead of 4th deg

[hollow geyser]

So then if theta is between 0° and 90° then the scalar projection will give me the norm of the projected vector?

[fallen moat]

without the condition you proposed, this is still valid

[hollow geyser]

hm

so which of the formulas should I use with the information I have available?

[fallen moat]

scalar projection of course because the question requires the projection to have a norm of 2

[hollow geyser]

I'll give it a try

thank you

[hollow geyser]

Alright

I got k= 2+sqrt(50) and k= 2-sqrt(50)

does it check out?

I didn't know about scalar projection, by the way, though it's derived from things that I do know

another tool in my toolkit I suppose

[cedar kiln]

@hollow geyser Has your question been resolved?

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[lofty schooner]

Hello can I have some help with this question

[hollow timber]

by slope does it mean gradient

[lofty schooner]

yes

i think

[hollow timber]

so you know what the gradient is

[lofty schooner]

so the answer is b

[hollow timber]

nothats not it

it has a gradient of -0.5

[lofty schooner]

ok

so its a

[hollow timber]

no

the gradient tells you how much it changes in y for every 1 in x

[lofty schooner]

ok

[hollow timber]

if that makes any sense

so you are looking for a negative gradient where for every 1 moved in x 2 is moved in y

[lofty schooner]

so its not a or b

[hollow timber]

a and b have gradienr of 0.5 and -0.5 so no

yeah mb

[lofty schooner]

so is it c

[hollow timber]

yeah

[lofty schooner]

ok so on to my next question

[hollow timber]

first of all look at the gradient

then look at the y intercept

[lofty schooner]

so this a positive slope

[hollow timber]

yeah

[lofty schooner]

knew because of this image

[hollow timber]

yeah now you need the y intercept and gradient

[hollow timber]

yeah thats right

[lofty schooner]

so that means this is a positive slope

[hollow timber]

yeah it has a positive gradient and thus is a positive slope#

[lofty schooner]

ok so since it only with the 2 digits 4 and 3

[hollow timber]

when x is 0 what is y

[lofty schooner]

y=b

[hollow timber]

yeah

[lofty schooner]

oh

[hollow timber]

this will help you narrow it down and then you wont need any further working

yeah

when x=2 then y=5 when x=1 y=1 when x=0 y=?

[lofty schooner]

and what is that potential answer might i ask

[hollow timber]

do you understand what the y intercept is

[lofty schooner]

is it y=-1/4x+3

[hollow timber]

no

you found a positive gradient earlier

and when x=0 y is not 3

[lofty schooner]

so there are no negatives in the answer

[hollow timber]

im not going to tell you but a positive gradient means the coefficient or a in ax=y is positive

look at where the line crosses the y axis

what is that co ordinate

[lofty schooner]

3 and 4

it crosses -1

[hollow timber]

no where the line crossed the y axis

[lofty schooner]

2,3

[hollow timber]

where x is 0 what is y

[lofty schooner]

-3

[hollow timber]

yeah thats the y intercept

now look for positive gradient with y interceot of -3

[lofty schooner]

4

[hollow timber]

if you mean answer 4 then no that has a negative gradient youre looking for a positive

[lofty schooner]

okay so is it y=-4x-3

or y=4 x -3

the first or last one

[hollow timber]

positive gradient

y=ax+b where a is positive

[lofty schooner]

so y=4x-3

[hollow timber]

yeah

[lofty schooner]

here is the 3rd question

[hollow timber]

so i im pretty sure direct variation function means the y intercept is 0

[lofty schooner]

ok

[hollow timber]

lines cross through origin (0,0)

theres 2 lines which do this

[lofty schooner]

so i might be wrong but the answer might be b and a

[hollow timber]

yeah thats right

[lofty schooner]

wait what the answer is b and a

[hollow timber]

yeah y intercept is 0 they pass through the origin they are in the form y=ax

[lofty schooner]

ok so next question

umm hello

.close

[cedar kiln]

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[crimson sedge]

Can someone help me with this

[dire geode]

,rotate

[dire geode]

!show

[cedar kiln]

Show your work, and if possible, explain where you are stuck.

[shell smelt]

$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} sin(cos(x)) sin^{99}(x) dx$

[wraith dagger]

Oğuzhan

[shell smelt]

Have I read it correctly?

[crimson sedge]

Yes

[shell smelt]

$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} sin(cos(x)) sin^{99}(x) dx$

Let $u = cos(x)$

$\frac{du}{dx} = -sin(x)$

$- \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} sin(u) (sin(arccos(u)))^{98} dx$

$- \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} sin(u) (\sqrt{1 - u^2})^{98} dx$

$- \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} sin(u) (1 - u^2)^{49} dx$

[wraith dagger]

Oğuzhan

[shell smelt]

I might be going in the wrong direction

[dire geode]

did you write down the problem wrong

[shell smelt]

No I asked it

[dire geode]

are you @crimson sedge ?

[shell smelt]

Technically solveable if you use binomial expansion and integration by parts but I don't think anyone will/should do that

[crimson sedge]

This is what I got in class so I think its right.

[dire geode]

what is "this"

[shell smelt]

this @dire geode

[dire geode]

are you @crimson sedge ?

[crimson sedge]

What?

[dire geode]

are you @shell smelt ?

[crimson sedge]

@dire geode so you can help me with the integral?

[dire geode]

no. i think you wrote it down wrong

,w int sin(cos(x)) sin^99(x) dx

[wraith dagger]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by riemann#2529

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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Closed due to the original message being deleted

[cedar kiln]

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[stark bronze]

[\text{Find }\frac{d^2y}{dx^2} \text{ for}]
[y=-3x^6 + 7]
[\frac{dy}{dx} = -18x^5]
[\frac{d^2y}{dx^2} = -90x^4]

[wraith dagger]

dopediscorduser

[stark bronze]

Does this look right?

[dire geode]

yes

[stark bronze]

thanks

.close

[cedar kiln]

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[lethal bison]

how did they do this simplification?

[dire geode]

do you know how to take determinants?

[lethal bison]

yes

ad-bc

I'm familiar with linear algebra already it's just how did the top part get to 4 and bottom to (x^2+y^2)^2

[dire geode]

did you do this calculation?

[lethal bison]

yes

[dire geode]

show