help-28

[bleak furnace]

Yeah

[hot herald]

would you be as to identify the roots and/or locations of the intercepts with the horizontal axis (usually x-axis but in this case you have k) of such a parabola

[bleak furnace]

Is like finding x in a x^2+bx+c right?

I used to factorize to find my intercepts

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[terse siren]

Someone said to me that Pythagoras theorem is the same as sin, cos, and tan as well as learning common values on the unit circle and I didn't believe them so is this true and if not how do I do sin, cos, and tan as well as learning common values on the unit circle?

[ivory cairn]

sine, cosine, tangent, and pythagorean theorem all relate to a right triangle, Yes those things can all be used together, there are a few versions of pythagorean theorem that relate the trig functions.

I don't know what you mean by "pythagorean theorem is the same and sin, cos, and tan"

[terse siren]

that's just what they said or maybe they said they all figure out the same thing.

[terse siren]

so what's the same thing about them?

[ivory cairn]

If you have a right triangle with legs $a$, $b$, and hypotenuse $c$, then pythagorean theorem says $a^2+b^2=c^2$. you know $\sin(\theta) = \frac ac$, $\cos(\theta) = \frac bc$. So dividing the pythagorean theorem by $c^2$ you get $\frac{a^2}{c^2} + \frac{b^2}{c^2} = 1$, which in turn gives you $\sin^2(\theta) +\cos^2(\theta) = 1$.

[glossy valve]

Zybikron

[ivory cairn]

you can get similar equations dividing by a^2 or b^2 that would involve tangent.

[terse siren]

I guess I was asking so if I know Pythagorean theorem do I still need to learn cos, and tan, learning common values on the unit circle?

[ivory cairn]

yes

[terse siren]

ok will it be easier though?

[ivory cairn]

yes

$\sin^2(\theta) + \cos^2(\theta) = 1$ is true, but you ened to know either sin or cos to find the other

[glossy valve]

Zybikron

[terse siren]

@ivory cairn Thanks

https://tenor.com/view/you-are-literally-the-nicest-person-you-are-so-nice-freaked-out-hughie-campbell-jack-quaid-gif-15588419

https://tenor.com/view/thanks-thank-you-gif-20870035

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[brittle iris]

Do we use quadratic equation for questions like such??

[brittle iris]

So to detemrine how long it takes for it to land in question 1

We use quadratic formula

Correct?

[cunning stump]

sure, as the rocket will have landed when it is at height 0 off the ground

quad formula will tell you where any 0s are

[brittle iris]

Ty so much

Do we also use it to find anything else?

Maximum height etc

[cunning stump]

no, quad formula will just tell you where the zeros are

but you may find the maximum or minimum of a quadratic formula in other ways, such as writing it in vertex form

that being said you can use the quad formula to derive things like where the axis of symmetry is, which you can then use to find a max/min point

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[warm oyster]

I would try taking the sin of both sides, should be able to cancel the trig with a few identities

[sage thorn]

How do you do that?

With the double angle formula or smth?

[warm oyster]

sin(u+v) = cos(u)sin(v)+cos(v)sin(u)

After simplifying, there are identities for cos(sin^-1(x)) and sin(2cos^-1(x))

Should just be a cubic equation at this point

[sage thorn]

Ohhh okayyy

[sage thorn]

How do I change it to sin when it is in arc sin?

[warm oyster]

At what point

[sage thorn]

In the wuestion

[warm oyster]

Do sin( sin^-1(x-1) + 2cos^-1(x-1)) = sin(pi/2)

Sin^-1 and 2cos^-1 are treated as u and v

[warm oyster]

In this

And the RHS is just 1

[sage thorn]

Okay

I’ll try it

I’ll ask again if I need help

Thank you

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[agile vigil]

i need to find a sum

integration term by term or differentiation term by term ?

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[sage thorn]

Still confused :^

Can someone show their working and and axplain for me? Thank yiu

[warm oyster]

Did you expand it yet after taking the sin of both sides

[sage thorn]

I didn’t know how to do that

Canbyou show the working out?

<@&286206848099549185>

!!!

<@&286206848099549185>

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[agile vigil]

can someone help with finding a sum

differentiation term by term

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@coral cape Has your question been resolved?

[coral cape]

<@&286206848099549185>

[fleet grove]

im not sure how to solve this but i see some circular edges, perhaps go and find the circumference of those

radius of each circle is 5 cms

[coral cape]

well im not sure how to split up the figure

[fleet grove]

ok wait lemme draw a diagram

i labeled the vertical surface as A

there are 2 A's

for B, use the 10 cms labeled on the picture times the length of it which u can obtain by 2x pi x r

for side A use pi x r^2

also don't forget the underside

the underside has an area of 10 times 30

[coral cape]

tysm ill go try it now

[fleet grove]

yw

[coral cape]

how do i use 2 x pi x r to find b? can i just use 2 x pi times r times h for 5 seperate quarter curved sa's?

well 2 x pi x r x h works

tysm for the rest of the help tho

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[torn jolt]

What is the meaning of this symbols

[gray hound]

x belongs to U right

[torn jolt]

I believe it's typo 🙂

[torn jolt]

Yep ig

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@faint raptor Has your question been resolved?

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[trail canyon]

i am trying to determine whether 47 is a quadratic residue of 97 or not

[trail canyon]

i initially thought of using $(\frac{47}{97}) = (\frac{-50}{97}) = (\frac{-10}{97}) * (\frac{5}{97})$ , but this seems to only make it worse, i am somewhat lost on where is a good place to start with this

[glossy valve]

Conor Sheridan

[trail canyon]

never mind i was being dumb and forgot about the quadratic reciprosity theorem

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[umbral plinth]

Hello

[umbral plinth]

i have a problem :

f can be derive from R to R

We assume that there is b in R such as f(b) = 0 and lim(f(x)) = 0 (x -> +inf)

I have to prove that there is c \in ]b; + inf[ such that f'(c) = 0

but how to proceed ?

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@umbral plinth Has your question been resolved?

[mellow niche]

@umbral plinth just use LMVT

[umbral plinth]

LMVT ?

[mellow niche]

Lagrange mean value theorem

[umbral plinth]

i already use that

but it leads means nowhere

[mellow niche]

f'c= lim n tends to inf f(n)-f(b)/(n-b)

Just zero

[umbral plinth]

ah

okkkkkk

i understand i'm stupid sry

thx

[mellow niche]

No worries

[umbral plinth]

but

[umbral plinth]

there is a problem

[mellow niche]

Yes?

[umbral plinth]

I can use the theorem only on a segment [a,b] where f(a) = f(b)

otherwise I will not have f'(c) = 0

I can study 2 cases

• case 1: there exists d in ]b;+inf[ such that f(d) = 0
• case 2: there is no such d

[mellow niche]

Yep

f(inf) tends to zero right?

[umbral plinth]

yes

[mellow niche]

So apply it b/w b to n where n tends to inf

There has to be some b±h where fx equals the value at inf.

h tends to zero*

[umbral plinth]

ok i understand thx

[mellow niche]

Use .close to free the channel

You can always take up an empty one later

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[torn jolt]

h

[torn jolt]

not understanding how they get from the first step to the second

[dim dragon]

they took y common in denominator

and then split it

[elfin stump]

That is called partial fractions

Google it, its a neat tool

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[torn jolt]

NO

wait im sorry

.reopen

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✅

[torn jolt]

did i do this right?

its been a while

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[limber forge]

could anyone help me with where to start here? im not sure what to do

[torn jolt]

it wants you to create the pattern as a summation then find the sum of that pattern

[limber forge]

should i stick with just 4 tho? or do more

and how do i find the sum without a value for x

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[torn jolt]

the 4 can be separated from the pattern and then you can make a summation like $4-\Sigma_1^n pattern here$

[glossy valve]

Critzzzy

[limber forge]

ah

im still a little confused, where would i go from there?

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[chrome idol]

If G is Abelian and H is a subgroup of G, does H have to also be Abelian

[brave blaze]

yea it is

[chrome idol]

I just realized what I asked wasnt relavant to the proof Im writing LOL

but good to know

thank you

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[cobalt kettle]

hello

[cobalt kettle]

How would you find the Particular Integral for something like this:

Given that I already have the complementary function as such:

[left glen]

Let x = Acos2t + Bsin2t

[left glen]

Or lambda and mu replaced by A and B respectively

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[stiff radish]

i don't know how this is done...

[left glen]

Try writing z in the form a + bi

Then compare real and imaginary parts

2a + 2bi + a - bi on the LHS

[stiff radish]

oh so what is z defined as? a+bi?

[left glen]

Yea

[stiff radish]

ohhh ok makes sense thanks man

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[unreal tide]

Need help on precalc homework not sure what to do; can someone explain thanks

[stiff musk]

what did you get for k?

[burnt pendant]

sin^2 + cos^2 = 1
since cos = 3/7, we get sin^2 + 9/49 = 1
so sin(A) = +/- sqrt(40)/7
since this is a triangle we assume positive sides, so sqrt(40)/7
sin(A) = 2 sqrt(10)/7

now what is the ratio for sin in a triangle

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[burnt pendant]

@unreal tide

[unreal tide]

i dont get it

[burnt pendant]

don’t get what

[unreal tide]

your explanation

[burnt pendant]

what don’t u get

[stiff musk]

i don't really see why the sine calculation is needed, you already know that l is 4

you need to find k in order to get e

[burnt pendant]

you don’t need k

you can find e without finding k

[stiff musk]

and to get k you know that cos(theta) = 3/7 = k/l = k/4

so can you solve 3/7 = k/4 for k?

[unreal tide]

we haven't learned the sine calculation yet so i think thats why im confused

[burnt pendant]

how does cos = k/l

[stiff musk]

oops, need more coffee 😆

[burnt pendant]

cos = k/e

you use the sine calculation to find l/e

you know l is 4 so 4/e = sin(A)

[burnt pendant]

and you do know the sine calculation if you know pythagorean theorem

[unreal tide]

oh ok

maybe my teacher just never gave us that term

ok thank u

[burnt pendant]

np

[unreal tide]

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[dull pumice]

I have z = 5 - 10i, w = -4 + 3i
asked for z * w

what I currently have: (5 -10i) * (-4 + 3i) = -20 + 15i -40i -30i^2
if this is even correct, how do I get to a form like x +/- yi ?

[hushed briar]

Be careful with the signs, -10i * (-4) = 40i.

You can use the fact that i^2=-1 so -i^2=1.

[dull pumice]

for example, I got a solution from a different task, that ends with 2 + 10i -4i -20i^2
= 22 + 6i

and I dont get why or how

[hushed briar]

(5-10i) * (-4+3i) = -20+15i+40i-30i^2 = -20+15i+40i+30

See this: -30i^2 = 30i

[dull pumice]

but why?

[hushed briar]

Because i^2=-1

That is the definition of i, the imaginary unit is the complex number that satisfies i^2=-1

It is the positive square root of -1.

[dull pumice]

oh I guess that makes sense.
so.. give me a sec to sum that all up

so that would be:
z * w = 10 + 55i
??

no wait

[hushed briar]

No

Yes, now yes

[dull pumice]

c:

and ehm the other task

[hushed briar]

I hope I could help you

[dull pumice]

so 2 + 10i -4i -20i^2
that means 2 + 10i -4i +20
then that would be 22 + 6i

ahhhhhhhhhhh

now I got it. thank u so much c:

great help!!!

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[craggy plinth]

So this was the answer to the problem i talked about earlier

What is the largest shelf we could make from the wood, if we wanted our shelf to be an isosceles triangle and still fit in the corner of the room perfectly?

This ^^^

Was the question

Amiright?

[echo lance]

where do we want this triangle to fit?

it should have hypotenuse < 130?

Isosceles so sides are equal
$$x^2 + x^2 < 130$$
$$2x^2 < 130$$
$$x^2 < 65$$

Assuming $x \geq 1$ so we don't have to switch the sign, we are looking for the largest one anyways

$$1 \leq x < \sqrt{65}$$

[glossy valve]

Doggo

[echo lance]

you can just do this..?

hm, can the hypotenuse be 130 though?

because you can't really give a number this way

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[torn jolt]

Uh, so, I just found out this... Why does this happen? Can anybody explain?

[spice orchid]

https://youtu.be/IMY2_yzDm9I

[torn jolt]

If you put a bunch of 5s together, let's say 13, and get its -1st power and get the sine of it, we get an approximation of pi

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[sweet pumice]

This is simple math in comparison to the other stuff I see here, but we all gotta learn.

[sacred fog]

"simple" is relative

[sweet pumice]

Isnt this lad just fabricating numbers or do I suck as much as I thought

[sacred fog]

what some may find simple is complex to others

[sacred fog]

wdym "fabricating numbers"

[sweet pumice]

where did he for instance get the 2 * 3 * x from?

[sacred fog]

good question

cuz x = 1 is wrong

[sweet pumice]

Yeah, we are supposed to anaylse and figure why this is wrong and argue for it.

But I can just argue that this lad is bringing numbers out of nowhere 🤷

[sacred fog]

they meant to multiply both sides by 3 * x

but forgot to multiply 5 by it as well

they just multiplied the thing on the rhs of the - by it

but forgot the left side

[sweet pumice]

Are you not supposed to multiply the top and bottom, sorry I dont know the words for it. Not a native speaker

[hot herald]

$$a \times \frac bc = \frac{ab}{c}$$
but that isn't really an issue here

[glossy valve]

ℝamonov

[hot herald]

the issue is they didn't apply the same operation to both sides of the equation

[sweet pumice]

So you cant cross out x with x on one side and do something opposite on the other?

[hot herald]

too vague

and an oversimplified representation of what's going on

if you have x/x simplifying that to 1 would be valid

however the issue is with how they got that in the first place

the issue is they didn't apply the same operation to both sides of the equation
one of the basic principles of algebraic manipulation is applying the same operations to both sides of the equation

here they multiplied the right side of the equation by 3x,
but multiplied whatever they felt like on the left; only the term -2/3

(ignoring multiplication to the 5, so they didn't actually multiply both sides of the equation by the same value)

[sweet pumice]

Ahh, so the simple issues lies in the fact that he didnt multiply the 5 ( left side ) with the same amount aka 3 * x?

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[sweet pumice]

<@&286206848099549185>

[sweet pumice]

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[light sonnet]

Sounds like physics, try asking in the physics server, linked in #old-network

No

[gritty rose]

https://discord.gg/physics

[light sonnet]

You can also look up STEM related servers

Because this one is specifically math

And not physics

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@torn jolt Has your question been resolved?

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@torn jolt Has your question been resolved?

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[rocky hare]

calculate the double integral over the region enclosed by 1<=y<=2 and y = x, y = x/2

[rocky hare]

I don't get the jump professor did here, where did the y in the denominator go?

[fast peak]

they evaluated the inner integral. it's just a constant with respect to x, so the value is length of the interval times the constant

[rocky hare]

yes with respect to x, so siny/y acts like a constant there

still doesn't explain how denominator disappeared?

[fast peak]

and the length of the interval is y

so the value of the integal is siny/y*y=siny

[rocky hare]

so this is some kind of geometrical/logical argument rather than straight up algebra/doing the integral?

aaaa

no x in there

[fast peak]

well you can do it the slow way. integral of c is cx

plug in upper bound and lower bound

subtract

[rocky hare]

integral of 1 = x

then put in lower and upper bound, x = 2y then x = y

and that then cancels the y in the denominator?

[fast peak]

yes

[rocky hare]

ahh wow, haha missed that

thank you

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[crimson falcon]

A bird on the tree that is 13 ft. above the ground finds that the angle of depression of a worm is 38°. How far is the worm from the base of the tree?

[crimson falcon]

Is this correct

I'm asking if that's where the 38° should be

[hot herald]

no

[crimson falcon]

At angle B?

It says angle of depression

[hot herald]

well <B will be 38° but it isn't quite the angle of depression

[crimson falcon]

Yeah that's what I thought

it would've been angle of elevation

[hot herald]

you can still use it and reach the same result

[last lynx]

I think A is 90-38

or is it

[hot herald]

angles of depression (and elevation) are angles made with the horizontal

consider drawing a horizontal line at the bird/A

the angle of depression will be the angle between the horizontal line and the line of sight to the worm

[crimson falcon]

Like this?

hello?

[hot herald]

the 38° is still incorrectly marked

[crimson falcon]

Above the current one right

[hot herald]

mark it properly and show me

[crimson falcon]

Like this?

[hot herald]

yes

[crimson falcon]

When it said "How far is the worm from the base of the tree?" was it asking for side Y or side X

I'm pretty sure side x

[hot herald]

do you know what base of the tree means

[crimson falcon]

The bottom?

[hot herald]

yeh, pretty much

[crimson falcon]

so side x

[hot herald]

yes

[crimson falcon]

Ok thnx

[crimson falcon]

Which trigonometric ratio am I gonna use to get for side x since the 38° is outside the right triangle of the worm and bird

[last lynx]

How do you figure out the angle next to 38°

[crimson falcon]

what

[last lynx]

or you could just tan38° and it gets you the same answer

[crimson falcon]

what?

[last lynx]

Ok so

What is angle A equal to

in degrees

[crimson falcon]

38°?

[last lynx]

No

Thats a part of it

But the full angle

[crimson falcon]

Do I do 90-38

[last lynx]

yes

[crimson falcon]

52?

[last lynx]

yes

Now that you are in the original triangle how do you get x

[crimson falcon]

which x

[last lynx]

There is one x?

[crimson falcon]

The x from this img?

[last lynx]

yes

[crimson falcon]

ok

I'll use tan right

Because the hypotenuse isn't given

[last lynx]

Well thats easier yes

[crimson falcon]

tan52 (x/13)?

[last lynx]

yes

[hot herald]

missing = sign

[crimson falcon]

what

[last lynx]

tan52=x/13

[crimson falcon]

Ok

[crimson falcon]

My scientific calculator doesn't have button for x

[hot herald]

nor should you be entering x into your caklc

[crimson falcon]

What do i put

[last lynx]

You dont need it

$tan52=\frac{x}{13}$

[hot herald]

ideally you should first isolate in your equation algebraically

[glossy valve]

.,..

[last lynx]

Get x alone

[hot herald]

to get an equation in the form
x = stuff

and then enter "stuff" into your calc to get the approximate value for x

[crimson falcon]

Oh i got it

Thanks a lot

I'll close now

[last lynx]

no

what did you gey

get

[crimson falcon]

16.639

Can round to 16.64

[last lynx]

yes

[crimson falcon]

My teacher needs nearest hundredths

[last lynx]

ok

[crimson falcon]

can I close now?

[last lynx]

yes

[crimson falcon]

Ok thnx

.close

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[cedar bear]

Hi! I think I'm supposed to use the central limit theorem (statistics) in the following question, but I haven't used it a lot before this and I'm not sure what I'm getting wrong.

So the question gives the following: Y = X_1 + ... + X_100, where X_i € Exp(2). They ask for following: Find a in P(Y <= a) ~= 0.99

So from what I've read it says that if you have a large enough stochastic variables (with the same distribution) then it will eventually approach N([mu]n, [sigma]sqrt(n)), where [mu] = E(X_i) = 1/[lambda] = 1/2 and [sigma] = sqrt(V(X_i)) = sqrt(1/[lambda]^2) = 1/[lambda] = 1/2.

So I eventually get ((100 * (1/2)) - a) / (10 * (1/2)) = [what I assume is lambda_0.01] = 2.3263, which then gives me a = 38.3685, but the correct answer is a = 61.6*. I'm very confused and it doesn't feel like anything is right 😅 Any help would be much appreciated <3

[cedar bear]

<@&286206848099549185>

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[cunning light]

Hello I need help understanding sigma notation questions

[cunning light]

i dont have any working because i dont know where to start and i wanted to get some help on understanding the question

[dreamy sorrel]

,rotate -90

[cunning light]

would it be possible for you to go with me and progress on the questions because this is a thing i struggle iwth and yeah

[dreamy sorrel]

i kinda suck at these questions ngl

[cunning light]

yeah i hate functions

doesnt make any sense

[dreamy sorrel]

for the first bit tho u use the infinite geometric series formula

[fathom saddle]

You don't need to understand sigma notation here. The regular notation is to the right of it

Σ (-1)ⁿxⁿ is the exact same as:
1 - x + x² - x³ + ...

[cunning light]

yeah i tried finding the r of that so i could plug in on geometric infinite formula but didn't work out

[fathom saddle]

r = -x

You multiply the previous term by -x to get the next one

[cunning light]

oh wow i was right then

ill try the formula give me a second

its 1-x/ 1+x

[maiden sable]

Noh

[cunning light]

ill try using r on the nth term of geometric formula and get a result for x

yeah no

these questions dont make sense to me

[maiden sable]

The infinite sum is 1/1+x no?

[cunning light]

no because u1 is x+1

[maiden sable]

For x between -1 to +1

It's 1

1+(1)*(-x)+(-x) (-x)...

[cunning light]

i dont undertand

[maiden sable]

Sum of an infinite gp?

[cunning light]

yes iinfinite geometric sequence

question a wants me to find x when the series go to 0 right

[maiden sable]

No, when it converges

It's asking when it equals a finite value

[cunning light]

so cant we use one of the numbers to the left to find this

because they are finite

[maiden sable]

Sum of infinite finite numbers is?

[cunning light]

let me take a pic of the formula for finite

,rotate -90

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[cunning light]

anything is fine i just need some help on how to work with the question

cause i have little to no idea

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[burnt kindle]

Suppose m = 1.9 and p = 4.25. Create two algebraic expressions, each with a value of 16.55. Use the variables m and p in each expression.

[full forum]

@burnt kindle Has your question been resolved?

[burnt kindle]

no

[fathom saddle]

(p/p) + (m/m) + 14.55 = 16.55

[burnt kindle]

thank you so much

[fathom saddle]

Actually? You're okay with that answer?

[burnt kindle]

yes

[fathom saddle]

Can do something similar with
(m - m) + (p - p) + 16.55 = 16.55

[burnt kindle]

okay thank you

[torn jolt]

shes rly thankful she said

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[granite linden]

apparently if

[granite linden]

then

can someone explain how this is true? i dont understand

[spice orchid]

What's the definition of inverse function?

[granite linden]

reflection in y = x?

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[grizzled lance]

Just apply the inverse to both sides

What is

$f^{-1}(f(f(x)))$

[glossy valve]

azeem321

[granite linden]

uhhhhh i dont know

i dont know if i dont know just because its late and im tired or i actually dont know

i seem to have neglected normal maths during my further maths revision

yh apparently you can do this

but im tryna comprehend how (f^-1(f(x)) = x

[sick cypress]

the definition of f^-1 is that it is a function that undoes f

so if f(x) = 2x then f^-1(x) = 1/2x

if a function undoes itself when applied twice, then it is its own opposite

for example: f(x) = -x

the negative of a negative is a positive, thus f(f(x)) = x

[granite linden]

oh hmm

lemme think about this for a few mins

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[storm olive]

I’m a bit confused with this one. I already found the interval for the 94% CI and the width was much narrower compared to 98% meaning the 94% is more precise as it has a smaller margin of error. However 98% had a greater level of confidence compared to 94% so which one is better?

[fallow frost]

no

98% is wider interval

what's the definition of confidence interval?

[storm olive]

range of estimates for an unknown parameter.

Yes I know 94% CI is much narrower meaning it’s more precise

<@&286206848099549185>

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[torn jolt]

How do I find the equation of this line?

[torn jolt]

Y= mx+c

[last lynx]

And what is m

[torn jolt]

M is the gradient

[last lynx]

and c

[torn jolt]

The intercept in the y axis

So where the line crosses “Y”

[last lynx]

And where does it do that

[torn jolt]

2

[last lynx]

And how would you find the gradient

[torn jolt]

That’s my question

[last lynx]

rise/run

[torn jolt]

Yeah

[last lynx]

Lets say we choose points (1, 3) and (2, 4)

the change in x is run and the change in y is rise

so $m=\frac{y_2-y_1}{x_2-x_1}$

[glossy valve]

.,..

[last lynx]

So you fill in the coordinates

You could count the squares but this helps you find the slope when you know 2 points

@torn jolt

Also it doesnt matter which coordinates go first

[torn jolt]

Ok then. Thanks for the clarification on the steps

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[lofty sail]

https://prnt.sc/0J7Q94X0kHVX

[lofty sail]

can someone explain why the cross multiply thing is right but when i multiply it with the normal formula it is wrong ?

[torn jolt]

it's +12i

and +9i

since -12i^2

is essentially

[lofty sail]

I did +9i

[torn jolt]

i got i

just i

[lofty sail]

how is it +12i

[torn jolt]

well because it is a double negative

i got i

tho

[true marlin]

i think what you got wrong is -9i^2

in the denominator

16 - 12i + 12i - 9i^2 = 16 + 9 = 25

not 7

[lofty sail]

ok so what do i put in the brackets

instead of what i did

[true marlin]

if what is written inside the red circle is correct

it should be i

[lofty sail]

the cross multiply thing is correct

but i don't think the brackets is

[true marlin]

why

it's just fractions right?

[lofty sail]

ok

what am i doing wrong in the 4 + 3i x 4 - 3i thing since that seems to be correct in the second part ?

16 - 12i + 12i - 9i^2

[true marlin]

no that's correct

i don't think there's anything wrong in the red circle

the only mistake i spotted is the 25i/-7 part

[lofty sail]

but it is marked as wrong lol

[true marlin]

$\frac{-12+9i+16i-12i^2}{16-12i+12i-9i^2}=\frac{-12+25i+12}{16+9}=\frac{25i}{25}=i$

[glossy valve]

Wa3Wa

[true marlin]

not 25i/-7

[lofty sail]

Yeah but I got - 3 deduction of marks not just one

[true marlin]

i suppose the alpha thing marks a mistake?

[lofty sail]

Also it says find it in the form a + bi

[true marlin]

yeah

so 0+1i

=i

[lofty sail]

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[torn jolt]

No idea on how to start on the following question:

Vertices A and D of the square ABCD lie on the line
u: 4x+3y-12 = 0 and, moreover, A lies on the y-axis and D on the x-axis.
Determine the general equations of the sides AB, BC, and CD of
the square.

[torn jolt]

any help greatly appreciated

[hollow wyvern]

Make a rough sketch first

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[tawdry fossil]

Don't have problem with the solution, but more like how confusing is using Pythagorean Theorem

[light sonnet]

But that's not Pythagorean theorem

[tawdry fossil]

I'm just at the beginning, already saw the class 2 times, still kinda bad at this

[light sonnet]

That's the quadratic formula

[safe trench]

What's your confusion of Pythagoras theorem?

[light sonnet]

I mean the first step is Pythagorean theorem, but everything else is quadratic formula

[safe trench]

They said they only have confusion with the Pythagoras

[tawdry fossil]

How to fit to the formula

[light sonnet]

Yes, except the only step involving Pythagorean theorem is just setting it up, everything else is quadratic formula

[tawdry fossil]

Yeah, but quadratic formula is kinda hard too, but I'm already used to it

[light sonnet]

All that happened with Pythagorean was plugging in numbers into the formula

Pythagorean theorem is a^2 + b^2 = c^2

Where a and b are the legs of the right triangle and c is the hypotenuse

[tawdry fossil]

But how can I find the hypotenuse?

I know is the biggest part

[light sonnet]

The side across the right angle

[tawdry fossil]

The 90 angle?

[light sonnet]

Yes

That is a right angle

https://www.khanacademy.org/math/cc-eighth-grade-math/cc-8th-geometry/cc-8th-pythagorean-theorem/v/the-pythagorean-theorem

[tawdry fossil]

Thanks, that solve it

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[terse crystal]

hey, can someone help me with a question of matrices?

[wide sundial]

just ask the question

[terse crystal]

here you go

[onyx glen]

,rcw

[spice orchid]

Have you found AB?

[terse crystal]

yeah i did

its the standard I

the 100 010 001

column wise

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@terse crystal Has your question been resolved?

[nocturne creek]

You can try writing the Coefficient Matrix. Variable Matrix = Constant matrix, im not entirely sure about my wording of the terms tho

And then if you pre multiply with the inverse of the coefficient matrix, on the lhs you get I3. Variable matrix and since I3= AB you can say that u did use AB💀

[torn jolt]

had your question been solved??

or do u need help??

@terse crystal

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@terse crystal Has your question been resolved?

[terse crystal]

nahh it aint working like that

weird values of xyz😐

[nocturne creek]

But it should work tho, there is nothing wrong 🤔

[terse crystal]

got the answer 🙂

we had to multiply it with the inverse on both sides

getting B=A inverse and then the solution

[nocturne creek]

I did mean both sides, i was specifically concentrating on the lhs to show that we did get AB

[terse crystal]

but we just gotta use A B and X

[nocturne creek]

So u had AB = I and then just pre multiply A inverse on both sides?
B= A inv?

Ah well as long as you got the answer!

[terse crystal]

yeah

thanks to you as well 🙂

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@terse crystal Has your question been resolved?

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[humble plaza]

hello

[humble plaza]

there's this problem my prof was working on

and he said this solid of revolution touches the y axis which doesnt make sense to me

[eager violet]

Hard to tell what he meant without the complete context

[full forum]

@humble plaza Has your question been resolved?

[humble plaza]

nah it's all good I got it

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[kind jay]

.... is a vectorfield of the class..
Proof that..

Isnt F1 a vector

You dont take the gradient of a vector do u

[candid ice]

if you mean $\nabla^2F_1,$ then that's $\nabla \cdot \nabla F$, where $\nabla$ is the grad operator. $\nabla^2$ is the laplacian

[glossy valve]

1345631

[kind jay]

Isnt it grad(grad(F1))

[full forum]

@kind jay Has your question been resolved?

[daring citrus]

its divergence of gradient

[kind jay]

ahhahahahahahahahhaha

me thought laplacian was just two times gradient

F1 isnt a vector? the components of a vectorfield are scalars?

[daring citrus]

if i understand it correctly then a vector field outputs a vector for a each (x₁, x₂, x₃)
so yes (F₁, F₂, F₃) should be scalar (idk if always)

also in books if a letter is bold then it means vector and F₁ is not bold so yeh 😅

[kind jay]

alrighty thxx

[main marten]

Pls help with this

[full forum]

@kind jay Has your question been resolved?

[nocturne pilot]

#❓how-to-get-help this is apparently occupied

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[torn jolt]

how do i know what x and y are equal to in the second part of the equation?

[loud mica]

you don't really need to know the value of x and y to solve the problem

[loud mica]

but to solve the problem you use the distributive property and then you combine like terms

[hot herald]

why did you close the other channel

I was preparing to gear a response based on your answers

and just to reinforce what I said earlier,
x is x and y is y
they're just variables and have no inherent value

[brazen yarrow]

what is 4x -x

[hot herald]

can you expand -(x+y)
yes/no/do something else/abandon channel again

[torn jolt]

idek what that means

i cant answer if i dont know LMAO

dont be rude

[loud mica]

basically

4(x + 2y)

is 4*x + 4*2y

you're distributing the 4 to both of the terms in parentheses

and once you have that

you can combine like terms

• (x + y) becomes - x - y

[hot herald]

if you can't just say no

[hot herald]

there's no shame if you're genuinely trying to learn

and we can adjust our approach

dont be rude
what's rude is to just abandon the channel you had implying that you're unsatisfied with the help you're getting only to reopen minutes later

[brazen yarrow]

we're trying to help but you just closed the other channel to have someone else explain it to you

[torn jolt]

because the way you are explaining it is confusing me

[brazen yarrow]

there is no simpler way to explain it

[loud mica]

which part is confusing you?

[hot herald]

the first step is fully understanding the distributive property
a(b+c) = ab + ac
do you understand this property/equation?

[torn jolt]

yes i understand that

[hot herald]

did you fully understand how applying that property to expand
4(x+2y) gets you 4x + 8y

[torn jolt]

yes

i dont understand

how 3x

and 8y

7y

[brazen yarrow]

what is 4x -x

[torn jolt]

i dont understand how to get those numbers

i thought x had no value

[brazen yarrow]

it does

[loud mica]

you need to do the same thing to the second part of the equation

[brazen yarrow]

well the value of x doesn't matter

it's what's before the x that matters

4x - x = 4x - 1x = x(4-1) = x(3) = 3x

-(x+y) = -1(x+y) = -1x -1y

now 4x + 8y -1x -1y

add up the parts with x and parts with y

[hot herald]

the next part to focus on was whether you understand
-(x+y) = -x-y

[torn jolt]

no i dont lmao

[loud mica]

distributing a - is distributing a -1

you’re multiplying the terms in the parentheses by -1

which is just making them negative

[hot herald]

many have this issue if they don't explicitly have a value on the outside

which was why I explicitly introduced a 1 earlier

applying the principle that multiplying something by 1 is itself

k * 1 = k

do you agree with that?

[torn jolt]

yea that makes sense

[hot herald]

and applying that principle to
-(x+y) gets you
-(x+y) * 1 = -1(x+y)

[torn jolt]

so because theres no number stated its equal to 1?

idk

[hot herald]

applying the principle that multiplying something by 1 is itself

you just don't have to write it

but it can be explicitly written if you really want

and can help you see things more clearly

now that you explicitly have a value next to the parentheses,
are you able to expand -1(x+y) yourself

and/or see how that's equal to what others have typed earlier?

[full forum]

@torn jolt Has your question been resolved?

[torn jolt]

nah im still confused as fuck

im playing fortnite now wish me luck <3

[brazen yarrow]

rather pay attention to class

[torn jolt]

ew

[brazen yarrow]

then don't bother us

[torn jolt]

tuff

[safe trench]

4(x+2y) - (x+y)

= 4x+8y-x-y

=3x+7y

Sometimes it helps to just observe the steps

[brazen yarrow]

they don't understand that 4x - x = 3x

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[stark prism]

how do i find range

[rugged harbor]

Look for horizontal asymptotes

[stark prism]

what about doing it without a graph

[sharp lagoon]

Well, look where it doesn't exist.

Like if you take a square root of a negative, that doesn't exist.

If you divide by zero, that doesn't exist.

You're not taking a square root, but you are dividing.

So, when will what you're dividing by be zero?

[stark prism]

its range so

[sharp lagoon]

Yes, this is for that.

[stark prism]

ill have to find g-1(x) first right

[sharp lagoon]

No, you can do it from the definition of g.

At the top.

No need to do g⁻¹ first.

[stark prism]

i can put x as 3

[sharp lagoon]

Right, so x can't be 3.

[stark prism]

but thatll give me domain right

[sharp lagoon]

Yes, but you need the domain to find the range.

[stark prism]

ohh ok

[sharp lagoon]

So, x can't be 3. Also, x ≥ 5.

So, x ≥ 5 already covers x not being 3.

So, we have the domain: [5, ∞).

[stark prism]

so i sub 5 into x to find range?

[sharp lagoon]

Yes, that will be the first step.

[stark prism]

15/2

[sharp lagoon]

So, that's 15/2.

Right.

[stark prism]

is it always the same < > sign

[sharp lagoon]

Now, what happens when you increase x?

What about x = 6?

[stark prism]

it decreases

[sharp lagoon]

Right. Let's try a really high number.

What happens with x = 1 milliion?

[stark prism]

oooh u get nearly 2

2.000011

[sharp lagoon]

Right.

[stark prism]

so range is between 2 and 15/2?

[sharp lagoon]

And notice the leading terms of the top and bottom.

2x and x, right?

[stark prism]

yeah

[sharp lagoon]

To figure out what it goes to as x goes higher and higher, you just divide those two.

2x/x = 2.

The small addition and small subtraction get very very very small compared to the 2x and x parts when x is huge.

So, they pretty much don't matter after a while.

[stark prism]

ohh i see

[sharp lagoon]

Now, can it get to exactly 2?

[stark prism]

dont think so

[sharp lagoon]

Right, 2x + 5 is always slightly higher than what you need.

And x - 3 is always slightly lower than what you need.

2x/x would be perfect, but those small things make it where the fraction will be just too large.

The top is too big.

The bottom is too small.

[stark prism]

oh i get it

[sharp lagoon]

So, the range is (2, 15/2].

[stark prism]

is the ( bracket <

[sharp lagoon]

( means it doesn't touch that endpoint exactly.

[ means it includes the endpoint.

[stark prism]

oooh ok

[sharp lagoon]

[ looks like it's grabbing on.

( looks like it could slide off.

That's how I remember it, at least.

[stark prism]

ahaha thats nice

thank you so much

ur a really good teacher

[sharp lagoon]

You're welcome.

Thanks.

[stark prism]

.close

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[torn jolt]

Is 5(5) or higher order than 5*5?

[tepid raven]

Anybody good at geometric series?

[sharp lagoon]

Sorry, Tommy got in a little bit earlier.

[sharp lagoon]

What do you mean about higher order?

[tepid raven]

Oh my apologies.

[torn jolt]

I don’t know how to explain it but given 5*5(2) would I first have to 5*2 because it’s in brackets? I know this is complete irrelevant because it’s multiplication.

[sharp lagoon]

Well, if there was something inside the brackets to do, you'd do that first, but once it's just a simple number, you can replace the parentheses with that number.

5 · 5(2) = 5 · 5 · 2.

The parentheses are done.

The parentheses rule is about doing the stuff inside the parentheses first.

[torn jolt]

Exactly. But if there was more in it. Let’s say something complex like an x. 5*5(2x) I’d first have to do 5(2x) = 10+10x? Right?

[sharp lagoon]

No, you could do it the other way.

The parentheses are fully simplified inside.

And multiplication/division is generally done left to right.

So, it would be like 5 · 5(2x) = 25(2x) = 50x.

Or you could just say multiplication can be done in any order and then do that.

The rule isn't to do the multiplication of the parentheses with something outside it first.

That's only about what happens inside the parentheses.

[torn jolt]

I know commutative law. But wouldn’t I still have to solve 5() first?

[sharp lagoon]

Nope.

The parentheses first rule isn't about multiplying the parentheses with what's outside it.

It's about simplifying what's inside it, which is already done there.

Then there's no rule that says to do the multiplication involving parentheses first.

[torn jolt]

Ok. Thanks. I can’t remember the last time this was important tho. I always did the number() thing first. 😂

[sharp lagoon]

No problem. It's fine to do that, if you like that way.

[atomic blade]

In 5*5(2), the 5*5 takes priority

Tf

There

[torn jolt]

.close

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[visual matrix]

part d

do i have to sub in x-5?

its only 1 mark

[spice orchid]

Yeah you can just sub it into the (x+3) factor

[atomic blade]

Or any roots

But it says that (x+3) is the only root so

[visual matrix]

i got the one real root for part c

can i just do x-5 = -3

and get 2

[spice orchid]

No point doing it to the rest, f(x-5) will only have 1 real root aswell

[atomic blade]

No

Replace x with "x-5" in the root

[visual matrix]

ok will do

[spice orchid]

It is going to be 2

[visual matrix]

.close

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[molten wadi]

Can someone help with this problem. Sin(θ) = ?, θ is in the Quadrant 1, Cos(θ) = sqrt(3)/2

[molten wadi]

i think it's asking what is θ

[spice orchid]

Draw a triangle

[molten wadi]

ok

done

I can't post, my phones dead

here

[crystal elbow]

Bruh

[molten wadi]

what

[crystal elbow]

The other guy will tell you

[molten wadi]

ok

[spice orchid]

Lol

Not just any triangle

[molten wadi]

what triangle then

[spice orchid]

The triangle that is relevant to the fact that cosθ = sqrt3/2

[molten wadi]

ok. dose it matter where i put the angle in the triangle?

[spice orchid]

Yes

That will determine which sides you say are what lengths

[molten wadi]

then where could i put the angle?

[spice orchid]

Just put it whereever, but obviously not the 90 degree angle

[molten wadi]

this work

[spice orchid]

Yep perfect

Now to find sinθ you need that missing side

[molten wadi]

so sqrt(3)^2 + b^2 = 2^2

[spice orchid]

Good

Square the 2