help-10

[feral prawn]

What is a, n, and d in relation to the diagram?

[high granite]

a is the first number in the sequence

n is the term number so 6th term means n=6 and d is the common difference

so 1,2,3

common differenve is 1 since the pattern increases by 1

[feral prawn]

I meant the number of sticks

Like this:

[high granite]

oh my bad

[feral prawn]

np

[high granite]

4,12,24

oh

so its a geometric sequence

[feral prawn]

For this

And I need the nth term rule

that connects the shape number and the number of sticks

like shape 1 = 6

sticks

shape 2 = 13

etc

Any idea?

.close

[obtuse pebble]

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[marsh coyote]

How do I find angle DAE?

[trail cloak]

Hmm

What did you get so far?

[marsh coyote]

I know what r and theta is

[rain forge]

2*theta

[marsh coyote]

?

why 2 theta?

[rain forge]

do you know the inscribed angle theorem

[marsh coyote]

btw answer says theta/2

[trail cloak]

You mean half of it?

[rain forge]

yes sorry theta/2

mb i wrote it fast

so D B and A are in the same circle

and the inscribed theorem say that the angle DAE = theta/2

[marsh coyote]

so central angle is theta

so inscribed is theta/2

?

[rain forge]

yes it's the center of the circle because we have on each part a radius r

[marsh coyote]

we have what?

[rain forge]

we have our angle theta which is 2 times the angle DAE

I will try to prove it

so we need to prove that DPE = 2 DAE

[marsh coyote]

yup

[rain forge]

so 2 DAE = 2EAP + 2 PAD right ?

it's chasles relation don't if it's like this in english

[marsh coyote]

ok

[rain forge]

EAP and DPA are isocele in P right so 2 DAE = EAP + PEA + DAP + PDA

[marsh coyote]

ye

[rain forge]

so we know that the sum of angles in triangle are 180 so we have = 180 - APE + 180 - DPA

I take DPA in my proof but we can supress him because his value is 180 the angle is flattened

so 2 DAE = 180 - APE

so it's equal also to theta

[marsh coyote]

ohhhhh

[rain forge]

so DAE = theta/2

[marsh coyote]

180 - (180 - theta)/2

thats theta/2

tysm

[rain forge]

yes like theta is equal to 180 - APE because the angle in a line is 180

and 180 - APE = 2 DAE

so theta = 2 DAE

u can even do it better, the angle EPA = 180 - theta right, and the triangle EPA is isocele and the sum of angles in a triangle is 180 so the sum of the 2 missing angles are theta but EPA is isocele so PEA = EAP= theta/2

[obtuse pebble]

@marsh coyote Has your question been resolved?

[marsh coyote]

got it

tysm

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[sour wave]

A girl bought a pair of skis. When she checked her bank she had been charged 3289,50 NOK. How much did the skis cost in sweden. The currency for swedish coins SEK is 104,43.

So the answer is 850 SEK. But i cant get to this answer. Im not sure how to do this. Im bad at math. Ive tried taking 3289.50 : 100 * 104,43. That equals 3435,22. 850 SEK makes no sense to me 😭

[ruby path]

,w solve 1/104.43 = 3289.5/x

[warm shale]

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[ruby path]

,calc 68704497/200

[warm shale]

Result:

3.43522485e+5

[ruby path]

How useful

[gleaming ridge]

Lmaoo

[ruby path]

Wdym the currency for Swedish coins SEK is 104.43

[sour wave]

The assigment gives me that information

[ruby path]

Irl the NOK and SEK are pretty much equal in value

[ruby path]

No I meant like is that the exchange rate

[sour wave]

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[ruby path]

[obtuse pebble]

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[bold siren]

not quite sure what to do here

is it asking me what a is?

or is it asking what log0.4^4 is?

<@&286206848099549185>

,n,

.close

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[royal karma]

Hey! I need help calculating a maclaurin polynom for parametric function.

[royal karma]

$$\begin{Bmatrix}

y=t^2+ln(t)
\ x = t^3-1

\end{Bmatrix}$$

[warm shale]

Kingo

[royal karma]

i need the second order

[obtuse pebble]

@royal karma Has your question been resolved?

[obtuse pebble]

@royal karma Has your question been resolved?

[tardy epoch]

are you supposed to find the maclaurin of $y(x)$?

[warm shale]

riemann

[royal karma]

Yes

[tardy epoch]

Do you know chain rule?

[royal karma]

Yes

I have an answer

But im not sure if it's correct

[tardy epoch]

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

[royal karma]

I did it already

[warm shale]

riemann

[tardy epoch]

and a very similar tedious one for y''

show your work

[royal karma]

Y't/X't

I don't have the work
It was on my calculus test today

I got 1+x-5/18x^2

[tardy epoch]

work it out again?

[royal karma]

Aah

Ok 1 min

@tardy epoch

Here

that's how i did, but im afraid im wrong somewhere

you just made my heart

going from 0-200

[tardy epoch]

i don't follow this step

[obtuse pebble]

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[tardy epoch]

$\frac{d^2 y}{dx^2} = \frac{\frac{dy}{dx}}{\frac{dx}{dt}}$?

[warm shale]

riemann

[royal karma]

.reopen

[obtuse pebble]

✅

[ruby path]

God what is that formatting

[royal karma]

@tardy epoch

that's the same?

[tardy epoch]

oh is it? i hadn't seen that before

[royal karma]

here i'm missing a tag on the y'(x)

it should be (y'(x))'/xt

[tardy epoch]

ah yea it is

[royal karma]

phew

[tardy epoch]

https://www.nagwa.com/en/explainers/613184581056/

[royal karma]

maybe i can put it in wolfram, but i couldnt find the right notation for it

i think mine is good, but i always want to be sure

[tardy epoch]

yea it looks right

you can also find the answer by solving for $t = (x+1)^{1/3}$ then plug that into $y$ to get a function of $x$

[warm shale]

riemann

[tardy epoch]

it should give the same answer

,w second order taylor ((x+1)^(1/3))^2 + log((x+1)^(1/3))

[warm shale]

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[tardy epoch]

matches

[royal karma]

YEAAAA

[royal karma]

Great idea!

Thank you so much!

[royal karma]

Here, you solve for x right?

oh for t

ok

[obtuse pebble]

@royal karma Has your question been resolved?

[obtuse pebble]

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[low frost]

does this equation have one solution no solution or infinite solutions

[slender nebula]

infinite

[low frost]

because it’s the same equation?

[slender nebula]

yes

[low frost]

okay

ty

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[narrow island]

i know its 5th degree root of 3, but can we write this without root?

[short spruce]

you already did

[royal basin]

well you already have

[narrow island]

actually what would it equal :) i mean, the result of it? like we know 3 power 2 equals 9. But what 3 power 1/2 equals?

[short spruce]

type it into a calculator

it'll give you a decimal approximation

[rigid lintel]

,calc 3^(0.2)

[warm shale]

Result:

1.2457309396155

[timid silo]

it is not a perfect square

power 1/2 that is

well 3 power 1/5 is not a perfect...quint?

[dark stirrup]

It's an irrational number

[narrow island]

ohh

[dark stirrup]

You can't represent it as a fraction if that's what you were trying to arrive it

[rigid lintel]

hypercube

[dark stirrup]

There's a well-known proof that sqrt(2) is irrational

You should look it up and you might figure out how to prove that 3^(1/5) is also irrational

[narrow island]

thanks a lot for the answers :)

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[tawny mesa]

i need help with the answe and how to show it

[timid silo]

going from step 1 to step 2?

[tawny mesa]

yeah i need to show my work on how to find the answer

[gray jungle]

do you know basic properties like distributive

[tawny mesa]

yeah

[gray jungle]

how would you start

the fraction complicates things, it will be easier to remove it

[tawny mesa]

ik but how would i remove it?

[gray jungle]

what is the inverse of division?

[tawny mesa]

any thing divided by itself=0?

[gray jungle]

no no no

the inverse operation

the inverse of addition is substraction

so the inverse of division is

[tawny mesa]

idk?

[gray jungle]

multiplication

[tawny mesa]

oh

[gray jungle]

so multiply both sides by 5 to get rid of the fraction

[tawny mesa]

i feel so dumb

XD

[gray jungle]

you get now $10x-30=300$

[warm shale]

꧁╭⊱尺αιηωα𝗋𝖾⊱╮꧂

[gray jungle]

do you know how to solve this now

[tawny mesa]

yeah

[tawny mesa]

x=33
10x 33= 330-30=300

[gray jungle]

good

[tawny mesa]

ik that x =33 but i need to answer the questions on the side

[gray jungle]

It can be solved in different ways, either by multiplying both sides by 5 or by using the distributive property

If you want to use distributive, multiply both 10x and 30 by 1/5

$=2x-6 = 60$

[tawny mesa]

The equation in step one is correct because Marshall multiplied both sides to balance the equation To determine the equation in step 2 Marshall reasoned that blank: answers for "blank are 18-6=12 12-6=6 18 divided by 6=3 and 12divided by 6=2 is what i need to figure out

[warm shale]

꧁╭⊱尺αιηωα𝗋𝖾⊱╮꧂

[gray jungle]

first step is incorrect

you don't apply distributive to both sides

[tawny mesa]

so what would it be then? but i dont need to rlly solve the equation i dont think

[tawny mesa]

those are what i need to solve

[tawny mesa]

but i dont know if the first step he did is correct

[gray jungle]

it is not

[tawny mesa]

so if he wasnt correct would he should of multiplied 10 and -30 by 5?

[gray jungle]

by 1/5 not by 5

[tawny mesa]

so the answer would be should not have multiplied 60 by 1/5?
for B

[gray jungle]

yes

which is doing the distributive

[tawny mesa]

so why would thta be correct thogh?

like he cant distribute in this problem?

[tawny mesa]

why should he have not multiplied 60 by 1/5

[gray jungle]

because the distributive is only for one side

[tawny mesa]

ahh ok

[tawny mesa]

what about c it says To determine the equation in step 2 Marshall reasoned that blank: answers to blank is: 18-6=12 12-6=6 18 divided by 6=3 and 12 divided by 6=2

[tawny mesa]

?

[tawny mesa]

would it be 12 divided by 6=2?

uhh you here? @gray jungle

.close

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[timid silo]

Does Anyone undestand where i am wrong?

I mainly apply hopital

[tardy epoch]

the 3 doesn't go on the outside of (e^x + e^(3x))

[timid silo]

Hallo, welcome in my thread

[tardy epoch]

(e^x + e^(3x)) ' = e^x + 3e^(3x)

[timid silo]

Riemann you are a boss. Basically derivative of e^3x is e^3x × 3

But i made a wrong multiplication

Problem solved thank you very much @tardy epoch

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[timid silo]

$\frac{p}{\frac{p+2}{p}}$

[warm shale]

okokok

[timid silo]

ok so i get

simplify

$\frac{p^2}{p+2}$

[warm shale]

okokok

[timid silo]

the answer sheet is saying p+1 i the numerator how did they get that

[tardy epoch]

show the original problem

and your work

[hexed agate]

,w simplify p^2/(p+2)

[warm shale]

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[hexed agate]

Wolfram 💀

Anyway it cannot be simplified further

Answer sheet on crack

[timid silo]

this is the original problem

[timid silo]

i multiplied the denom p and the num p

[tardy epoch]

no. that's your typing

show the original statement. screenshot or picture

[timid silo]

ok it says determine the LCD of all the fractions appearing within the expression

and the one I posted first is identical to the complex fraction shown

[tardy epoch]

nah

screenshot or picture

[timid silo]

whoopsie daysie

in the physical text book it looked off

[tardy epoch]

this is not the same

[tardy epoch]

as this

[timid silo]

ik

so the two p denoms cancel each other out right

[spiral knot]

yes

[warm shale]

Mehdi_Moulati

[obtuse pebble]

@timid silo Has your question been resolved?

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[vast wren]

then what’s the slope of your line

if it’s perpendicular to the given line

hint: ||m1 * m2 = -1||

[warm plover]

Well the intercept at this point doesn't matter. All that matters is the slope. There is a relation between perpendicular line's slopes, do you know it?

[obtuse pebble]

@daring valve Has your question been resolved?

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[rain olive]

Hello

[rain olive]

I have gotten my test back

Quiz

And got 86

This is unlike my last year math grades

And I felt like I didn’t have much time to do it

As I chose not the go to my ARCH to do my test and get extra time

Because I wanted to do it like other kids

.

I made some grave mistakes that lost me marks

If I had more time, I would have been able to correct

Them

[sharp pecan]

yes, but whats the question

[rain olive]

Do you have any tips on doing stuff faster

Math

[latent walrus]

practice

[rain olive]

I did

I practiced a lot.

Or at least did all assigned homework.

[sharp pecan]

I guess you can learn specific question types more extensively, but in general its just practice

[onyx moss]

in an exam it’s useful to know how many marks the paper is worth, and how long you have to complete the exam. That way you can work out roughly how many minutes you should be spending on each question. It’s not a foolproof method but in general it can help if you find you’re running out of time

[obtuse pebble]

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[worldly oasis]

can someone help? i dont know how to start

[wise talon]

first make the "b" term the same number

[worldly oasis]

a : b = 9 : 6

b : c = 6 : 4

[wise talon]

yeah I think you got it lol

[worldly oasis]

oh

[wise talon]

you're really close

[worldly oasis]

and now just

[wise talon]

yep

[worldly oasis]

okay

[wise talon]

yeah yeah

[worldly oasis]

that was easy

thanks

[wise talon]

LOL

np

[worldly oasis]

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[fallow granite]

Does this look right?

A = 60
B = 19
C = 101
D = 79
E = 50

[unique depot]

I guess so

[fallow granite]

Which one are u not sure on?

[timid silo]

checked it through, looks perfect

[fallow granite]

bet ty

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[long holly]

Hi

[obtuse pebble]

@long holly Has your question been resolved?

[obtuse pebble]

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[dawn quarry]

Can someone help me please

[dawn quarry]

.close

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[short spruce]

if both roots are equal, what does that tell you about the quadratic?

[novel trench]

⬜

if both roots are equal in a quadr. eq. then it is a square

[nocturne minnow]

You mean discriminant?

[novel trench]

need to send k - 5 to the other side

how does that look

send k too

as minus

[nocturne minnow]

No need to do this at all

Do you know what is the discriminant is?

[novel trench]

you dont need it here at all?

what will you do with discriminant here dude

[nocturne minnow]

What is the equation for the discriminant ?

[nocturne minnow]

You use it

x^2 + 3x = k - 5
You have this

You know a and b

You just need to get it into the form of ax^2 + bx + c = 0

[nocturne minnow]

Oh btw my bad, I didn't realize k - 5 was on the right side, I thought it was on the left

[novel trench]

is k = 2,75

do you have the answer

got the answer or

[novel trench]

.

[nocturne minnow]

No

[novel trench]

i was gonna tell if u say yes

ok now listen

[nocturne minnow]

You want to move everything to one side, to get it in the form of ax^2 + bx + c = 0

[novel trench]

why do you act like im wrong

just move everything to one side now what do you get

[nocturne minnow]

It's better keep it as a fraction, fractions are more exact, and better looking

[novel trench]

okay doesnt change the fact that its 2,75

now, if a quadratic equation has 2 equal real roots

it is a square of something

which means you can write it as

a(x+b)^2

[nocturne minnow]

Now that part, you don't need to do, just use the discriminant

Plug in what you know, solve for k

[novel trench]

please shut up for a moment

[nocturne minnow]

You know b^2 - 4ac is the discriminant

You can plug in a, b, and c

I suggest grouping up the terms

[novel trench]

now you have this from there you know a is 1 and b is 3

so -b/a is the sum of the roots right

-b/a = -3 here then

[nocturne minnow]

This is unnecessary to do

So 0 = x^2 + 3x + 5-k is your equation, what is a, b, and c?

Use ax^2 + bx + c as a template to compare

[novel trench]

so if the sum is -3

we know the roots are equal

x + x

2x = -3

[nocturne minnow]

This is really not needed

[novel trench]

x = -1,5

can you ignore this guy btw im getting mad

[nocturne minnow]

Like completely unnecessary to do. All you need is the discriminant, that's it

No need for that extra stuff

[novel trench]

now you just need to write it as

[nocturne minnow]

Both work, but in your case, the discriminant is better to use

[novel trench]

(x+1,5)^2

[nocturne minnow]

Especially if you are learning/applying the concept of the discriminant

[novel trench]

from there you'll find c

yes

5-k = c

if you find c

then you find k

that would work too

whichever way you prefer

np

need to use a .

[obtuse pebble]

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[nocturne minnow]

Yes and no, depends on what math you need to apply. If you're learning the discriminant in class, and you should be applying that concept, you shouldn't deviate from what you should apply

[obtuse pebble]

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[novel trench]

@opaque lily by the way just keep in mind that in a quadratic equation in ax^2 + bx + c form
-b/a = sum of the roots
c/a = multiplication of the roots

[novel trench]

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[dark orchid]

Do i use the diamond method to factor problem 17?

[teal turret]

U can use whatever u want, it may be easier to factor out a 2 from the start, however

[dark orchid]

Huh

Alright then.

Ty

.close

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[fickle root]

Systems of inequalities graphs

[fierce lagoon]

What about it is confusing

To you

[fickle root]

i dont understand it like how to solve it

[fierce lagoon]

Well first identify which line is which function

[fickle root]

what do i do next

[fierce lagoon]

Well you gotta me which lines are which

There's a green line and a blue line

[fickle root]

uh how i do that

[fierce lagoon]

Look at the slopes

[obtuse pebble]

@fickle root Has your question been resolved?

[obtuse pebble]

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[hidden rain]

Envision Algebra 2 Unit 4 Rational Functions Test B

[hidden rain]

does anyone have that test? i took to today and wanted to see the problems

[timber fox]

You want to see the problems for a test youve already taken or are going to take?

[hidden rain]

already taken

i know what i put on them i just want to know the score that i would get on it

my teacher takes a very long time to grade tests so im nervous about what i got

[timber fox]

Then use an online calculator or post the questions if the test has already been taken there is no harm ig

[hidden rain]

ah

[wise talon]

a more helpful thing may be to

tell us what questions you felt unconfident in

[obtuse pebble]

@hidden rain Has your question been resolved?

[obtuse pebble]

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[fluid gulch]

i need help

[modern pond]

When I worked it out I got 168 and Symbolab got 147 which my homework says 147 is correct

So I got the answer and worked it out but am having trouble knowing how we got rid of the denominator h or do we even have to do it?

[fluid gulch]

why r u in my help

[fossil mountain]

So, for letter a)

Do you think you can further develop that equation?

[fluid gulch]

idk

[fossil mountain]

m + m
— — = m
m m

[fluid gulch]

idk

[fossil mountain]

You gotta play around with this.

[fluid gulch]

what to do

[fossil mountain]

So, do you agree that is the same as:

m + m
——— = m
....m

?

[fluid gulch]

ye

2m/m=m?

[fossil mountain]

Can you get rid of the denominator?

Isn't it annoying?

[fluid gulch]

2m=m^2?

[fossil mountain]

Yes.

Now bring both to the same side.

[fluid gulch]

2m-m^2

m^2-2m

[fossil mountain]

= 0

Do you know how to solve it from here?

[fluid gulch]

m(2-m)

m=2 m=0

?

[fossil mountain]

Almost

M can not be 0.

Because m is the denominator.

[fluid gulch]

so m=2?

[fossil mountain]

Yes.

[fluid gulch]

oh ok ty

[fossil mountain]

Try messing around with the equations.

See how you can make it look better.

[fluid gulch]

oke

ty

[fossil mountain]

Most math problems at this point are about rewriting the equations.

Let's do b)

[fluid gulch]

m+1/m=5/2

[fossil mountain]

........1 1
m + — = 2 —
.......m 2

[fluid gulch]

m^2+1=5m/2

2m^2+2=5m

2m^2-5m+2

(2m-1)(m-2)

?

m=2

m=1/2

is that correct?

[fossil mountain]

Seems like it.

You're fast typing.

Yes.

[fluid gulch]

how to do c

[fossil mountain]

Let me check it.

You're doing great, by the way.

Fast learner.

[fluid gulch]

nvm i got c

[fluid gulch]

ty

i got d

[fossil mountain]

Yeah, C is a situation in which you gotta rewrite the equation doing the opposite: making it more complicated.

[fluid gulch]

this one is complicated

i need help on thsi

[fossil mountain]

Is this series?

[supple granite]

Look at what numbers u can cancel out

[fossil mountain]

I just had a test on this.

[fluid gulch]

wait what?

[fossil mountain]

Oh yeah.

Look closely.

[fluid gulch]

idk

[fossil mountain]

You can cancel the numbers out.

[fluid gulch]

how

why

[fossil mountain]

Nominators with denominators.

[fluid gulch]

oh

[supple granite]

For ex, there's 4 on both the numerous and denominator, so you can cancel that out

[fluid gulch]

ye

[supple granite]

If you keep doing that with 5, 6, 7...., you'll find a pattern

[fluid gulch]

but after that

what to do

[supple granite]

So you have a/3 left in the sqrt, right?

[fluid gulch]

ye

[supple granite]

Can you find a?

[fluid gulch]

12?

[supple granite]

Yup

Now you need to find b

[fossil mountain]

. a n
..Σ ———
i = 4 n - 1

[supple granite]

If you look at the other fractions, like 4/3, 5/4, 6/5.... what pattern can you find?

[fluid gulch]

11?

[supple granite]

Great

[fluid gulch]

ty

[supple granite]

So a=12 and b=11

[fluid gulch]

ty

[supple granite]

Yw

[fluid gulch]

.close

[obtuse pebble]

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[obtuse pebble]

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[fluid gulch]

how to do b

[wise talon]

25/11 = (things on the right)

[fluid gulch]

ik

how to do b

[wise talon]

literally what I said

lmao

[fluid gulch]

What?

What did u say

[nocturne minnow]

.

[fluid gulch]

Still dont get it

[nocturne minnow]

Set 25/11 equal to what it gave for part b

[wise talon]

i was trying to avoid having to typeset this but

$\f25/11. = 2 + \f1/{3+\f{x}/y.}.$

[fluid gulch]

Ye ik

But how to solve it

[tardy epoch]

start by subtracting 2 from both sides

type your result here

[hidden garnet]

Yes

Go step by step

[obtuse pebble]

@fluid gulch Has your question been resolved?

[fluid gulch]

how to do d

[fluid gulch]

.

<@&286206848099549185>

@tardy epoch

[frigid prism]

dont ping helpers unless you wait 15 min

just move the stuff to the other side

get rid of the fractions

same concept as the other ones

[fluid gulch]

i dont get it

it has been 15mins

[frigid prism]

subtract by 2 and then multiply by (1 + x/y)

[frigid prism]

you asked it here

[fluid gulch]

ok

i still dont get it

[frigid prism]

wait you only have one equation

is it asking for the final value of x/y

[fluid gulch]

ye

[frigid prism]

oh

[frigid prism]

what do you not understand about this then

you want to isolate x/y

[fluid gulch]

no question d

[frigid prism]

its the same thing

[fluid gulch]

i dont know how to do d

[frigid prism]

just -2 and multipy by (something + x/y)

you want to get x/y out of the fraction

[fluid gulch]

i still dont get what ur tryna say

[frigid prism]

treat x/y as a single variable

[fluid gulch]

i got it

x=1

y=2

[frigid prism]

idk i didnt solve it

plug it back in the equation to see if it works

[fluid gulch]

.close

[obtuse pebble]

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[obtuse pebble]

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[fickle wing]

hey all, could someone show me the answer to this

[thick gyro]

use integration by parts

[fickle wing]

uh

how do i integrate cos^ (n-1) (9t)

please can someone show me the answer

pls pls

[thick gyro]

isnt that just n/9 * sin^n (9t)

[fickle wing]

mm i plugged that in and it didnt work

heres the full thing for more potential context

idk

[thick gyro]

didnt i say do by parts

[fickle wing]

HANDOKYO

[obtuse pebble]

@fickle wing Has your question been resolved?

[obtuse pebble]

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[lament jungle]

Hi could someone help me with this

[lament jungle]

I did 750^2 + 1000^2 - 2(750)(1000)Cos(140)

Sqrroot of that

And that is not the answer

Idk what I am doing wrong

<@&286206848099549185>

Got it nvm

!close

.close

[obtuse pebble]

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[fallow atlas]

how would i know when the particle is at its leftmost position?

[faint cave]

integrate the modulus of v(t)

with the points it travels from and two as the bounds

[fallow atlas]

im sorry i dont really know what this means

would it just be

setting up an integral from 2 to 7?

[faint cave]

thats ok

[faint cave]

um im guessing that the particle starts where its velocity is 0

so you need to find its starting position

can you use calc?

[fallow atlas]

oh here ill j set the velocity equation to 0

[faint cave]

\int_{t(2)}^{t(1)} |v(t)|$

$\int_{t(2)}^{t(1)} |v(t)|$

[warm shale]

soupy

[faint cave]

we dont know what t is yet

[fallow atlas]

ah okay

for v(t)=0 i got sqrt(4pi+8pin) - sqrt(4pi+8pin)

[faint cave]

can u use calc?

it seems to start, when velocity is 0, at 3.54,

[faint cave]

what is pin?

[fallow atlas]

pi like the symbol

and the letter n

[faint cave]

sorry where is n from

[fallow atlas]

because its cos

so on the unit circle

you multiply by n

[faint cave]

ahh

[fallow atlas]

yess

[faint cave]

are you not allowed to use calculator?

[fallow atlas]

we can use a calculator

but i still have to show work

[faint cave]

what kind? GDC or normal?

[fallow atlas]

graphing

[faint cave]

ahhh you can use n solve then

[fallow atlas]

oh rlly?

[faint cave]

for ap im not certain but yes i think so

[fallow atlas]

ok ok

[faint cave]

but even without it

the first thing i would think is that either e^t/2 or cos(t^2/8) must be equal to 0

e^? cannot equal to zero

but cosine(pi/2) is equal to zero

so t^2/8 must be pi/2

[fallow atlas]

oooh ok

[faint cave]

so t^2 must be 4pi

and then we find that t is the square root of 4pi

(3.54)

now we can do the integral $\int_{2}^{sqrt(4pi)} |v(t)|$

[warm shale]

soupy

[faint cave]

because 2 is the leftmost position i believ

[fallow atlas]

ah okay

[fallow atlas]

how did you know that t^2/8 would equal pi/2

[faint cave]

because cos(pi/2) is the only scenario where v(t) = 0

[fallow atlas]

oooooh

[faint cave]

and the equation given to us has t^2/8

so to make it equal to 0, t^2/8 must be pi/2 so that cos can equal 0

[fallow atlas]

yes that makes sense

and then i just plugged the integral into my graphing calculator

and i got approx 2.977

[faint cave]

yepp :)

but im not sure

if the particle goes to 7 and then back to 2

or if it goes straight to 2

[fallow atlas]

hmm

i have an example from a past problem that my teacher did and it was similar to how you just explained it

so i think its right

but anyways thank you sm

[faint cave]

okayy :)

[fallow atlas]

this helped a lot :)

[faint cave]

ofcc

[fallow atlas]

.close

[obtuse pebble]

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[timid silo]

yes

[timid silo]

why is this true ?

how do we arrive at the condition from this expression like in above situation

<@&286206848099549185>

[gleaming ridge]

$|x| \le a \iff -a \le x \le a$

[warm shale]

numbpy

[fierce wind]

This is true because the absolute value of any number is that number positive. x-1 no matter what the x is will be a positive. Depending on what you solve for the absolute value of x the a value has to be greater than or equal to the absolute value and not the x. -2 <= x-1 <= 2 is true because -2 <= 3-1 = 2, -2 <= 2-1 < 2, -2 <= 0-1 <= 2, -2 <= -1-1 <= 2.

You have four possible values for x and the statement is true because you have solvable x values

@timid silo

[gleaming ridge]

If you wanna know why it works, just expand the definition of |x|

[timid silo]

wait

yes

let me

oh yes

i got it

thanks

nice

how do i proceed towards solving this

can we conclude anything if

[timid silo]

can we not conclude (x+y)(x-y) = 0 from this expression

@gleaming ridge

@fierce wind

[sage geode]

|x| = |y| -> (x + y)(x - y) = 0, yeah

[timid silo]

do you think i can use this to solve this

[sage geode]

That will lead to considering cases |x - 1| - 2 = x - 3 and |x - 1| - 2 = 3 - x anyways, so yeah

[timid silo]

i am not very good with graphs of absolute values

[fierce wind]

Think about it X has to be 0 because it would be the absolute value of 0-1 = 1 and the absolute value of -2 = 2 and the absolute value of 0-3 = 3 then both sides are equal and you have your x. If the absolute value of x = the absolute value of y then (0+0)(0-0) = 0

[timid silo]

.close

[obtuse pebble]

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[obtuse pebble]

@edgy briar Has your question been resolved?

[obtuse pebble]

@edgy briar Has your question been resolved?

[obtuse pebble]

Please read #❓how-to-get-help

[median dome]

!show

[obtuse pebble]

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

[median dome]

@edgy briar

[edgy briar]

the whole thing

[obtuse pebble]

@edgy briar Has your question been resolved?

[obtuse pebble]

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[tawny fog]

It is easy to prove but if z and x are the mid points

Do you have any more data?

[obtuse pebble]

@delicate grail Has your question been resolved?

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[fleet pond]

help

[autumn adder]

Hi @fleet pond

Have you tried anything yet?

[obtuse pebble]

@fleet pond Has your question been resolved?

[obtuse pebble]

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[proven zealot]

do u put a dx behind this? I think he forgot it

[ruby path]

Yes you do

[obtuse pebble]

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[bright meadow]

idk how to do this

[timid silo]

ratio will be same equate the ratio

[bright meadow]

so 25/11 = x/10?

[timid silo]

yeah

its theorem

[bright meadow]

alright thanks]

.close

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[wheat trout]

i wanna find n

wait

[obtuse pebble]

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[hard minnow]

Can someone help me understand how my professor got this answer?

[hard minnow]

I marked in red what I’m confused on..first off, how did he get l/20= x/30…

[sage geode]

The two triangles that you have on the right are similar, right?

[hard minnow]

Not sure why they are. If I’m only given the problem without a diagram of shapes like in the example above, how would I know if the two triangles are similar?

[sage geode]

The bases (that have lengths 20 and l) are parallel

This makes the triangles similar

[hard minnow]

Why do we have two shapes though if the question is asking for a pyramid..Why do we have to use another shape to solve this problem?

[sage geode]

We're taking a cross-section of the pyramid in order to gain information about the variables that we have (in this case it gave us the relationship between l and x in the form of an equation, l/20 = x/30)

The equation itself comes from the fact that the triangles are similar

Meaning the corresponding line segments should have the same ratio

It counts height too

The smaller triangle's height is x, the bigger triangle's height is 30

The smaller triangle's base is l, the bigger triangle's base is 20

Thus l/20 = x/30

[hard minnow]

Will questions like these always come in this format? l/(number)=x/(number)?

[sage geode]

In the context of similar triangles, no, you may have one of the variables in a denominator meanwhile the other will be in a nominator

[hard minnow]

Ok..I understand how they got 2/3x..by simplifying, the why did they raise it to the 2nd power? Is there a formula I’m missing here?

[sage geode]

Do you see the square there highlighted with light blue?

Inside the pyramid

[hard minnow]

Yes

[hard minnow]

I do

[sage geode]

It's side length is l

Meaning its area should be l^2

And, since we know l = 2x/3, we get that the area is (2x/3)^2

[sage geode]

Basically, the area is to add up all of the areas of those squares inside the pyramid

To get the volume

[hard minnow]

How do we know this from the problem? Like, what if the depiction of the labeled shape isn’t available?

[sage geode]

It is helpful to draw a diagram if you are not given one

[hard minnow]

How would I know how to draw this diagram from the problem? For example, this question,.

[sage geode]

I don't think you should have trouble with drawing a pyramid, I guess you are asking how would you come up with dividing the pyramid into those squares that we talked about?

[hard minnow]

Yes

[sage geode]

Generally you have to think about what would be the way to divide the volume into the simpler possible shapes

[sage geode]

In this case, since the base of a pyramid is a square, we can take the horizontal cross-sections of the pyramid

Because they are simply squares

And it is easy to deal with squares

[sage geode]

If you took any vertical cross-sections

Then you would have a triangle whose base and height will be really hard to determine

So it is basically better to go with squares

Especially consider the fact that all of those squares will be similar to the base of the pyramid

Let's say you were asked to do the same problem except with a cone with height H and the radius of the base circle R

How would you divide the cone?

[hard minnow]

into a circle at it’s base?

[sage geode]

Yes

In fact, you will be able to calculate the radius of a circle at some height h (measuring from the top of the cone) using the same method involving similarity of triangles

You should get that r/h = R/H

Meaning r = R * h/H

[warm shale]

A Lonely Bean

[hard minnow]

This is difficult to understand, we great r/h=R/H why?

[sage geode]

One moment, let me draw

From a side, a cone looks like this, right?

[hard minnow]

yes, like a triangle

[sage geode]

Okay, let's say we cut it somewhere and got a circle

[hard minnow]

ok

[sage geode]

The colours do not match, there are no labels, I know, I am doing it in desmos

[hard minnow]

No problem

[sage geode]

Alright, you see that these triangles are similar, right?

Because the orange and cyan line segments are parallel

[hard minnow]

As in the two triangles that make up the full triangle in that graph?

[sage geode]

Yes

One inside another

[hard minnow]

Yes, I see that the two triangles that make up the overall triangle

I see that they are similar.

[sage geode]

Just like we did in the previous problem, let us consider what the heights and bases of these triangles are

The bigger triangle's height is H (the cone's height), the smaller triangle's height is h

The bigger triangle's base is 2R (because it is the diameter of the base circle of the cone), the smaller triangle's base is 2r

[hard minnow]

The bigger triangle height which is H is NOT the triangle you were just showing me from desmos, right? You're just speaking of the cone that was the original problem?

and the smaller triangle height which is h is the one you were showing me from desmos?

[sage geode]

They both have a height of H

The cone and the bigger triangle

[sage geode]

By the bigger triangle I mean the one with the cyan base and by the smaller triangle I mean the one with the orange base

In the desmos diagram

[hard minnow]

AHh okay..

I notice we always capitalize variables associated with bigger shape..is this on purpose?

[sage geode]

It feels more natural to do that, yeah

But it isn't necessary, you are free to call your variables however you want

Doing it this way will just not let you forget about what stuff are you talking about

Continuing, since the triangles are similar, we must have 2r/2R = h/H

So r/R = h/H and r = R * h/H

[hard minnow]

I’m trying to understand this equation…why is it that since the triangles are similar we derive this equation?

[sage geode]

It is the definition of similarity for triangles

The ratio of corresponding line segments (heights, bases, medians, so on) are equal

[hard minnow]

Thank you. I just think I need more practice with this..for a problem like this…wouldn’t the process be similar to when we integrate over x-axis? Wouldn’t I just set both equations equal to each other to find the bounds and then subtract the expressions?

Like this:

[sage geode]

I think here you need to integrave over y

Yup, that way

[hard minnow]

Then I just find the derivative of the new expression and plug in?

[sage geode]

It's supposed to be (-3 + y^2) - y + y^2 though

[sage geode]

Antiderivative*

Oh

[sage geode]

y - y^2 - (-3 + y^2) actually

You have to subtract the small from the big in order to get something positive

Hm, wait

I may be wrong

[hard minnow]

I’m always getting confused on what turns positive/negative on this step

[sage geode]

Ah okay nvm that is correct

[hard minnow]

The one you mentioned before?

[sage geode]

Actually yeah do y - y^2 - (-3 + y^2)

I kept confusing >= and <=

[hard minnow]

Why does y turn positive? I might be looking at something wrong here

[sage geode]

Generally the integrand should be |f(x) - g(x)|

[hard minnow]

Yeah

[sage geode]

But here we have y - y^2 and -3 + y^2

And y - y^2 >= -3 + y^2 (for -1 <= y <= 3/2, which is the interval that we want)

[sage geode]

So the integral becomes this - that

y - y^2 - (-3 + y^2)

[hard minnow]

Is that a rule? We subtract the 2nd expression by first expression?

[sage geode]

y - y^2 >= -3 + y^2 means that y - y^2 - (-3 + y^2) >= 0, right?

By moving everything to the left

[hard minnow]

Got it

This would be our anti derivative for expression?

[sage geode]

Yup, you could have also expanded the parenthesis before doing that

[hard minnow]

Now all I should do is substitute 3/2 and -1 in for y?

[sage geode]

Yup

Calculate them and subtract

[hard minnow]

So this final expression to compute?

[hard minnow]

Or did I mess anything up on this expression?

[sage geode]

Looks fine

[hard minnow]

Thanks…I have one final final question before I head out..it’s on a problem integrating over the y axis…

They said I’d have two integrals…

Is that because the solution to cos pheta =0.5 has two solutions?

My solutions are up above in red…if I have two solutions after setting the expressions equal to each other…how do I solve?

[obtuse pebble]

@hard minnow Has your question been resolved?

[hard minnow]

<@&286206848099549185>

[timid silo]

shift the origin and solve

[hard minnow]

How could I solve something like this?

[hard minnow]

What do you mean?

[timid silo]

you can change the equation such that the point(1,0.5) becomes the origin and then integrate

wait I am providing the solution

[hard minnow]

I don’t understand.. could you elaborate just a little?

[sage geode]

Because there's one subinterval where |cos(theta) - 0.5| = cos(theta) - 0.5 and there's another subinterval where |cos(theta) - 0.5| = 0.5 - cos(theta)

[sage geode]

The point (1, 0.5) doesn't lie on y = cos graph though

[timid silo]

just integrate cos x-0.5

[timid silo]

I was not trying to say that

[sage geode]

[ \int_0^\pi |\cos{x} - 0.5| dx = \int_0^\frac{\pi}{3} (\cos{x} - 0.5) dx + \int_\frac{\pi}{3}^\pi (0.5 - \cos{x}) dx ]

[warm shale]

A Lonely Bean

[timid silo]

integrate y=cosx -0.5 within the limits provided

Is the answer -pi/2?

[hard minnow]

So how would I solve this in the manner that I solved the last problem?

[sage geode]

Just do this

[sage geode]

Answer this btw

[hard minnow]

How did you get pi/3 as the upper bound?

[sage geode]

That's the intersection point

cos(pi/3) = 0.5

[obtuse pebble]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[timid silo]

.reopen

[obtuse pebble]

✅

[timid silo]

I think that the intersection point does not matter

[hard minnow]

How did you get it? By setting cos pheta = 0.5?

[sage geode]

Yes

What is the answer sheet saying?

[hard minnow]

There is none..that’s why I’m constantly here ☹️

[hard minnow]

I got two solutions when I set those terms equal to each other though…

[sage geode]

You are asked to calculate the shaded area, right?

[hard minnow]

This is what it is asking me:

[sage geode]

Yeah go this way

[hard minnow]

I think I’ve seen something similar... Let me ask, why does one integral have bounds 0 to pi/3 and the other has bounds pi/3 to pi?

[sage geode]

Because for x between 0 and pi/3 we have |cos(x) - 0.5| = cos(x) - 0.5 and for x between pi/3 and pi we have |cos(x) - 0.5| = 0.5 - cos(x)

[obtuse pebble]

@hard minnow Has your question been resolved?

[hard minnow]

So one value results in a negative solution when substituted and another results in a positive solution when substituted in?

[sage geode]

Yeah

[hard minnow]

We take anti derivative of these expressions?