help-23

[hallow moss]

And in the case of 3 3, none of them are divisible by 8

9, 1 works

But 3 3 doesnt

Am i misunderstanding the question here

?

[cold aurora]

How does 3 and 3 not work?

2 is divisible by 2.

[hallow moss]

But not 8

Same power of 2 no?

2/8 =/= it

[cold aurora]

Yes. 2 is divisible by 2 which is 2^1.

[hallow moss]

But not the same power of 2? Its not 2^3?

[cold aurora]

I think what they mean is that whichever power of 2 divides x-1, it will also divide y-1.

[cold aurora]

They both are 2. None of them is divisible by 8.

[hallow moss]

Ye ik ik, weird wording for a question then

Couldve just said divisible by 2

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[worldly mango]

I got it

Red=wrong

do u know the answer to the second one?

ik i asked other ppl and they agreed with 72.6 but it marked it as wrong

so the teacher is wrong?

it is 72

i check it and the quiz says it is 50

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[gusty trench]

l'hopitals

you get 0/0 for the limit and you're given the tangent lines so you can directly find f' and g'

[clever ridge]

wait but what is the equation?

i only see the equations of the tangent lines

also how do you get 0/0 for the limit i just dont see what you're seeing plz hlep

[gusty trench]

you have an x-y plot of f and g, you can see that they're continuous and their function values equal 0 at x = 2

so the limit of f at 2 is 0, and the limit of g at 2 is 0

since you get 0/0, the limit as the same as the limit of f' / g' at 2

f' is exactly the slope of the tangent line to f at 2

[clever ridge]

oh

so f' = 1.5(x - 2)

and g' = 2 -x

so the limit is still 0/0

idk y i dont get this bc y isn't f' 1.5x - 2.5????

[clever ridge]

wait so it's f' = 1.5x - 2.5? and g' = 2 - x?

because that is 3 - 2.5 = .5 / 0

undefined.

i dont get this

can someone helpme im going crazy

<@&286206848099549185>

anyone when you're ready

@shy lynx why do you keep typing and then go away

sketchy kid @shy lynx

been doing it for 20 minutes

??? say something already then

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[lilac osprey]

can someone help me understand this question

[lilac osprey]

im not sure what theyre asking for when they say trig sub in the form of x=f(t)?

i thought trig subs were in the form of x=a tan(theta) for example

[quiet plume]

The variable you choose for inside the trigonometric function is irrelevant.
You'll just have something like x = a tan(t)

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[lilac osprey]

oh i see, so essentially its in the same form as the example i listed above except using the variable t instead of theta?

[quiet plume]

Yes, that's what they mean

[lilac osprey]

that makes sense, thanks!

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[viscid glade]

i asked my professor for extra problems to practice, but i'm having a lot of trouble with this question? i get that you have to take the derivative of both sides, but i keep getting lost after that,,,any chance someone could break it down like you're explaining to a fetus LMAO

[tardy wigeon]

In this context, we know that the radius is 15 centimeters: r=15

We also know that the radius is expanding 20 centimeters per minute: dr/dt = 20

Now we take the derivative of both sides to get an an equation for dV/dt

Then sub in the constants for r and dr/dt and solve for dV/dt

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[viscid glade]

i ended up with 18000π and that feels,,very wrong 😭

[hard crest]

especially wrong because there are no units

show your work

[viscid glade]

derivatives are,,not my strong suit clearly lmao

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@viscid glade Has your question been resolved?

[hushed trail]

Your step looks good

[viscid glade]

wait, so was it just the units-?

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[torpid fossil]

Need some help on this

Isn't it just getting the sum of the x and y components of E = kq (assume r = 1)? then arctan y/x

[torpid fossil]

i tried pythoning it and got 1.83 rad or around 100 degrees
but using some online interactive simulator, its supposedly around 160 degrees

from math import pi, cos, sin, atan2

k = 1 / (4 * pi * (8.854 * 10**-12))

def find_field_x(q, theta):
    return -q * k * cos(theta)

def find_field_y(q, theta):
    return -q * k * sin(theta)

q_values = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
theta_values = [pi/6, (2*pi)/6, (3*pi)/6, (4*pi)/6, (5*pi)/6, pi, (7*pi)/6, (8*pi)/6, (9*pi)/6, (10*pi)/6, (11*pi)/6, 2*pi]

result_x = [find_field_x(q_values[i], theta_values[i]) for i in range(len(q_values))]
result_y = [find_field_y(q_values[i], theta_values[i]) for i in range(len(q_values))]

print("x:", result_x)
print("y:", result_y)

sum_x = sum(result_x)
sum_y = sum(result_y)

print("Total sum of x:", sum_x)
print("Total sum of y:", sum_y)

angle = round(atan2(sum_y, sum_x), 2)
print("Angle of electric field relative to the x-axis in radians:", angle)

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[torpid fossil]

<@&286206848099549185>

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@torpid fossil Has your question been resolved?

[dreamy elk]

The problem I see is that you paired the angles and numbers wrong

so you paired $\frac{\pi}{6}$ with 1, but it should be paired with 2.

[flat frigate]

Crystopher

[dreamy elk]

I can see this since the sum of x in your program is -6, but it should actually be 6 on y.

so it is a problem of rotation so to speak, it is like you rotated the whole thing by 90 degrees or something.

[torpid fossil]

theta_values = [pi/3, pi/6, 0, 11*pi/6, 5*pi/3, 3*pi/2, 4*pi/3, 7*pi/6, pi, 5*pi/6, 2*pi/3, pi/2]

like this? @dreamy elk

i just followed unit circle
idk why i didnt think of that first lol

gave -0.26rad
seems like 180 the other way tho

[dreamy elk]

yes, apparently

the other problem is the value of x, even if you flip it 180 degrees, it would be wrong, I did it by hand and it gave around -16.3923

[torpid fossil]

weird..

[dreamy elk]

the sign/180 degree problem disappears if you change your functions to this

I also removed the k, it doesn't really affect the outcome.

now it gives around $165^{\circ}$ (2.88 rads)

[flat frigate]

Crystopher

[torpid fossil]

but charges are negative tho
would that mean anything?

or does it only become -q in the code if its repelling??

[dreamy elk]

how I did in paper I interpreted negative charges as repelling, so pointing outwards from the clock

[torpid fossil]

icic

thanks!!

[dreamy elk]

I made error when doing it by hand, it is actually x=-22.3923..., giving exactly 165 degrees, so the program should be correct.

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[abstract comet]

The position vectors of four points A, B, C, D are a, b, 4a, 7b.

The position vector of the point of intersection of lines AB and CD is K1a+K₂b. Find K1 and K2

Can someone please tell me where i should start from?

[abstract comet]

I found AB (b - a) and CD (7b - 4a)

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[slate carbon]

You can start by expanding the 4 position vectors into components, like a =(a1, a2). Then express the line AB & CD in terms of a1, a2, b1 & b2. You can use the representation of lines from 2 points, first by finding the slope from the respective 2 points and then using one point as pivot. Then just equate the 2 equations to find the common point.

[slate carbon]

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[lean otter]

can someone pls explain what is angle in standard position

[fathom adder]

if its vertex is located at the origin and one ray is on the positive x-axis

[lean otter]

ah and coterminal is like two rays coinciding on each other so for angles u just add 360 and for rad u do 2pi

right

yay

[lean otter]

thanks

👍🏻

[lean otter]

thank you

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[gloomy veldt]

if [ cot(θ) + cos(θ) = p ] and [ cot(θ) - cos(θ) = q ] , then { (p^2 - q^2)^2 } in terms of p and q is

[gloomy veldt]

Answer is [ 16pq ]

<@&286206848099549185>

[proven merlin]

Yes

[gloomy veldt]

no

i mean how to get the answer

i put the values is eq

[proven merlin]

Oh nvm

[gloomy veldt]

then i got ( 4cotθcosθ)^2

idk how to solve further

[ ( cotθ + cosθ)^2 ] - [ (cotθ - cosθ) ^2 ]

[ 4cotθcosθ ] ^2

any idea how to further solve this

to get the answer

[grim plover]

simplify it (convet cot to cos)

[lime dust]

Expand (cot θ+cos θ)^2

This is p^2

Do the same with q^2

[gloomy veldt]

ya i do it

[lime dust]

What u got

[grim plover]

this ig?

[lime dust]

No

[gloomy veldt]

ya

[grim plover]

also remember $pq = ( \cot \theta + cos \theta)( \cot \theta - \cos \theta)$ which can be simplfied using $a^2 - b^2$

[lime dust]

Cot^2 θ+cos^2 θ+2cot θcos θ = p^2

[flat frigate]

rak³en

[gloomy veldt]

yes

[lime dust]

Do this with q^2 and subtract

[gloomy veldt]

cot^2θ + cos^2θ -2cotθcosθ

[lime dust]

Now i get 4cot θcos θ= p^2-q^2

[gloomy veldt]

this what i also got

[lime dust]

p * q = (cot θ + cos θ)(cot θ - cos θ)

cos^2/sin^2 -2cos^2

[gloomy veldt]

but why p * q

[lime dust]

Because that is the diff of sq we are gonna use

[lime dust]

Can u continue from here?

[gloomy veldt]

ok thanks

but i do not understand

why we are using

( p^2)^2 + (q^2)^2 - 2pq

you are saying this

[lime dust]

Because we are using p^2-q^2

[gloomy veldt]

ya

so we just put values

of p and q

[lime dust]

p^2 and q^2 are those

[gloomy veldt]

then got some eq

[lime dust]

Then we use p * q

[gloomy veldt]

?

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[neon yarrow]

how do i solve these

[burnt notch]

Area under f - area under g

For ex. 2 I mean

[twilit heath]

You will do integral of f(x) - g(x) then:

x^2 = 2x - x^2
2x^2 - 2x = 0
2x(x - 1)
x=1,0

The complete integral will be
Integral from 0 to 1 of f(x) - g(x)

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[twilit heath]

This is the explanation for the two remaining questions.

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[glass field]

i do not understand how change of base was used

context

[velvet flax]

$\frac{\log_a \frac{m}{2}}{2}=\frac{\log\frac{m}{2}}{2\log a}=\frac{\log\frac{m}{2}}{\log a^2}=\log_{a^2} \frac{m}{2}$

[flat frigate]

Ham

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[umbral ginkgo]

A - Quantity A is bigger
B - Quantity B is bigger
C - Quantity A and Quantity B are Equal
D - Cannot be determined

[west haven]

what did you try?

if it says rains on atleast one of the days, it means total - rains on none of the days

total is 1, and you can calculate rain on none of the days

[umbral ginkgo]

here is what I did I calculated the probability that it would not raid for 6 days

which is (1/2)^6 = 1/64

then I do half minus 1/64, I think. Frankly I am not sure how to narrow it down to just a single day

[fathom adder]

the probability that it rain on one days is 1 - probability that it doesnt rain so it's :
1 - (1/2)^7

which is equal to 127/128

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@umbral ginkgo Has your question been resolved?

[umbral ginkgo]

but isn't that also the probability that it doesn't rain just 1 day?

[fathom adder]

so it rains for 6 days ?

ig since it's 1/2 there will be some strange thing like this

But the quantity are equals ! since Quantity A is an “at least” problem, use the 1 – x shortcut. Rather than calculate the probability of rain on exactly 1 day next week, and then the probability of rain on exactly 2 days next week, and so on (after which you would still have to add all of the probabilities together!), instead calculate the probability of no rain at all on any day, and then subtract that number from 1.

[umbral ginkgo]

That makes sense

Thank you

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[umbral ginkgo]

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[lean otter]

does someone have a really simple explanation for the derivation of the quadratic formula?

[thin bridge]

its the end result of solving
ax^2 + bx + c = 0
using completing the square

[cunning pasture]

simple twistings can get you done.

[lean otter]

complete the square on ax^2 + bx + c

idk how to complete the square lol. that concept is alien to me. i watched videos about how to derive the quadratic formula and that was the part i couldnt understand

[tepid walrus]

completing the square, is essentially adding and subtracting an expression, where that expression is the piece that's missing for ax²+bx to be a perfect square

[thin bridge]

then look up vids on completing the square

[lean otter]

thx

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[misty elbow]

What is the correct answer to this question

[misty elbow]

I think the slope would be the same and the length is 6.6 but that is not a option

please ping me on response

[hardy lion]

@misty elbow how long is a grid square?

[misty elbow]

OH

OK

THANKS

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[trim musk]

I need help with my math hw

[fathom adder]

Sure, post the problem

[empty gyro]

You need to derscribe a problem to receive help on it

[limber shore]

bro just said hi and ran away

[fathom adder]

He flew math

[limber shore]

lol=))

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[green mica]

Hey

[green mica]

A line passes through the point (1, -4) and is perpendicular to the line 3x + 2y – 6 = 0. Determine a scalar equation for the line.

The scalar equation for the line is 2x-3y-14 = 0 please check

My work

[icy lance]

seems alright

[green mica]

Thanks

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[trim musk]

OH shoot

mb

.reopen

😭

Its too late now nvm

I actually did it by msyelf n forgot I had dis open

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[queen yarrow]

how would i solve this?

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@queen yarrow Has your question been resolved?

[queen yarrow]

<@&286206848099549185>

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[queen yarrow]

<@&286206848099549185>

[lean otter]

moments

[queen yarrow]

yes but do i have to do something with the 90° or the 30°?

[lean otter]

yes

moments act perpendicular to centre of rotation

The force of P doesnt act perpendicular

youll have to take the component of P that does act perpendicular

[queen yarrow]

like this right?

[lean otter]

no

[lean otter]

The force on P isnt 120

120(a+b) is the moment around the pivot from the F force

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[stark vortex]

Can anyone help me out on the next step for this problem? (Pre Calc) Converting polar equations to rectangular. The orginal problem is in black and my work is in pink.

[quiet plume]

You should complete the square for the terms $(x^2 + 6x)$ and $(y^2-2y)$.

[flat frigate]

Azyrashacorki

[stark vortex]

Ok, if I did the x problem first, would I do
6/2 = 3 and then square the 3?

[quiet plume]

Yes

[stark vortex]

Ok let me write that down, and I put the 3 on the other side of the problem?

[quiet plume]

You want to write it like x^2 + 6x + 9 - 9, and then you have a perfect square and some trailing constant.

And yes at the end you can bring those to the other side

[stark vortex]

Why do I add the -9 though?

[quiet plume]

You don't want to change the initial expression (x^2 + 6x), so you need to add 0 in a clever way.
That's why we do (x^2 + 6x) + 0 = (x^2 + 6x) + 9 - 9 = (x^2 + 6x + 9) - 9

[stark vortex]

Oh ok

Let me send a picture

Is this right?

Hello?

?

.close

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[dull shell]

How should I explain this?

[proper wedge]

In desmos the region looks like this

lattice points look like this, i.e (1, 1) (2, 1) and so on are lattice points

So all they ask is to count integer lattice points in that region (points on the triangle's lines do not count)

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@dull shell Has your question been resolved?

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[runic heath]

Hey guys, I was just wondering if anyone could explain why we substitute x=radius×sin(theta) when integrating circles/semicircles?

I've been watching a few videos on it and they all just say we use sin, no one explains why. I would assume we use cos because it represents the x on the unit circle, but I'm not sure why that is not the case and would really appreciate if someone could explain it

[inner parrot]

its because of the pythagorean identity

[flat frigate]

y0shi

[inner parrot]

semicircle sorry

if we want to integrate it

[flat frigate]

y0shi

[inner parrot]

itll cancel out the square root

cos also works here

[flat frigate]

y0shi

[inner parrot]

its just that the derivative of cos is -sin, and we prefer not to deal with it

[runic heath]

So I could have used x=cos if I wanted too? It just means there would be negatives

[inner parrot]

mhm yep

it would give you the same thing

[runic heath]

OK, thank you very much for that explanation

[inner parrot]

yw!

[runic heath]

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[alpine aurora]

I’m having trouble finding the integral

Basically

I found it out

5 would be the bigger one

So it’s 5-4/x

I’m having trouble though

[cerulean coyote]

with finding the indefinite integral?

[alpine aurora]

Not rly…??

Um isn’t it just

5x-4ln|x|

[cerulean coyote]

correct, so just fundamental theorem it, unless i forgot something.

[alpine aurora]

Yes but

5(-6)-4 ln(6)

(5(-2)-4 ln(2))

94 just

Or just

10+4ln(2)

[cerulean coyote]

should be other way, since -6 < -2, so it's F(-2) - F(-6).

[alpine aurora]

Oh

💀

Wait lmao

I forgot negs

💀

Thanks

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[modern bloom]

why $\text{dim}(\mathcal{L}(V, W)) = $\text{dimV} \times \text{dimW}$

[flat frigate]

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

[languid siren]

Hmm maybe this is only true in finite dimensions…?

Like say - linear maps from $F(\mathbb{N})$ to $F(\mathbb{N})$ correspond to functions from $\mathbb{N}$ to $UF(\mathbb{N})$

[flat frigate]

Pseudonium

[languid siren]

There are definitely uncountably many such functions

But the product of the dimensions is the |N| x |N| = |N|

So yeah I’m convinced this only holds in finite dimensions

In that case, it follows from looking at bases

[modern bloom]

btw if we say V can be seen as a space of m-dimensional column vectors and W can be seen as a space of n-dimensional column vectors. Then the linear transformation T:V→W can be represented by an n×m matrix A.
if V is the space of 3×2 matrices and W is the space of all 2×3 matrices , A will be 2×3, right?

[reef shadow]

I want to ask a question,can you solve it?

[modern bloom]

no open another channel

[languid siren]

If V is the space of 3x2 matrices, it is 6 dimensional

And if W is the space of 2x3 matrices, it is also 6 dimensional

So the matrix A will be 6x6

[modern bloom]

wait......what

[languid siren]

At least, here I’m assuming you add matrices component wise, and scale matrices componentwise

Not sure how else you’re suggesting to define a vector space structure on a space of matrices

[steep lily]

For a more intuitive explanation, imagine unfolding these matrices out into single row/column vectors

because essentially as vector spaces that's the same structure

[languid siren]

Yep

[steep lily]

(obviously you can e.g. multiply matrices too and what not, but that's separate from their vector space structure which is all we care about in this case)

[languid siren]

Mhm, although even then it’s not clear to me how one multiplies a 3x2 matrix by another 3x2 matrix

[gusty trench]

if you understand the isomorphism between linear maps and matrices then this is the simplest explanation, the space of linear maps from a space V of dimension n to a space W of dimension m is isomorphic to the space of mxn matrices

each entry in the mxn matrix is "free" to be whatever it wants, so that space has dimension mn

[modern bloom]

my confusion is, The dimension of a vector space V, denoted as dim(V), is the number of vectors in any basis of V, right? then like we say a 3×2 matrix, then its dim should be the number of vectors of its basis, right？how can its basis vectors be 6?

[gusty trench]

you're mixing two things up

[languid siren]

Well, there’s a basis for V that has 6 vectors

[gusty trench]

the space generated by the columns of a matrix vs the space of all matrices of that dimension

those are two different spaces

[languid siren]

They’re the 6 elementary matrices

The first of these is $\begin{pmatrix} 1 & 0 \ 0 & 0 \ 0 & 0 \end{pmatrix}$

[flat frigate]

Pseudonium

[languid siren]

This is a 3x2 matrix

So an element of V

The next of these could be $\begin{pmatrix} 0 & 1 \ 0 & 0 \ 0 & 0 \end{pmatrix}$

[flat frigate]

Pseudonium

[languid siren]

Again, this is a 3x2 matrix

So an element of V

You can probably guess what the other 4 are

And you can verify that, indeed, this collection of 6 matrices is linearly independent, and spans V

So is a basis for V

Hence, dim(V) = 6

[gusty trench]

$\begin{pmatrix} a & b \ c & d \ e & f \end{pmatrix} = a \cdot \begin{pmatrix} 1 & 0 \ 0 & 0 \ 0 & 0 \end{pmatrix} + b \cdot \begin{pmatrix} 0 & 1 \ 0 & 0 \ 0 & 0 \end{pmatrix} + c \cdot \begin{pmatrix} 0 & 0 \ 1 & 0 \ 0 & 0 \end{pmatrix} + ...$

crap what did i mess up

[steep lily]

two \?

[languid siren]

You should use double \

[gusty trench]

oh

[flat frigate]

neil

[modern bloom]

aah

[languid siren]

Yes, so that shows how they span

It’s also not hard to show they’re linearly independent

This just follows from the definition of matrix equality

[languid siren]

One confusion here - you say “its dim should be the number of vectors of its basis”, but I’m not sure what you mean by “the basis of a 3x2 matrix”

A vector space can have a basis

But a 3x2 matrix is not a vector space

The collection of all 3x2 matrices is a vector space though

[modern bloom]

I was thinking the basis of 3×2 matrix are the indep col vectors that spanning the space, yea i know i might mix it up

[gusty trench]

yeah this is what i was mentioning earlier, mixing up the space generated by the columns of a matrix with the space of all such matrices

[languid siren]

Yes, so that’s different

A matrix doesn’t have a basis, a vector space is allowed to have a basis

Now, given a matrix, one can define lots of vector spaces

The kernel of the matrix

The span of the columns

The span of the rows

All of these are vector spaces, so it makes sense to ask if they have bases, and if so how many vectors are in that basis

But the matrix itself is not a vector space

So it doesn’t make sense to ask if the matrix has a basis

[modern bloom]

i see! Finally cleared it up

[languid siren]

Cool

It’s important to remember the precise definitions of things so that you avoid confusions like these

[modern bloom]

Thank you all, need to reorganize these definitions now

.close

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[oak violet]

are these correct?

[oak violet]

.close

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[waxen rampart]

Need help with part B

[waxen rampart]

I have absolutely no idea how to solve it and im running low on time

🙏

[quiet plume]

You will need to integrate the absolute value of this function over the interval.

[waxen rampart]

What I think you are saying is do this:

[quiet plume]

No you've already integrated it now.

[waxen rampart]

im confused

oh you mean:

[quiet plume]

You need to compute this $$\int_a^b |v(t)|dt$$

[flat frigate]

Azyrashacorki

[waxen rampart]

oh so:

[quiet plume]

This will mean that you have to find the intervals in there where the velocity is negative

and where it's positive

[quiet plume]

and change the sign accordingly

[waxen rampart]

this accurate?

[quiet plume]

no

[waxen rampart]

just positive signs?

oh

[quiet plume]

As I said you'll need to break it up into two picece

[waxen rampart]

How would I do that

[quiet plume]

.

[waxen rampart]

Im sorry im really lost

i dont know how to do that

Like foil it into (x-x)(x+x)

sorta thing

[quiet plume]

No. The integral as it is gives net area under the curve.

But since you need the total distance, you need to consider the distance travelled when the particle is moving to the left (negative velocity)

[waxen rampart]

I think I got it

its 72 thank you!

I took out the negative and re-integrated everything and it worked

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[viral cave]

i thought the answer was e, but it turns out it was d? idk how that works tho

[viral cave]

like how can i find the perimieter since most of the horizontal edges dont have a length wirtten

omg nvm

i just realised

sorry i have been working for like 2 hours so it slipped my mind

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[rugged cargo]

Please check anyone?

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[steep magnet]

I know that at -infinity its 'none' because you cant have a negative inside the square root, but what about at infinity

[polar wren]

let’s expand the formula to get 2/x*sqrt(x) + 14/sqrt(x) first

analyze each term separately as x approaches positive infinity

[steep magnet]

The first term goes to 0

[polar wren]

yup

[steep magnet]

The second one im not sure

[polar wren]

alright so let’s say x is some really large number like 10^1000

and we take the square root

which is 10^500, still a massively huge number

[steep magnet]

Aaaaah I see

[polar wren]

basically sqrt(x) continues to increase as x increases

until infinity

[steep magnet]

so it also goes to zero

[polar wren]

yup

[steep magnet]

I get you

[polar wren]

👍

[steep magnet]

😄

Thank you, you explained it really well!!

❤️

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[smoky tide]

For this question, why does this working not produce the correct answer?

The correct answer is instead pi(1200+600ln3)

[split fulcrum]

hsc q!

[smoky tide]

yep

on that trial prep grind

[split fulcrum]

same same

[smoky tide]

valid

u doin x2?

[split fulcrum]

naww

u?

[smoky tide]

<@&286206848099549185>

[smoky tide]

yeah i am

its painful

[split fulcrum]

is this from 3u?

[smoky tide]

yeah it is

[split fulcrum]

rite

[smoky tide]

2023 paper

[split fulcrum]

ahh okay

hold on

ima do this q rn

[safe radish]

@smoky tide Has your question been resolved?

[smoky tide]

can anyone help with why my method didnt work?

for reference i used this formula

for the volume of revolution

[vale oriole]

im too lazy to read this, but u should be able to find the volume from y=4 to y=12 then add that to the volume the rectangle would create

[split fulcrum]

also isnt it -600ln3 not + btw?

[split fulcrum]

also u didnt need to do a limit

[vale oriole]

agreed this seems outside the scope

[smoky tide]

yea it is

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[safe radish]

@celest musk Has your question been resolved?

[celest musk]

<@&286206848099549185> 🙏

[remote spruce]

hi

there?

[celest musk]

yep

[remote spruce]

AK=KD

angle of BAD = 60

so BK = root(3)*AK

right?

so 24 = root(3)* AK* AK

and BD = AB = 2 * AK

clear?

[celest musk]

so far yes

[remote spruce]

BD = 2 * AK = 2 * root(24 / root(3))

make sense?

[celest musk]

BD is equal to 2xAK is equal to the root of 24/root of 3

yes

[remote spruce]

in the triangle of ABK, cuz angle of ABK = 60, BA is equal to BK / cos(60)

cos(60) is 1/2

so AB = 2*BK

[celest musk]

so the answer is what?

[remote spruce]

BD = AB = 2 * AK = 2 * root(24 / root(3))

[celest musk]

let me try to input that

see if it works

@remote spruce its telling me thats incorrect

do you have any other approaches in mind?

?

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[uneven pagoda]

can you help me

[uneven pagoda]

Can anyone help me with this?

[vale oriole]

first determine the gradient of the line

[uneven pagoda]

I dont know what that is

please explain it as easy as possible

[vale oriole]

rise/run?

[uneven pagoda]

I dont know anything at all dms pls

[dull shell]

Right but how should i explain this

is there anny way to explain this?

[lean otter]

im solving it on my paper hold on

[safe radish]

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.

[lean otter]

my bad, im not solving it, im just writing the steps down

[uneven pagoda]

is y = (25/3)x + 10 good

[lean otter]

where you get 25

it's basically

y = mx + b, where m is the slope (tilt of the line) and b is the y-intercept (where the line passed in the y-axis)

but first, determine the rise over run

[uneven pagoda]

i dont know any of dis dm me pls

[lean otter]

i think you skipped the basics

[safe radish]

@uneven pagoda Has your question been resolved?

[safe radish]

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[halcyon light]

How do I do this

max value usually occurs at 1

on sin graph

but its stretched by x 3

wouldn't this imply that k is 7

so - 3 + 7 for minimum

which is 4

but then whats meant by the smallest positive x coordinate

[polar wren]

well, the smallest x for which 3sinx + 7 = 4

if you isolate sinx, you get sinx = -1

the first positive angle for which this is true is x = 3pi/2

[safe radish]

@halcyon light Has your question been resolved?

[halcyon light]

wait isn't it pi/ 90 degrees though

since x = arcisin(-1)?

ohhh wait i got it now

thanks

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[lean otter]

is this relation coming from the doing the magnitude

of

on both sides

bruh who is adding reactions lmao

💀

[lean otter]

doing magnitude of this will give
1 = cos^2 x + cos^2 y + cos^ z

is this a valid proof

[safe radish]

@lean otter Has your question been resolved?

[empty gyro]

Is this is how alpha, beta, and gamma are defined, then yes

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[rough bear]

Ive tried so many times and cant get this solved😂

[oak tangle]

first, have you figured out which all angles are equal?

[rough bear]

its about finding the similarity between triangles that overlap, i didnt think angles matter. or am i wrong?

[oak tangle]

two triangles are similar when all their angles are equal

the difference is that one is scaled to some other size

[rough bear]

ohh okay imma try to figure out the angles real quick

i can only find the 90 degree angles

[oak tangle]

you dont have to find the exact measures, just which all are equal

[oak tangle]

say that angle MKL is x, what would angle MLK be?

[rough bear]

also x?

[oak tangle]

no, do you know the angle sum property?

[rough bear]

ive learned it yes, i dont remember it tho

[oak tangle]

its that the sum of all angles is 180 in a triangle

in the triangle MKL, this means angle MKL + KML + MLK = 180

you know that MKL is x and KML is 90, what would MLK be in terms of x?

[rough bear]

90 - x?

[oak tangle]

yes

similarly, what would angle KJL be?

[rough bear]

also 90 - x?

[oak tangle]

correct

the last angle is MLJ

[rough bear]

which is also 90 - x?

[oak tangle]

no, try again

[rough bear]

180 - x

[oak tangle]

one angle is 90 (LMJ), the other angle is 90-x (MJL), what would the third one be?

[rough bear]

180 - 90?

[oak tangle]

180-(90+90-x)

[rough bear]

oh okay

[oak tangle]

which simplifies to x

now you know all the angles

so you just need to find which vertice corresponds to which in all triangles

start with triangle JKL and KML

angle K is x in JKL and in KML too

KML is 90 and JLK too

so K corresponds to K, M corresponds to L

which means L (KLM) in KML corresponds to J in JLK

so JKL ~ LKM

if the letters are too confusing, you can also think of it in the order of the angles

JKL is (90-x) -> x -> 90

LKM is also 90-x -> x -> 90

can you figure out the same order in the third triangle?

[rough bear]

so JKL ~ LKM and LKM ~ JLM?

so JKL ~ LKM ~ JLM?

[oak tangle]

yes

[rough bear]

ohh okay i see

im confused about
JM/JL = JL/? and JK/LK = ?/KM

wait so would
JM/JL = JL/JK?

[oak tangle]

thats correct

[rough bear]

okay and
JK/LK = LK/KM?

[oak tangle]

its correct but would be better to write it as KL/KM

[rough bear]

okay thank you so much sir🙏

[oak tangle]

np

[rough bear]

okay one more thing

i know that to find the missing angle ( m<B) i have to do 90+46
then take 180-136 to get 44
but idk how to find Leg a and Leg b

[rough bear]

<@&286206848099549185>

[edgy pasture]

sup

[rough bear]

how do i find leg a and leg b?

like what steps do i need to take to figure these out?

[edgy pasture]

by that u mean length?

[rough bear]

yes

< B = 44

[edgy pasture]

u shd use trigonometry

[rough bear]

is there a specific theorem i should use?

[lavish wharf]

a/sin46 = 20/sin90

[edgy pasture]

sin 46 = a/20

yes

that

[lavish wharf]

Then pyth

[rough bear]

okay thank you!🙏🙏

[edgy pasture]

main thing is here to solve the sin 46 which is p easy

[rough bear]

okay thanks!!

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[weak grail]

If a chord 4y=3x-48 subtends an angle theta at the vertex of the parabola y^2=64x then the tan theta=?

[weak grail]

so basically I took 2 parametric points (at1^2,2at1) and (at2^2,2at2) and then using the slope of the line they're passing through I got 8=3(t1+t2)
now idk what to do after that

for some reason i almost never get help

[safe radish]

@weak grail Has your question been resolved?

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@weak grail Has your question been resolved?

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[visual anchor]

i'm used to it being framed like -4*x_1 + x_2 + x+3 =0 and that being the end, i'm not familiar with what to do next with this question

[safe radish]

@visual anchor Has your question been resolved?

[gusty trench]

are you familiar with the four fundamental subspaces?

mainly the idea that N(A) and C(A^T) are orthogonal?

[visual anchor]

yes somewhat

[gusty trench]

okay, you can write W as the null space of a matrix

you have

-4x1 + x2 + x3 = 0
3x1 - x2 = 0

so any vector x = (x1 x2 x3) in W satisfies Ax = 0 for A =
-4 1 1
3 -1 0

does that make sense?

[visual anchor]

yes yes

[gusty trench]

then do you see why W is exactly N(A)?

[visual anchor]

no

[gusty trench]

^

the vectors in W satisfy the property that A multiplied by that vector gives the zero vector

where A is those coefficients

so those vectors in W are exactly the vectors in N(A)

[visual anchor]

okay i see

[gusty trench]

so if W = N(A) and you want to find W_perp

you know that N(A) is orthogonal to C(A^T)

which means W_perp is exactly C(A^T)

C(A^T) is also sometimes called the row space of A

[visual anchor]

ok

[gusty trench]

so you have A from earlier, all you need to do is find C(A^T)

specifically you want a basis for C(A^T)

so you can transpose A to get A^T, and then row reduce that, and the pivoted columns will be the basis vectors for C(A^T)

which are exactly the basis vectors for W_perp

[visual anchor]

ok perfect thank you

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[twin fulcrum]

can anyone help with this? i've been stuck on it for like 20 minutes. gotta find the surface area

[ripe scaffold]

assuming the bottom (right side) has 12m on all sides

you can know the SA of that side

[twin fulcrum]

the what?

[ripe scaffold]

surface area

[twin fulcrum]

oh

[ripe scaffold]

by pythagoras you could get the length of the peak to half of 12

[twin fulcrum]

i've been trying for a while but all my answers have been wrong

[ripe scaffold]

in that case you have this

find the area of this triangle

x4 + the 12m2 gives you the total surface area

[twin fulcrum]

62 m^2?

[ripe scaffold]

||after you do pythagoras to find out x, you know that a secondary triangle has sides xm, 12/2 = 6m, and you can do pythagoras again to find the third side||

||from that you can calculate the area of one of the 4 triangles

[ripe scaffold]

for? 1 triangle?

[twin fulcrum]

i think so

[ripe scaffold]

let me try to calculate

x=13.4 i got

[jovial gale]

No its 36(root 2)

Per triangular side

[ripe scaffold]

14.68 i got for the longer side on a secondary triangle

[twin fulcrum]

rip

[ripe scaffold]

in that case

wait acc you don't even have to get 14.68

i'm stupid

you just do bxh/2

it gives me 80.4

$80.4m^2\times4+12m^2=465.6m^2$

is what i got

[flat frigate]

Bridge

[twin fulcrum]

hm

[jovial gale]

Isn't the height of the triangle 6(root 2)??

[ripe scaffold]

uhh

6√2?

uh

i do x^2=12^2+6^2

[jovial gale]

So area of each triangle = 1/2 126(root 2)

[ripe scaffold]

√(36)

wait something isn't right

yeah

√(180)

mb

[jovial gale]

Nope you cannot take 12 square you need to take 6 the half of it

[ripe scaffold]

why can't i?

[jovial gale]

Because in the right angle triangle

[ripe scaffold]

^

[jovial gale]

That you have drawn

The yellow one

[ripe scaffold]

wait wait

wait hold up

i'm stupid

you're right

it's 6

[ripe scaffold]

let me revise this then

[jovial gale]

The base is half of the side length of square

[jovial gale]

It's alr we all make such mistakes

But wait did the person who's question it was understand this??

I think he left

[twin fulcrum]

😭

[jovial gale]

🙂

Okay wait let me draw you a diagram

[ripe scaffold]

$\sqrt{(\frac{12}{2})^2+6^2 } for\ the\ isosceles\ triangle's\ height$

[jovial gale]

A large one

[ripe scaffold]

ok yeah shuddap

[flat frigate]

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

[ripe scaffold]

that you plug into uh

$\frac{(b\times h)}{2}$

what

[flat frigate]

Bridge

[ripe scaffold]

which gives you the area of ONE triangle

that x4 + the base gives you total surface area

[twin fulcrum]

ic

[ripe scaffold]

i cna't draw sorry

[ripe scaffold]

this is what you get expanding the shape thingy

yeah

yeah

that

[jovial gale]

Look carefully at the diagram @twin fulcrum

[twin fulcrum]

yeah im looking

[ripe scaffold]

crucial you understand how to get a slice of the triangle for pythagoras

you need to see the shape with depth and imagine the space between lines

so that you can search for a plane where you have enough data to manipulate and calculate

[jovial gale]

Yes, imagination is the key in many concepts of mathematics.

[twin fulcrum]

oh i got it

so its 348m^2

thx

[ripe scaffold]

🙏

[safe radish]

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[full agate]

let n be a natural number bigger than 3. Calculate the sum

how do you even start?

'n-ori' means n times

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[acoustic glacier]

I'd first try to parameterize the sequence $22\ldots2$ in terms of natural numbers $n$

[flat frigate]

KirPlop

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[acoustic glacier]

Let $a_n:=22\ldots2,(\text{n-ori})$. It follows that $\frac{9}{2}a_n = 99\ldots9,(\text{n-ori})=10^{n}-1$.

[flat frigate]

KirPlop

[acoustic glacier]

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[restive laurel]

did i do something wrong?

[restive laurel]

when i put it in a differential equation calc its

[quiet plume]

This

It's $$e^{\ln|\frac{x}{x+1}| + C}$$

[flat frigate]

Azyrashacorki

[restive laurel]

i thought u could cancel out logs by rasing everythin to its base

[quiet plume]

Yes, but this becomes $$e^{\ln|\frac{x}{x+1}|}\cdot e^C$$

[flat frigate]

Azyrashacorki

[quiet plume]

Which now you can cancel like you have, but you get a multiplying constant, not one added after

[restive laurel]

ohh

its e raised to everyhting in the same exponent

[quiet plume]

Yes

[restive laurel]

then exponent rule from there

[quiet plume]

Yup

[restive laurel]

makes sense

thanks!

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[jaunty raft]

Question:Decide if the vector lies in the span of the set, inside of the space. I got to this point and I don’t know exactly where to go. Any direction is appreciated. Thanks!

[jaunty raft]

Ah

Read the problem wrong. Should be x+x^3

Not x+3

But book says to solve it as a system

Not sure how to do that if we only have 1 eq

[woven hound]

One way you can do this is write in vector form

The Standard Basis of $\mathcal{P}_3$ is $1, x, x^2, x^3$.

[flat frigate]

Ari

[woven hound]

So the set of vectors you're given is ${(0,0,1,0), (0,2,1,0), (3,1,0,0)}$. And you want to know if $(0,1,0,-1)$ is in the span of this set.

[flat frigate]

Ari

[woven hound]

You also have a mistake here. 2x and x^2 need to have the same coefficient since 2x+x^2 is a single vector in the set

[jaunty raft]

Good catch Ty

One minute working your responses out rn! Thanks

[jaunty raft]

How is this

[woven hound]

looks good

[jaunty raft]

Awesome
Tyvm!

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[split anvil]

Ok I am asking a third time because I have spent over 3 hours on this problem and I want to die now.

Here is the problem and my latest attempt at an answer. I am not sure where I am going wrong, but I am desperate for an answer at this point. I don't even need the credit, I just want to know the answer at this point so I can understand

[violet stratus]

!status

[safe radish]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[split anvil]

I dont know. I have tried solving it countless times at this point. I think I have submitted 6 answers all wrong. I don't know where I am going wrong

[violet stratus]

what was your first step?

[thin bridge]

the -1 isn't part of the power

[split anvil]

Ok lets walk through from the beginning... Let me know where I am going wrong please:

So first off I am using chain rule to take f'(g(x))*g'(x) so I get:

[thin bridge]

you have an issue already

its the one i stated

at a glance the issue looks like you're viewing this as
$$\br{4^{\underline{\cos(x^6)-1}}}^5$$
instead of
$$\br{\underline{4^{\cos(x^6)}-1}}^5$$

[flat frigate]

ℝαμΩℕωⅤ

[thin bridge]

-1 isn't part of the power

[split anvil]

oh I see what you're saying... wtf why they have to make it so hard to tell?

[split anvil]

uhg fml

but still, I think I am doing it wrong. would it be this then?

[thin bridge]

not quite

[split anvil]

or no, my bad, this

[thin bridge]

where g(x) = , not x

[split anvil]

ok right , so I need g'(x) where g(x) is 4^(cos(x^6))-1

[thin bridge]

yes

[split anvil]

ok so this is where the biggest issue is I think. trying to take the derivative of 4^(cos(x^6))-1

so I think I got something like

or this rather

or is it this?

[thin bridge]

h(x) would be the entire power of e

[e^(h(x))]' = e^(h(x)) * h'(x)

[split anvil]

liek this?

[thin bridge]

no

[split anvil]

am I combining 2 steps or something?

[thin bridge]

you're misapplying exponent laws

[split anvil]

of course

[thin bridge]

${(a^b)}^c = a^{bc}$ is not the same as $a^{\br{b^c}}$

[flat frigate]

ℝαμΩℕωⅤ

[split anvil]

hmm ok

so... not sure how I do this then... is it

[thin bridge]

no

$e^{\text{what is here?}}$

[flat frigate]

ℝαμΩℕωⅤ

[split anvil]

I have no idea

[thin bridge]

don't overthink

[split anvil]

I can barely think none the less overthink

[thin bridge]

$e^{\ln(4)\cdot \cos(x^6)} = e^{\what}$

[flat frigate]

ℝαμΩℕωⅤ

[split anvil]

I don't know. I think you might underestimate how terrible I am at math lol. I've been out of school for almost 20 years just now trying to learn this stuff

[thin bridge]

you're WAY overthinkiing this

[split anvil]

is it an algebra thing?

[thin bridge]

not really no