help-23

[dreamy salmon]

to solve the problem

[crude nest]

eh that would just make it more complicated

[dreamy salmon]

oh

[crude nest]

because you would have negative terms to deal with

use division law for a and multiplication law for b

[dreamy salmon]

so log_2(16)?

[crude nest]

for a? no

[dreamy salmon]

b)

[crude nest]

yes but it's log base 4, not 2

so it will be asking what power of 4 is equal to 16

[dreamy salmon]

aa answer page says

2 for both

a and b

[crude nest]

correctamundo

[dreamy salmon]

but its log_4(16)

not log_4(2)

[crude nest]

log_4(16) is equal to 2

which is the answer

the final answer is log_4(16)

[dreamy salmon]

what the heck

then why doesnt it say

answer log_4(16)

[crude nest]

because its also a very simple log that can be solved with ur eyes closed

[dreamy salmon]

😭

its so over

[crude nest]

for log notation, if its asking log_4(16), that is the same thing as asking 4^x = 16

and 4^2 = 16 so x = 2

[dreamy salmon]

okay i get that but

a)?

theres nothing to do with 2

[crude nest]

the answer for a was log_2(4)

which is the same as asking 2^x = 4

and 2^2 = 4, so x = 2

[dreamy salmon]

aa okay

i get it

ill try to remember whilst doing rest of the chapter problems

thank you for being patient w/ me

[crude nest]

ofc, math is tough

keep chugging along on those problems, and next time maybe try to study a little earlier 😂

[dreamy salmon]

defo

thank u sm fr fr

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[lean otter]

Is this correct

[lean otter]

.close

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[candid mirage]

Is the operation $0_2$ associative?

\begin{table}[h!]
\begin{tabular}{c|c}
(x, y) & $0_2$ \
\hline
(a, a) & a \
(a, b) & a \
(b, a) & a \
(b, b) & b \
\end{tabular}
\end{table}

I also did $0_5$ which looks like:

\begin{table}[h!]
\begin{tabular}{c|c}
(x, y) & $0_5$ \
\hline
(a, a) & b \
(a, b) & a \
(b, a) & a \
(b, b) & a \
\end{tabular}
\end{table}

Which was NOT associative, so I’m unsure.

[flat frigate]

VexedRumble

[safe radish]

@candid mirage Has your question been resolved?

[candid mirage]

I plugged it in all eight possible ways to set it up, and I got it is associative but I want to double check because it seems weird to me 0_5 isn’t associative but 0_2 is

Oh right, this is for Abstract Algebra, specifically from Chapter 2 of A Book of Abstract Algebra by Pinter

<@&286206848099549185>

I’m sorry, am on my phone but I’m gonna write down what I checked for it

[safe radish]

@candid mirage Has your question been resolved?

[safe radish]

@candid mirage Has your question been resolved?

[candid mirage]

For Abstract Algebra, covering Operations

Unsure of what else to add on

[candid mirage]

.

[candid mirage]

I think it is associative but I’m just looking to confirm

When I plug it into every one of the associative possibilities it turns out to be true

But I still feel like I’m messing something up since 0_5 isn’t associative

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[dull sequoia]

What’s this?

[twilit spindle]

I dont know fs it could be some fancy math symbol but sometimes I've seen people write that for "and"

[safe radish]

@dull sequoia Has your question been resolved?

[dull sequoia]

It might be and, but what’s the (3) doing after

He says it’s about the variance being a constant but I’ve never seen this before

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@dull sequoia Has your question been resolved?

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@dull sequoia Has your question been resolved?

[plucky elk]

Assumptions number (2) and (3)

[dull sequoia]

Oh

Im dumb 😞

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[pine burrow]

Can someone explain this top problem?

[pine burrow]

Why is f(x) being set equal to b?

[reef estuary]

if you set f(x) and solve for b, you get f^-1 (b)

[pine burrow]

Why do you find x=pi and f(pi) = pi + 1

I just don’t understand the steps being taking

[reef estuary]

well pi - cos(pi) = pi + 1 = b

so f(pi) = b

[pine burrow]

But how do you know to take that step

Is that just standard

.end

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[lean otter]

can someone explain what b means

thanks!!!!!

[reef estuary]

it's just one of the ways, you can of course find the inverse first and then plug in b into it, but this is faster

[reef estuary]

for part (i), look at the graph and see where is intersects y = 0

[lean otter]

4

oh

wait

y=4 (intercept) i mean

i read it wrong

[lean otter]

has 2
x=-2.528 or 0.528

[reef estuary]

so those are your solutions

[lean otter]

yep

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

<@&286206848099549185>

[nocturne tapir]

x = -2.528; x = 0.528

[reef estuary]

you got your solutions

[lean otter]

i already got that

[reef estuary]

you said them yourself

[lean otter]

yeah

[reef estuary]

yeah that's the answer

[lean otter]

how does that help b

?

[reef estuary]

you mean b(ii)

?

[lean otter]

ye

uh both

[reef estuary]

one sec let me read it

[lean otter]

what does it mean to use the graph

[nocturne tapir]

?

[nocturne tapir]

you get the coordinates

[lean otter]

i already have them

[nocturne tapir]

(x,y)

right?

[reef estuary]

it means you look at the function graphically to determine the answers

[nocturne tapir]

so use those values to solve the equations

[lean otter]

so ultimately, whats my goal in b i and ii

just the solutions?

[reef estuary]

for b(ii) i dont know how you would do it graphically. My closest guess would be to plug in the value of 3x^2 given to you into your original equation and to solve for x, but that isnt graphical

[reef estuary]

yes

[nocturne tapir]

(i) - find all the coordinates for -4 <= x <= 2 for P(y)
(ii) - Use those values of the graph of the first part to solve each of the equations

[safe radish]

@lean otter Has your question been resolved?

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[rough venture]

pls help with the following question. i dont even know how to begin

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[ancient marsh]

how to find equaiton of the line that passes through (-2, 1) and makes 13 degree with x axis?

[granite compass]

uh, theres no line that both forms an angle of 13°, passes (-2, 1) and the origin

[granite compass]

but if you ignore the (-2, 1) part and only look at a line which makes an angle of 13°, recall your trig that $\tan \theta = \frac{y}{x} 😒 slope, so your line would be $\y = (\tan \theta)x)$. however bear in mind that $\theta$ is the angle made with the positive x-axis, so you'll needa add 90°

[ancient marsh]

U+1f612

[granite compass]

sry one sec

but if you ignore the (-2, 1) part and only look at a line which makes an angle of 13°, recall your trig that $\tan \theta = \frac{y}{x} = $ slope, so your line would be $y = (\tan \theta)x$. however bear in mind that $\theta$ is the angle made with the positive x-axis, so you'll needa add 90°

[flat frigate]

PhenomPlasma

[ancient marsh]

Well I wrote that y=xtanx in the screenshot

I don't understand adding pi/2

[granite compass]

wait sry

you need to do pi - a. we measure angles from the positive x-axis counterclockwise

[ancient marsh]

That'd be a reference angle?

[granite compass]

sorry i dont remember the term anymore, but just rmb to take angles starting from +ve x-axis and going counterclockwise

[ancient marsh]

With y axis it makes 13 degree though

And passes -1,2

Let me share the question

And answer

Its not possible.

so answer is y = tan(13)(x- (-1) + 2
I think gradient/slope is m = tan(13)
and used the line equation I guess y = m(x-x1)+y1
@granite compass

[olive pecan]

Can make $13^{\circ}$ in the positive direction or in the negative direction, which gives 2 lines with two slopes: $\tan(13^{\circ})$ and $\tan(167^{\circ})=-\tan(13^{\circ})$. Then use the point-slope formula $y-y_1=m(x-x_1)$.

[flat frigate]

adzetto

[ancient marsh]

right; that's why we have two answers.

Thanks

how to close the post?

.close

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[ancient marsh]

/close

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[river thicket]

Working on this inequality where a + b + c = 1 and a + b + c are all positive real numbers

[river thicket]

correction: 4/9 should be 2/3

I have tried multiple ways of applying AM-GM

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[west hedge]

throw a dice 3 times what is the probability when the sums are prime numbers

[west hedge]

Is there any clever way to do it?

[peak estuary]

not really

[dull sequoia]

I’d try generating function then sum all the prime results

[peak estuary]

I would say that for three dice that is not really worth it yet

[dull sequoia]

Look at his working

;-;

[west hedge]

that sounds interesting, even though i do not know how to do it

Also speak now is the best album

[peak estuary]

how is a generating function better. you would have to go through the same stuff, just different notation. unless you let a computer multiply it out

[dull sequoia]

The expansion of $\left(\sum_{i=1}^6x^i\right)^3$ gives you all the ways to roll any sum, just look at the coefficients of the corresponding power with the sum you want

[flat frigate]

Frosst

[dull sequoia]

I suppose it doesn’t, I’d use a computer for sure

So for example the expanded form has no x or x^2 since you can’t roll a sum of 1 or 2 from 3 dice

But there is exactly 1 way to roll a sum of 3

(The coefficient of the x³ term is 1)

[dull sequoia]

it is a close second

[west hedge]

is there any website or program that i can used to calculate sth like this.

[dull sequoia]

wolfram alpha

[west hedge]

terrfic

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[indigo veldt]

I tried 2a+b+27=180)-(2b+a=180

But it doesn’t solve

[tidal imp]

Try expressing b in terms of a

[indigo veldt]

So I tried 4a+27=180)-(3a=180

[tidal imp]

Why 4a+27?

[indigo veldt]

I add the whole triangle like from a+b+c=180

[tidal imp]

And…?

[indigo veldt]

If you said try expressing b to a

[tidal imp]

I didn’t say b=a, because it’s not

[indigo veldt]

Oh…

So what do mean by expressing

[tidal imp]

Look at the figure

And see if you can relate a to b somehow

Maybe some supplementary angles, external angles…

[indigo veldt]

27+a=b

[tidal imp]

Yes!

You should be able to solve for a now using sum of angles of a triangle

[indigo veldt]

Okay thanks

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[charred lark]

On chercher à démontrer l’inéquation de haut, et on peut utiliser les information du dessous ? Si quelqu’un a des idées ?

[drifting mountain]

,rccw

[charred lark]

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[lean otter]

Why is he wrong

I got 2x(x+3)

But that seems like it works too

[dapper venture]

fully

factorize fully

[lean otter]

What does factorise fully imply?

[placid oak]

that you can't break any of the factors into any smaller parts

(2x+6) can be split into 2 and (x+3), so it's not factorised fully

[lean otter]

How do I know that x(2x + 6) is not fully factorized?

[cosmic grove]

2x and 6 have a common divisor

[lean otter]

Ohhh okay thanks I get it now

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[earnest finch]

A

[earnest finch]

ANYONE HERE

TO HELP ME

FOR MY MATHS

[quasi bison]

MAYBE IF YOU EASED UP ON THE YELLING AND ACTUALLY POSTED A QUESTION SOMEBODY COULD COME HELP YOU!

@earnest finch

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[zinc flax]

need help with 5 word problems i can do the problem just have no idea how to set them up

[icy lance]

what do you think you may do with the above sentence?

[zinc flax]

1/3x+1/2x=25?

[icy lance]

yup

[zinc flax]

x+60 divided by x =4?

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[lean otter]

I’m a little confused about moments

For the like perpendicular distance I thought you’d take sin(30) but apparently you take cos(30) and I don’t understand why

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[true yoke]

need help

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[little cipher]

the ratio between the length and height is 5:3

the width of the red cross is 1/5 of the flags height

[feral relic]

ok that's written Norwegian right

[little cipher]

yep

wait

hol

[feral relic]

Jeg taler dansk

[little cipher]

so you understand?

[feral relic]

jeg forstår lidt

[little cipher]

ait

uhm

never mind the first task

[feral relic]

ok so question saids the breadth of the stripe

is 1/5

of the height

[little cipher]

yes

w=7.5

[feral relic]

and 5*length = 3*height

[little cipher]

7.5*5=3x

37.5/3

[feral relic]

5*7.5=3*h

[little cipher]

?

ok

[feral relic]

yep

the height

and now the width is

1/5

so time that by 1/5

and you good

[little cipher]

uhm

[feral relic]

also 你说中文吗?

[little cipher]

会一点

and

[feral relic]

哦

[little cipher]

i am the same dude

[little cipher]

from that group chat

[feral relic]

xd

whoops didn't recognize

sorry

[little cipher]

lmao

[feral relic]

[little cipher]

no problem

[feral relic]

to better understand ratios

think it as yx/zx=y/z

[little cipher]

hmm give me a sec

[feral relic]

uhh how should I explain

so basically something times a number divides by another thing timing the same number crosses off that same number

[little cipher]

so linear equation

[feral relic]

yeah

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[near mango]

yesss sssssssirrrr

im here to save u

no probleme

for the 6 ?

Its 7

oh ok

idkn

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[worn musk]

This expression is equal to 2, and I've been able to prove so by making it into an equation and simplifying. My question is, without knowing what it it is equal to beforehand, how would I go about simplifying this expression?

[peak estuary]

call its value x, make an equation, simplify. hopefully get something nice

[worn musk]

I tried with x but it doesn't seem to get me anywhere, it's only when I already know its numeric value can I progress

[glass carbon]

in fact you can also start from the most nested square root and do it step by step

using short multiplication formulas and abs. value

like

[worn musk]

What do you mean by that?

I'm not a native so I may not know the name of formulas or operations

[flat frigate]

Modus

[worn musk]

I don't see how you get from the second to the third expression

[glass carbon]

and now it's pretty easy, because sqrt(5) cancels out

and you'll stay with 6 - 4sqrt(2) under the square root

apply that trick once again, that's it

13 + 4sqrt(10) = (a+b)^2 = a^2 + 2ab + b^2 for some a, b

2ab has to be 4sqrt(10) ---> ab = 2sqrt(10)
a^2 + b^2 = 13

we can guess a = sqrt(5), b = sqrt(8)

[worn musk]

Ohh I see

So just apply square of a sum

But in reverse

[glass carbon]

now
6 - 4sqrt(2) = (a-b)^2 = a^2 - 2ab + b^2
-2ab = -4sqrt(2) ---> ab = 2sqrt(2)
a^2 + b^2 = 6
a = ..., b = ...

[worn musk]

Alright I'll try to solve it with that and see if I get stuck again

How do I close the channel?

[glass carbon]

.close

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[distant bay]

HI BIG BRAINS Thursday i have a test abous primitives and integral and i need some help to exercise to learn the rules of the "game" and yes i prefer to say that is a game better than a something boringly complicated.

[drifting mountain]

,tex .int rules

[flat frigate]

ils

[quiet juniper]

https://dontasktoask.com/

You'll be very hard pressed to find somebody who makes up a whole game for you so it'd be better if you just found some exercises in the internet and asked questions about them 🙃

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[pure grove]

How do you prove surjectivity

[stoic dune]

Show that every output has an input that can reach it

[pure grove]

so for 1

id take an arbitrary y of Q

and show how x = y-1/2

[stoic dune]

Yeah basically

In other words, if you plug in (y-1)/2, you get y.

Does this reach all y in Q?

[pure grove]

wait what

so this?

[stoic dune]

The "since every y in Q..." line is weird

[pure grove]

what would you put instead

[stoic dune]

Take arbitrary y in Q. f((y - 1)/2) outputs this y. Function is surjective

[pure grove]

ty

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[toxic isle]

Is this proof/work valid?

[light shoal]

can you explain your reasoning?

[toxic isle]

Oh

Umm

For example 3/7

If we just put 3 and 7 as power to let’s say 2

2^3/2^7

It’s not equal per se

Like the ratio

[light shoal]

right

so definitely the = sign at the start of the second line isn't correct

[toxic isle]

So what can I do this make this proof/ work valid?

[light shoal]

well is there any rule that says you can exponentiate the numerator and denominator and get the same limit?

i.e. what rule are you appealing to with this argument

[toxic isle]

I don’t think there is a rule for this

But I feel like

If the numerator is > denominator

And if we take that as the power of a constant number like e

Either the numerator or denominator will grow ever so much bigger depending on which is greater initially

[light shoal]

^ is this true?

the log of n+1 is generally much smaller than n

[toxic isle]

Oh whoops

I mean the other way around

Mb

[light shoal]

nw

the idea of exponentiating isn't a bad one

but i don't think there's any valid argument that involves separately exponentiating the numerator and denominator

what if you were to try exponentiating the whole thing

[toxic isle]

Then it will turn into a radical question?

Since denominator is greater than numerator

[light shoal]

try it and see what happens

you can do some manipulations to get it into a useful form

[toxic isle]

Like e^(ln(n+1)/n)?

[light shoal]

hmm actually scratch that, we're taking n->infinity, not 0, so that's not gonna be the silver bullet

my thinking (flawed) was to observe that
ln(n+1)/n = ln((n+1)^1/n), then exponentiating you just get
(n+1)^(1/n), which would go to e if n were a continuous variable that approaches 0, but we're going the other way

what techniques are you allowed to use? L'Hospital's rule?

[toxic isle]

We actually only just begun ap calculus so we didn’t learn much or any rules

[dull sequoia]

Well is it not the same as $\lim_{k\to\infty} \left(\frac{1}{k} + 1\right)^k$

[flat frigate]

Frosst

[dull sequoia]

Set k = 1/n

[light shoal]

but in that change of variable, n->infinity transforms to k->0

not k->infinity

[dull sequoia]

Oh n is going to inf

[light shoal]

yea, the nice "e" limit doesn't work here

[dull sequoia]

$\lim_{n\to\infty} \left(n + 1\right)^{\frac{1}{n}}$

[flat frigate]

Frosst

[dull sequoia]

This just tends to inf no?

Actually that sounds wrong

[light shoal]

it's almost the same as the limit of n^(1/n) as n->infinity, which goes to 1 (although this also takes some work to show)

so presumably (n+1)^1/n also goes to 1

and that will imply that the original limit goes to 0

(by continuity of exp)

[dull sequoia]

The derivative here is 1/n (n+1)^(-(n-1)/n) which is always positive

Looks like that tends to 0 when n goes to inf

So I guess it just looks like a log function starting at 1?

Not even

That’s wrong

[light shoal]

the derivative is trickier than that since both the base and exponent depend on n

[dull sequoia]

Oh

[toxic isle]

Not to be rude or anything, but I’m not 100% sure about what just happened

[light shoal]

no, not rude, it's your channel after all 😁

we're just thinking aloud

[toxic isle]

Oh ok 😅

[light shoal]

here's an idea...

first of all let's do a change of variables to get just an integer as the argument of ln

let k = n-1

then you have ln(k)/(k-1)

let's take a fixed k value, say k=2

and then take its powers

k, k^2, k^3...

the fraction will be:

ah wait one moment let me scribble this down

and see if it's gonna work

[toxic isle]

Ok take your time

And thank you

[safe radish]

@toxic isle Has your question been resolved?

[light shoal]

sry got distracted for a moment, let's see..

[toxic isle]

No worries

[light shoal]

so after that transformation and considering say $k=2$ and its powers $2^2, 2^3, 2^4$, etc we have:
$$\ln(k)/(k-1) = \ln(2^n)/(2^n-1) = n\ln(2)/(2^n-1)$$

[flat frigate]

Bungo

[light shoal]

so we've transformed into checking whether $n/(2^n-1)$ converges

[flat frigate]

Bungo

[light shoal]

and that should be easy to check by e.g. the ratio test

now we've restricted ourselves to looking only at certain values of k, namely 2, 2^2, 2^3, 2^4

so does convergence of that subsequence imply convergence of the entire sequence ln(k)/(k-1)?

the answer is yes if you can show that ln(k)/(k-1) is monotone

which in fact it is

btw, this is probably not the most straightforward way to solve this, there's probably a quicker way

[toxic isle]

I see

I think I got a better grasp now

Thank you

.close

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[void steeple]

first is this right? and then i actually have some questions

[vital cairn]

no, theres still a discontinuity at x=-2 even though you divide the terms out

[void steeple]

so how would i implement that?

[vital cairn]

well the first option is correct

[void steeple]

ah cause it includes the -2?

[vital cairn]

yeah

[void steeple]

I see thanks!

[vital cairn]

this might be helpful

[void steeple]

this is my next

[void steeple]

ohhhhhhh yes def is

<@&286206848099549185>

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[rancid star]

could someone give me a hint as to how i can solve this q?

[rancid star]

i have the following notes from lecture but i cant pin how i should use the given to incorporate the different cosine transforms:

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[stray palm]

My final answer is wrong it’s supposed to be 15N

Anybody know where my mistake lies?

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@stray palm Has your question been resolved?

[stray palm]

<@&286206848099549185>

[stray palm]

<@&286206848099549185> please :D

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@stray palm Has your question been resolved?

[stray palm]

<@&286206848099549185>

[quasi ingot]

@stray palm

Do you still need this done?

[stray palm]

I do

@quasi ingot

[stray palm]

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[hazy pumice]

Evaluate

Stuck on 49 and 51

I know that I should do sin ( theta) = -4/3 for problem 49

.closer

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[devout steeple]

how do you solve equations like this:
x^x = 1.5

[devout steeple]

is the only method guessing an x value and then getting closer and closer approximations?

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[vivid trail]

What am I doing wrong?

A chicken yard is to be built up from a fence in their garden. The fence is 12 metres long. The girls have a 12 metre long fence that they can use to build the chicken yard.
How can you make the chicken yards area as large as possible?
Image

[vivid trail]

I know for a fact that x is not 1.5, it is ca. 3.

<@&286206848099549185>

I got it, I just for some reason wrote 4x^2 instead of 2x^2

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[high sun]

I am doing calculus 1 and I really need help learning how to graph limits/understanding what I am supposed to do. I got this far??? I don’t even know if it’s right but

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@high sun Has your question been resolved?

[safe radish]

@high sun Has your question been resolved?

[lean otter]

What part about limits is confusing?

[high sun]

everything 💀😭

[lean otter]

Limits at infinity might be a little confusing because of the whole infinity part, but really what you are doing is plugging in really really big numbers for x and seeing what happens to the result

As x gets bigger and bigger (or smaller and smaller), f(x) gets closer and closer to some number (or just keeps going up/down)

Does that make sense?

[lean otter]

Let's say I had a function y = x^2

I want to find what happens to the y value as the x value approaches 2.

While I could easily just plug in x = 2, I can take the limit too

Instead of plugging in x = 2, I could plug in values that get closer and closer to 2, like 1.9, 1.99, 1.9999, etc. or 2.1111, 2.0111, etc.

All values very close to 2, but never actually reaching 2

f(1.9) = 3.61
f(1.99) = 3.9601
f(1.999) = 3.996001

f(2) = ?

f(2.001) = 4.004001
f(2.01) = 4.0401
f(2.1) = 4.41

I am able to find out what happens at this point without actually plugging it into the function

You can dm me if you would like should the channel expire or something

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[rare vale]

How did this become $(10sqrt(3))/3$ ? we got $(5sqrt(3))/3$ but we do not know why we needed to multiply 2

[flat frigate]

ren

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[whole acorn]

!show

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

[whole acorn]

@rare vale

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[whole acorn]

Say: the coordinates of the points of the two upper corners is (x , 25 - x^2) and (-x, 25 - x^2)
From the coordinates we can find that:

Width = 2x
Height = 25 - x^2

So we get the equation for area:

Area = 2x(25 - x^2)
Area = 50x - 2x^3

Take derivative to find the max/min

Area' = 50 - 6x^2 = 0
x = ± √(25/3) = ± 5/√3

Since we're finding length, it can't be negative so:

x = 5/√3

And since the width is equal to 2x:

2x = 10/√3 = 10√3/3

Because the channel is closed ima just give the working

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[wary frost]

A point P is defined on the plane. Various equilateral triangles ABC are considered, such that PA=2 ,PB=3 . What is the maximum value that PC can take?

[wary frost]

@dapper venture can i receive your help?

[dapper venture]

maybe calculate PC in terms of A and B

while A lies on circle of radius 2 and B lies on circle on radius 3

[wary frost]

okay thanks lemme try

[harsh parcel]

do u have the answer? i got one but idk if its correct or not

i got root 21

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[wary frost]

i got 5

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[harsh parcel]

5 is the side of the triangle ig

[harsh parcel]

oh

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[lunar lagoon]

How do answer 1A

[lunar lagoon]

Not sure where that A supposed to come from

This is what I get

[dapper venture]

$e^{-kt+c}=e^{-kt}e^c$ not $e^{-kt}+e^c$

[flat frigate]

WhereWolf(ping if needed)

[dapper venture]

then we set e^c = A

[lunar lagoon]

Why would the E^c be multipled with the e^-kt?

When they was separate before

[dapper venture]

property of expoenential function

$a^{b+c}=a^ba^c$

[flat frigate]

WhereWolf(ping if needed)

[dapper venture]

for example $2^{1+2}=2^1*2^2$ not $2^1+2^2$

[flat frigate]

WhereWolf(ping if needed)

[lunar lagoon]

Oh ok thanks

For b

How would I answer?

Would I just draw a normal E^x graph?

But reflected in the x axis?

[dapper venture]

yeah

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[ruby ore]

doing this integral $\int_0^1 \cos (x+\cos x) dx$

[flat frigate]

bostock

[ruby ore]

via a substitution $u=\cos x$ and some rearranging, I get $\int_1^{\cos 1} \sin u du - \int_1^{\cos 1} \frac {u \cos u} {\sqrt{1-u^2}}du$

[flat frigate]

bostock

[ruby ore]

$=\cos 1 -\cos \cos 1 +\int_{\cos 1}^1 \frac {u \cos u} {\sqrt{1-u^2}} du$ and this is where I have a problem

[flat frigate]

bostock

[ruby ore]

Integration by parts seems tempting but if I differentiate $u$ and integrate the rest, then I need to solve $\int_{\cos 1}^1 \frac {\cos u} {\sqrt{1-u^2}} du$ which isn't much better

[flat frigate]

bostock

[ruby ore]

if I differentiate $\frac u {\sqrt{1-u^2}}$ and integrate $\cos u$, I can't construct the rebound integral so that's also useless

[flat frigate]

bostock

[ruby ore]

can anyone give me a hint without giving away the whole solution please

I should write the integral as $\cos 1 -\cos \cos 1 + \lim_{a\to 1}\int_{\cos 1}^a\frac {u \cos u} {\sqrt{1-u^2}}du$ to be more precise

[flat frigate]

bostock

[safe radish]

@ruby ore Has your question been resolved?

[peak estuary]

WA cannot integrate it so you probably have to use some symmetry in some way and probably get that (integral)=(some function of the integral)

first obvious symmetry would be u -> -u

[ruby ore]

let me see what I can do

you basically get that the integral from cos 1 to 1 = the same integral from -cos 1 to -1, not sure if that helps

I'm going to try $u\mapsto 1+u$ and see what i get

[flat frigate]

bostock

[ruby ore]

with $u \mapsto \sqrt{1-u^2}$ we get $\int_{\sin 1}^0 \cos\sqrt{1-u^2} du$

[flat frigate]

bostock

[safe radish]

@ruby ore Has your question been resolved?

[safe radish]

@ruby ore Has your question been resolved?

[ruby ore]

asking this q on math se, it seems that a series expansion is probably the only way to do it

specifically the jacobi-anger expansion

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[hollow haven]

Hello, I just need some clarification to be sure.
I got that the true answers to the equation showed are X = 1/5, 2 and also got a false answer of X=1. However the answer sheet says that X = 1 is actually a true answer. How can this be?

[safe radish]

@hollow haven Has your question been resolved?

[toxic shoal]

https://www.desmos.com/calculator/qbqopo1don

i only see the solution of x = 2

so it should be the last answer, if i can interpret it correctly. i can't read hebrew

Dalet

neither the left nor right hand side is defined at x = 1

[hollow haven]

That is what I am saying. However, there is also x =1/5, which you can get from comparing the exponents

[toxic shoal]

oh yeah, you're right. weird that desmos doesn't show it

[hollow haven]

I used desomos as well lol

[toxic shoal]

it's so good . but yeah i agree that x = 1 can't solve the equation

[hollow haven]

As far as I understand, it is because x with a power of x are weird functions that are composed of many dots.

Thank you!

[toxic shoal]

yeah, that makes sense

[hollow haven]

.close

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[rain trout]

we have f(x) = 2x-3 and g(x)= x/2 , x<= 1 and x+3 , x>1

[rain trout]

if i want to find the domain of the (gof)(x)

x belongs to the domain of f

and f(x) belongs to the domain of g

what is the domain of g tho ?

[stray socket]

The x <= 1 and x > 1 is a strong indicator on what is in the domain (and, by extension, what isn't)

So if you were to union those inequalities, what do you get

[rain trout]

R?

[stray socket]

Yeah

[rain trout]

ok ty

.close

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[void brook]

Let f(x) = 10 − 2x and determine
a) f(3)
b) f(a + 1)
c) The zeros of the polynomial. Answer exactly.

[gilded pagoda]

4
12-2a
5

[void brook]

a) should be -8

[torpid fable]

It is not

[void brook]

In a i should just plug in x as 3 right?

[torpid fable]

Did you mistype it? Seems like there should be a square term

[void brook]

I get 26 when i do that

[torpid fable]

10-2(3)

[void brook]

it looks like this, but its in swedish

[torpid fable]

Yeah the squared part wasnt appearing

[void brook]

So it's not 26?

[torpid fable]

No

How did you get 26

[void brook]

I plug in x as 3

[torpid fable]

Then what next

[void brook]

0 = 10 - 6^2

[torpid fable]

2x^2 != (2x)^2

Does not

You have to do the square first (exponents before multiplication)

[void brook]

How do i do that first?

[torpid fable]

10-2(9)

[void brook]

10-18?

[torpid fable]

Yeah

[void brook]

-8

[torpid fable]

Yeah

[void brook]

But where does 9 come from

[torpid fable]

3 squared

[void brook]

3 squared

[torpid fable]

I think you may be missing something because you copied the problem wrong when you first entered the channel

You wrote just 10-2x which seems to be what you are doing instead

[void brook]

Nah I just dumb cus I took 2x as one and ^2 as one

[torpid fable]

I see

[void brook]

instead of 2 and x^2

But now i know I should use 2 and then 3^2

And the b) is 10- 2(a+1)(a+1)

But do I solve the (a+1)(a+1) first or the 2(a+1)?

[safe radish]

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[tough monolith]

can someone help me with this i really need help

[quiet juniper]

SOH CAH TOA

[tough monolith]

can someone help me with this

[lean otter]

You dotn know what

[tough monolith]

nvm .close

[lean otter]

Like
sinx = Opposite / Hypotenuse?

[tough monolith]

.close

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[oblique idol]

I'm stuggling on how to think about this question relating to the identity theorem

[oblique idol]

can someone share a hint on how to start thinking about this question

[peak estuary]

there is an obvious candidate

[oblique idol]

then ill sit with it for a bit longer, thx

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[lean otter]

uhm

[icy lance]

whats up

[safe radish]

@upper fossil Has your question been resolved?

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[gilded viper]

.reopen

[rugged surge]

we have to sketch this graph

this is what I have done:

let x-1 = x' and y+2= y'

then it becomes

(x')^2/3^2 + (y'/4)^2 = 1

so i changed the coordinates to x' and y'

and this is what I got

and then i kind of just rubbed x' and y'

for the final ans

but its wrong according to desmos

so can anyone tell me what am i doing wrong

[safe radish]

@rugged surge Has your question been resolved?

[rugged surge]

<@&286206848099549185>

i know i can just plug in y=-2 and x=1 and just make a graph bit i wanted to know why my method is not working

[safe radish]

@rugged surge Has your question been resolved?

[safe radish]

@rugged surge Has your question been resolved?

[toxic shoal]

this is (x-2)^2/3^2 + (y+1)^2/4^2

you just messed up where the center is

(also the scale being different on the x-axis vs. the y-axis makes it look way different)

[rugged surge]

but how? isnt x axis going down by 1 unit and y going right by 2?

[rugged surge]

yeaa i am sorry about that

[toxic shoal]

sorry i can maybe explain when im home, but itll be 20 mins or so

[rugged surge]

yea sure

[toxic shoal]

so you want to graph the ellipse defined by the equation (x-1)^2/3^2 + (y+2)^2/4^2 = 1.

if you define f(x,y) = x^2/3^2 + y^2/4^2, then the set of points that satisfy f(x,y) = 1 is an ellipse of the same "shape" but centered at the origin. the ellipse you want to find is defined by f(x-1, y+2). the question is, how does replacing x with x-1 and y with y + 2 change the location of the center of the ellipse.

replacing x with x - 1 shifts the ellipse f(x,y) = 1 by 1 unit to the right. to see this, suppose that (x_0, y_0) is on the ellipse f(x,y) = 1. then the point (x_0 + 1, y_0) is on the ellipse defined by f(x-1,y) = 1, because f((x_0 + 1) -1, y_0) = f(x_0,y_0) = 1.

similarly, replacing y with y + 2 shifts the ellipse f(x,y) = 1 by 2 units down. indee, if (x_0, y_0) is on the ellipse f(x,y) = 1, then (x_0, y_0 - 2) is on the ellipse defined by f(x,y+2) = 1 for essentially the same reason as before.

therefore, the set of all points that satisfy f(x-1, y+2) = 1 is just the ellipse defined by f(x,y) = 1 but translated 1 unit to the right and 2 units down.

this might be a little too abstract. if so, you can play around on desmos if you want

https://www.desmos.com/calculator/mezmkqqcdn

so basically, changing coordinates shifts the ellipse in the opposite way you might expect. i think that you confused your x and y-axes, though, as well.

your ellipse was shifted 2 units to the right and 1 unit down instead of 1 unit to the right and 2 units down.

[rugged surge]

okay that makes perfect sense

thankyou so so so much!!

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[toxic shoal]

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[valid vapor]

Calculus Optimization Question: A rancher plans to make four identical and adjacent rectangular pens against a​ barn, each with an area of 225 m squared. What are the dimensions of each pen that minimize the amount of fence that must be​ used?

Here's what I tried:
Perimeter = 5y+x
Area = xy = 900
Solving Area for x you get x= 900/y

Perimeter in terms of x is 5y+900y^-1
Derivative of Perimeter is 5+-900/y^2
Solving derivative for roots = sqrt(180) or 6*sqrt(5)

Plugging that y value back into the x equation gets 900/6*sqrt(5), or 150/sqrt(5)

Where am I making a misstep here?

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[lean otter]

does anyone know an easy way to understand the degree of a zero in a polynomial graph

[final quartz]

I dont know if I'll answer your question as clearly as some experts here but if you don't have a graph with you, the amount of zeros or factors that you have are a good indication of the graph's number of zeros

If you have the graph in front of you, the graph's degree can be determined by counting the number of turns the polynomial graph makes

and then adding 1

so for example

a graph that has the degree of 3 will have two turns, eventually going the opposite direction from where it came from the left

so here

you got three factors

just the expressions in the paranthesis

and then you got 2 turns for a 3rd degree polynomial

[safe radish]

@lean otter Has your question been resolved?

[lean otter]

no like

i get you but

idk why i have trouble figuring out the degree w/o the equation of a specific zero

like how do i tell if it's 1, 2, 3, etc.

[safe radish]

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[wispy wraith]

how do I solve this system of linear differential equations by systematic elimination?

[safe radish]

@wispy wraith Has your question been resolved?

[wispy wraith]

<@&286206848099549185> i multiplied the top equation by -2 and applied a (D^2+5) to the bottom equation and added them together to eliminate x, which left me with a 4th order DE that I have no clue how to solve. eliminating y's would be result in a nearly identical 4th order DE to solve. how the heck am I supposed to solve this?

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@wispy wraith Has your question been resolved?

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.close

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[rain hare]

x star y=x^2-4xy+3y^2
Prove that theres an infinite amount of irrational numbers "a" for which a*1 is natural

[rain hare]

so i found theres an infinite amount of irrational numbers a but idk how to prove that they satisfy a*1 is natural

and i cant think of an irrational number that i could plug in a*1 that would give me a natural number

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@rain hare Has your question been resolved?

[rain hare]

<@&286206848099549185> 🙏

[toxic shoal]

a ⋆ 1 is natural if and only if a ⋆ 1 = n for some n in ℕ. This is the same as saying a² - 4a + 3 = n, or a² - 4a + (3 - n) = 0

now you have to show that there are infinitely many n ∈ ℕ such that x² - 4x + (3 - n) has irrational roots

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[brisk iris]

.close

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[celest elm]

any advice on whether i performed the differentiation correctly? I tried breaking it up into more manageable pieces before putting it all together, but i think i may have missed a step somewhere

[safe radish]

@celest elm Has your question been resolved?

[misty bay]

,w differentiate cos(sqrt(sin(tan(pi*x))))

[flat frigate]

Results provided by WolframAlpha

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Requested by Saccharine#3780

[misty bay]

seems like you're missing a term in the numerator

I suggest you become familiar with a computer tool that differentiates things for you

and then you can use it to check whether each of your steps follows from the last

[celest elm]

Im not sure where i missed the term, I was thinking its where i differentiated cos, because i applied the power rule and then put it straight into the differentiated cos function, should i multiply by the original again?

[misty bay]

oh this is not true

the sqrt is inside the cosine

neither is this

[celest elm]

so the exponent affects the sin function not the cos function?

[misty bay]

the parentheses mean something

just like how cos(x^2) is not the same as (cos x)^2

cos(sqrt(x)) is not the same as sqrt(cos(x))

but in any case, you should become familiar with some computer tooling to check every step for you

it'll save you a lot of time in the long run

[celest elm]

okay thank you

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[dusky harbor]

Physics Question: Two masses and tension.

[dusky harbor]

I believe I messed something up during the process of finding the answer for Qa.
This is what we know:
m_1 = 10kg
m_2 = 2kg.
W_1 = 98.1N
W_2 = 19.62N

Considering g = 9.81m/s^2

So I wanted to draw a free-body diagram of the two masses and the forces being applied

wait (fixing a potential error* (?))

So here's where I want to get a hint... The tension that the 10kg mass is experiencing, is that equal to the weight of the 2.0kg mass?

This is the free-body diagram of the situation.

[twilit flame]

That's only if the system is static

ie not moving

you manage to get the first part?

[safe radish]

@dusky harbor Has your question been resolved?

[dusky harbor]

Nope. And what do you mean by "system is static"? considering that there is no friction?

[twilit flame]

No that the system is not moving

The tension won't be equivalent if there's an acceleration

Newtons 2nd law

[dusky harbor]

Tension is just one of those things I confuse myself with.
But because question a) asks us to find the acceleration, we assume that the masses are in motion, correct?

[twilit flame]

Yes

[dusky harbor]

Hmm, if mass and weight is the only thing I know, how do I find the Force or acceleration or tension?

[twilit flame]

By using two simultaneous equations

If you derive equations for the two components of the system in their two components you can create a set of simultaneous equations

give me 1 sec to illustrate

(We don't need to consider the other components cause they are 0)

[dusky harbor]

hmm-- I was about to ask why you don't consider the Weight of A, but perhaps it's because we are ignoring friction?

Weight doesn't come into play when there is no friction, yes?

[twilit flame]

yes as it's a smooth the coefficient of friction would be 0

[dusky harbor]

aaha

[twilit flame]

So μR = 0

[dusky harbor]

Wait a sec-- The tension for equation A and equation B are not the same (?) because the masses are in motion

[dusky harbor]

I'm either stuck at a dead end or struggling to fit this understanding through my rusty brain cells...
The tensions aren't the same for the equations A and B, and we are trying to find a.

Mmm- if I move away with that idea, assuming that the tensions are the same, I'm getting 1.635ms^-2 or 1.6ms^-2 (which the answer sheet states the same)

[twilit flame]

they are teh same

As it's the same string (i think extensible should be added to the q)

[dusky harbor]

Yeah, I found the answers after this acknowledgement

thank you very much :3

[twilit flame]

nws

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[vital creek]

I'm back again still not the brightest

Consider the curve y=x²-3, find the gradient of the secant joining the points where x=1 and x=1+h. Hence deduce the gradient of the tangent to the curve at x=1

i did the problems before and after this one okay but im just really confused on this for some reason

[quasi bison]

!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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[vital creek]

sorry i thought the wording gave that i don't know where to start on this

[quasi bison]

never hurts to ask...

but ok like

you have the points (1, ___) and (1+h, ___) on your curve

fill in their y coords

[vital creek]

yea 1,-2, and 1+h, (1+h)²-3

[quasi bison]

missing parentheses

(1, -2) and (1+h, (1+h)^2 - 3)

ok now find the slope of this line

the line joining those two points

[vital creek]

ok i can y1-y2/x1-x2 that but i don't know what im supposed do with the result is it not just 2+h

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@vital creek Has your question been resolved?

[vital creek]

i don't understand why this would be any different than the one that is just finding the curve at x=1 on y=x² considering its the exact same thing just moved down a bit but it says it can't be

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.close

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[cyan flame]

Hello! I have a question about a quadratic equation with 1 variable.
ax²+bx+c=0
if we devide both sides by "a" we get
x²+bx/a+c/a=0
subtract "bx/a+c/a" from both sides
x²=-(bx+c)/a
therefore
-(bx+c)/a >= 0
As we can see there is a clear restriction for values of x. To make the equation true, it has to satisfy the linear inequality we got. Does it matter while solving a quadratic equation? Is there a possibility that quadratic formula produces the result which does not satisfy that inequality, and therefore the equation?

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[round tiger]

Hello

[round tiger]

if i wanted to create an equation that represents the population of a city, how would i go about that?

3 factors that i said affect the population of a city were

1. job opportunities
2. affordability
3. personal preference

how do i include these into my equation and make it calculus related?😭

[wheat cave]

they just affect the growth rate

[round tiger]

oh right

but it wouldn’t be exponential right?

[wheat cave]

if you only have those 3 factors yes

2 rabbits give birth to 2 rabbits, then those 4 rabbits make 4 new ones etc.
population growth is exponential if you look at it naively

[round tiger]

ohh

i’m looking at this equation here

[wheat cave]

that would be exponential

[round tiger]

yeahh

would i be able to add my 3 factors into this equation?

[wheat cave]

its just r

what else would those factors affect?

initial population?

[round tiger]

no i guess not

could i add stuff onto r

like change it up so that it incorporates my factors?

[wheat cave]

I don't see the issue

why not?

you can just rewrite r as anything

like r=a+b+c

[round tiger]

oh okay

could i do r= a • b • c?

wait but how do i know that it’ll actually calculate the approximate population

man i don’t get it

[wheat cave]

you dont because in all honestly those 3 factors give you absolutely not enough information about a population of a city

[round tiger]

yeah i understand

[wheat cave]

maybe it would help if you could provide your actual task/question you are supposed to solve

[round tiger]

sure one second

i probably should have started with that

this is basically the task