help-39

[gleaming prism]

It’ll be x/2+14+x/3(x/2+14)+14

The 14 remains outside the brackets cause he spends another 14$ after he spends 1/3 of his remaining money @olive rover

Is it okay??

[olive rover]

I just tried solving it and got 296

[pearl pond]

@olive rover Has your question been resolved?

[late moon]

@olive rover hey you're still there?

[olive rover]

Yes I’m stuck between the answers 112 and 70

[late moon]

Alright

Wanna go through the problem again one more time together?

[olive rover]

Yeah we can

[late moon]

Alright

Let's call the whole budget "T"

Okay?

Now it says that when Mr. Howell entered the first store, he paid half of his money (aka T) and added $14 on top of that.

So that means $\frac T2 + 14$

[jolly parrot]

VulcanOne

[late moon]

That covers the first part

Good so far?

@olive rover

[olive rover]

Yup I have that down for his first store visit

[late moon]

Alright

I need you to focus with me rn

Now after he paid that, what does he have left?

In terms of T?

[olive rover]

1/3 of his remaining budget?

[late moon]

Nope

So like

After the first store visit

He went in with T dollars and paid T/2 + 14 dollars

How much does he have now?

[olive rover]

28? Or 7?

[late moon]

Hmm

[olive rover]

Should I be equaling it to 0

[late moon]

Alright let's approach it from somewhere else

You have $100 dollars

You went to a store and the item you want was half your budget.

And you decided to buy it with an accessory that costed $5

How much are you left with now?

[olive rover]

45$

[late moon]

Yep

[olive rover]

Sorry I was away from my phone

[late moon]

No worries :)

So you did 100/2 first and got 50, then added 5 to it and lastly, you did 100-55 right?

[olive rover]

Yeah

[late moon]

Alrighty

Now let's go back to our example

Mr Howell had a budget of T dollars

He spent T/2 + 14 dollars in the first store

How much remains?

[olive rover]

I got -28

[late moon]

You should still have T in the mix

Hmm

Keep in mind how you got $45 earlier

100 -55

Same concept here

[olive rover]

But last time I knew the budget and now idk what I’m subtracting by

[karmic hound]

The budget is T

[late moon]

We called the whole budget T

It has a value, but we will work to find it

For now, it is a placeholder

So we act as if we know it so that we write our equation

[olive rover]

T-14/2

[late moon]

You're a bit confused but you got the spirit

[late moon]

This answer is close. It just needs a bit of rearranging.

How did you reach that btw?

[olive rover]

I just thought subtracting the 14 and added to the top of the fraction

Is it T+28/2

[late moon]

Nope remember

T is the whole

And the payment was half the budget and a bit more

So the remaining should be lower than the half by a bit

Wait I thought of pizzas just now

1 min

Alright let's think of pizzas instead

This is Mr Howell's whole pizza

We represent this pizza with the letter T

Okay?

Now Mr Howell ate this pizza on 2 days

On the first day, he ate half of the pizza and a bit more

So like, T/2 + 14

So first he ate half

Then ate a bit more

What remains is less than half

Right?

[olive rover]

Yes

[late moon]

So looking at our example

The first day, he ate T/2 and ate 14 more after that

And the whole pizza was T

So T - (T/2 + 14) should be the remaining portion right?

$T - \left( \frac T2 + 14\right)$

[jolly parrot]

VulcanOne

[late moon]

@olive rover

Did I lose you?

[olive rover]

Sorry I was trying to try it in my own and I set it up like this

[late moon]

It's almost correct that way

I just want you to understand where it is wrong

We agreed that $\frac 12 x + 14$ is slightly bigger than half right?

[jolly parrot]

VulcanOne

[olive rover]

Yes since we’re adding the 14

[late moon]

Yep

Now what happens when we subtract that from the entire whole?

[olive rover]

1/2x-14?

[late moon]

Exactly

Same half minus that 14

When he hate half the pizza and a bit more, what remains is the other half but missing that bit

[olive rover]

Correct

[late moon]

Now when he goes to the second store, he spends one third of the remaining money, and spends 14 more on top of it

So like

Exactly what you wrote but instead it is -14 inside the brackets with 1/2 x

So in the end he has no more money

Now this begs the question

If we added the first store, and the second store, wouldn't that give us the entire budget?

[olive rover]

So combine both equations and equal is to zero?

[late moon]

Why 0?

Let's go over it

So like

Let's call it the "spending" equation

So before any store trips, he spent nothing right?

[olive rover]

Yeah

[late moon]

So at first, he had 0 spending

Which means he had the entire budget

T

[olive rover]

Yup

[late moon]

Now after that, he spent T/2 + 14

So that he had T/2 -14 remaining

[olive rover]

Yeah

[late moon]

Lastly, he went and spent (x/2 -14)/3 +14

And he ends up with 0 money

Now 0 money means he spent how much again?

[olive rover]

7?

[late moon]

When you have a budget of 100$ and you come home with 0$, how much did you spend?

[olive rover]

Oh 100$

[late moon]

Yep

[late moon]

So returning to this question, can you answer it?

[olive rover]

No… like am I just subtracting instead of adding now

[late moon]

Hmm

Shouldn't the spending be equal to what he started with?

Let's have an example on that

[olive rover]

Am I setting the equation equal to itself?

[late moon]

Nope

Let's have an example on the side

So like

You are walking with your best friend

And then you feel hungry so you tell your friend that you wanna buy something to eat

So your friend gives your money and says: "Please buy me a bag of chips and a coca cola"

The money is totaling $10

So you go buy your stuff and buy the bag of chips which costs $6 and the coca cola costs $4

And you come back with 0 friend's money

So you give your friend the coca cola and the bag of chips and your friend asks: "how much does each item cost?"

[olive rover]

6+4=10

[late moon]

Exactly

Let's apply the same logic to our question

So Mr Howell spent so far x/2 + 14 and (x/2 -14)/3 + 14

$\frac x2 + 14$ and $\frac 13 \left( \frac x2 -14\right) + 14$

[olive rover]

So I set my equation equal to x since that’s the total budget?

[jolly parrot]

VulcanOne

[late moon]

Yepp

You got the key to the whole question now

All that remains is the algebra

@olive rover did you get there?

[olive rover]

When I solve it is that my answer?

[late moon]

Yep

The question wants how much he had when he entered the store so basically how much budget he started with

Aka the whole budget

[olive rover]

I got x=70!!

[coarse dawn]

355044260642859198243475901411974413130137600000000?

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[fringe juniper]

could anyone help me with this?

[spiral sierra]

is this a test?

[fringe juniper]

no its my hw

i had like 5 attempts for this question and i’m on my last one but i still dont get it

[feral sedge]

Where does csc(x) have discontinuities and what type are they

[pearl pond]

@fringe juniper Has your question been resolved?

[pearl pond]

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[agile needle]

My goal is to prove that for all real numbers $a$,
$$\lim_{x \to a}{|x|} = |a|$$
In other words, that the absolute value function is continuous.
I imagined that I break into three cases. Case 1: $a> 0$; Case 2: $a = 0$, Case 3: $a < 0$. In case 1 and 3, the absolute value function could be replaced by x or -x respectively. However, how do I justify this? I want to say “since $x > 0$, then $|x| = x$”, but how do I establish that $x > 0$?

[jolly parrot]

Rmoney

[hollow cobalt]

You can always restrict the values of x by choosing delta correctly

[static bloom]

If you proof something by cases, you do not need to justify your cases.

Instead you need to proof that the cases are exhaustive, i.e. that there is at least one case that is always applicable.

Instead of working with cases I also want to suggest using triangle inequality. I think this would result in a much more concise proof but if you prefer doing it with cases, you can also do that.

[agile needle]

Oh that’s an interesting idea. I will think about how to do it with the triangle inequality

[agile needle]

@hollow cobalt I think I comprehend what you mean

[hollow cobalt]

Yeah, in the general case you can choose delta to be min(|a|, somethingelse)

Which will guarantee that x > 0 or x < 0 depends on a

[agile needle]

okay thanks all. That answers my question

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[fossil lodge]

If $x$ is a real number, then either $x$ is an integer or there is an integer $n$ such that $n <x < n+1$

[jolly parrot]

KN

[fossil lodge]

Assuming there is no integer n such that n <x < n+1.

[pearl pond]

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[fossil lodge]

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[lime salmon]

can i get help with #3 please

[lime salmon]

im looking back at my notes and i never did a problem like this, i need a walkthrough on it

[unborn abyss]

if epsilon is 0.1 then that means we need an equation like

$|\blue{1x - 4} - \orange{1}| < 0.1$

[jolly parrot]

hayley!

[unborn abyss]

see if you can turn that into a range for x

[lime salmon]

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[grand perch]

@unborn abyss

i think one of the issues here is that unlike with addition and subtraction, where you start with two things of the same unit type (volume, mass, time, etc) and end up with something of that same type, the result in multiplication is a different type than what you started with. If you're thinking of multiplying two measuring cups partially full of water and arriving at a new measuring cup, then yeah that doesn't make any physical sense.

but if you think about a measuring cup that is 2/3 of a liter, and filling it up one and a half times, then it might make more sense that you end up with exactly one liter. (the units here are liters / cup, and cups)

Sorry, I had to do some stuff so I couldn't respond. You said this a moment ago.

You say "but if you think about a measuring cup that is 2/3 of a liter, and filling it up one and a half times, then it might make more sense that you end up with exactly one liter."

I guess I'm still confused on this and my brain just breaks.

If I have a measuring cup that's nearly full, then fill it up one and a half times, my head thinks I'm getting 1.5 times the volume of the cup (3 liters) and then adding that to the cup that's already 2/3 filled, which would obviously overflow.

Maybe I'm just making a very basic mistake and conflating addition and multiplication.

[pearl pond]

@grand perch Has your question been resolved?

[unborn abyss]

you are conflating addition and multiplication, yes;

remember that I said to think about a measuring cup that is 2/3 of a liter, like a small measuring cup that only carries 666 mL

if you were to take that cup, fill it up once and dump it into an urn, and then fill it up another halfway and dump it into the same urn, you'd end up with 1 liter of water in that urn

[grand perch]

Okay, I think this mostly fixed my issue. Thanks 🙂

[pearl pond]

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[hard locust]

im very confused on how too find the lower limit of this graph, its a rectangular hyperbole so its not on a line so i cant start from 0 like usual. The question for the problem is, "For the following exercises, draw the region bounded by the curves. Then, use the disk method to find the volume when the region is rotated around the x-axis."

[jaunty cedar]

Start from the intersection of the rectangle's top side and the function.

y = 3 and y = 1/x.

[hard locust]

Instead of going x to x we are going y to y?

[jaunty cedar]

No, simply find the x at which you start integrating.

So find the first intersection between the 2 graphs.

That's the point where you start integrating, right?

[hard locust]

So I find the inverse of 1/x?

[jaunty cedar]

Solving 3 = 1/x gives the intersection between y = 3 and y = 1/x, right? That will give you your starting point for the integration.

[hard locust]

Got x = 1/3

[pearl pond]

@hard locust Has your question been resolved?

[pearl pond]

@hard locust Has your question been resolved?

[pearl pond]

@hard locust Has your question been resolved?

[hard locust]

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[bleak cipher]

This is a product + quotient rule hw sheet. What is this 😭

[bleak cipher]

(ap calc ab)

[tranquil fjord]

What are you trying to do

[bleak cipher]

first derivative

[tranquil fjord]

Do you know what the quotient rule is

[bleak cipher]

yeah

if we're going with respect to x what do i do w the other variables?

[hybrid basin]

Haven't heard this course in a long time lol

[tranquil fjord]

If A is a variable isn’t the derivative of that just 0

[bleak cipher]

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[tribal compass]

Say I have an Image that has been transformed into the quadrilateral where ABCD are arbitrary points forming a convex quadrilateral.
If I put in a random coordinate, how can I calculate if its inside the quadrilateral, and if it is inside, what the Image coordinate would be?

[pearl pond]

@tribal compass Has your question been resolved?

[pearl pond]

@tribal compass Has your question been resolved?

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[toxic cliff]

I'm suppose to find a and b

[plush bramble]

What are the conditions on f

[toxic cliff]

I'm not sure what you mean, but it says to make f continuous everywhere

I'm not sure what I did wrong. It doesn't take my a value

for the systems of equations solving thing at the bottom, I set lim x ->2 to 3 when it should have been 1

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[urban tangle]

need help with quotient rule problems

[toxic lichen]

what's troubling you with this

do you have the quotient rule itself on hand?

[urban tangle]

nope

got the quotient rule

just don’t know what to do after i get this

it wants it simplified

[urban tangle]

^

f’g-g’f

over g2

[toxic lichen]

g^2.

[urban tangle]

g^2*

[toxic lichen]

simplify as it tells you to

[urban tangle]

i dunno how

[toxic lichen]

just carefully expand

ok, how does (1)(1-x^2) simplify?

[urban tangle]

just 1-x^2?

[toxic lichen]

yes of course.

and how does -(-2x)(x) simplify?

[urban tangle]

+2x^2?

i mean +2x

wait

idk actually

[midnight haven]

$x * x \neq x$

[jolly parrot]

Brandon H#1125

[urban tangle]

so just-2

[toxic lichen]

2x^2 is correct

[urban tangle]

how?

[midnight haven]

$x * x \neq 1$

[jolly parrot]

Brandon H#1125

[toxic lichen]

x * x = x^2

[urban tangle]

oh

ok

[toxic lichen]

1 - x^2 + 2x^2

how can you now simplify this

[urban tangle]

no idea

[toxic lichen]

do you know basic algebra

[urban tangle]

nope lmao

[toxic lichen]

bruh what

how tf

[urban tangle]

yup.

[toxic lichen]

you need to be reviewing/leerIng algebra, not calculus

this is such "run before crawling" energy

[urban tangle]

adhd autism girl power

[toxic lichen]

correlation dubious

but ok

ok let's try this one thing:
what does 6a + 11a simplify to

[urban tangle]

17a

[toxic lichen]

ok so this does come to you right

now try (-1)x^2 + 2x^2

[urban tangle]

all that but negative

[toxic lichen]

no

let's try again

2x^2 - x^2

[urban tangle]

uh

4x- 1^2

[toxic lichen]

no...

[urban tangle]

WAIR

4(1)-(1)

[toxic lichen]

no

what happens when you have two apples and take away one apple?

[urban tangle]

damn

[toxic lichen]

what do you have left?

[urban tangle]

uno

[toxic lichen]

you have one apple left

[urban tangle]

correct

[toxic lichen]

2 🍎 - 🍎 = 🍎

you need to be more precise about that

[urban tangle]

I LOVE ADVANCED CALUCLUS ⁉️⁉️🔥🔥

[toxic lichen]

because this is the same shit

and no it is not calculus it is algebra which you are missing

2x^2 - x^2 = x^2

the numerator ends up as 1+x^2 and at that point there is nothing else to do.

[urban tangle]

2-1 = 1?

[toxic lichen]

yes obv.

[urban tangle]

dope

forgor x^2 is 1

ok so the problem is

[toxic lichen]

x^2 is x^2

x^2 is not equal to 1

[urban tangle]

1-1 + 2x^2

yea but if there’s nothing to plug into it then it’s one

so it’s

1-1+2(1)

[toxic lichen]

no!!!

no!!!!

the numerator ends up as 1+x^2 !!!!

[urban tangle]

where duz the 2 go

also what happens to the denominator

[toxic lichen]

absolutely nothing. i am leaving it intact.

[toxic lichen]

the -x^2 and +2x^2 combine into +x^2

[urban tangle]

so 1+x^2?

col

cool

[toxic lichen]

that's what i said earlier twice already

[urban tangle]

heard

how do they combine

like where do the numbers go

[full badge]

hi

[full badge]

52461026391552+x^530153835382577269253795193581582518428519348934299439941893?

[toxic lichen]

don't troll

[toxic lichen]

when you combined 6a + 11a into 17a, where did the six and the eleven go?

[urban tangle]

together

but what value is x^2

[toxic lichen]

x^2 is x^2.

well the 2 and -1 also go "together" into +1.

[urban tangle]

ok

[toxic lichen]

it is the exact same thing only with different numbers, promise...

[urban tangle]

i trust lol

i news help simplifying this now

<@&286206848099549185>

need help simplifying this

[toxic lichen]

look up "distributive law", "collecting like terms" and maybe "FOIL"

[urban tangle]

its due in like 10 minutes fr

[toxic lichen]

then you are fucked!

[urban tangle]

yup

[toxic lichen]

take the L and catch up later.

[urban tangle]

booooo

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[pearl pond]

@wind harness Has your question been resolved?

[unborn abyss]

$pq\bar r + p\bar q \bar r + \bar p q \bar r$ \
$\bar r(pq + p\bar q + \bar pq)$ \
$\bar r(p + q)$

[jolly parrot]

hayley!

[unborn abyss]

yeah seems good

[wind harness]

alright thank you!

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[toxic cliff]

sorry for all the questions, but for some reason I cannot evaluate this even though it looks simple

here is my work

[toxic lichen]

0/0 doesn't mean DNE

also $\sqrt[3]{x} - 1 \neq \sqrt[3]{x-1}$ either so your second step is formally right only by coincidence anywaay.

[jolly parrot]

Ann

[toxic lichen]

so none of this work is even remotely salvageable, sorry.

[toxic cliff]

that's fine lol

[toxic lichen]

there is a substitution here that can make your life easier: $t := \sqrt[6]{x}$ turns the limit into $$\lim_{t \to 1} \frac{t^2 - 1}{t^3 - 1}.$$

[jolly parrot]

Ann

[toxic cliff]

t? is that like x

why didn't u just put x? just curious idk if there's any significant reason but I always notice that t is used for like substitution I think

but isn't t suppose to be 1

how can you substitute t for the square root thing if it's approaching 1

[toxic lichen]

it is a different variable, but there's no sacred meaning to my choice of the letter t over the other 24 options i had.

[toxic lichen]

what do you mean?

[toxic cliff]

like when you're evaluating a limit, don't you normally plug in what it is approaching for the variable

[toxic lichen]

also i am putting t as the sixth root of x.

and i'm not evaluating anything.

i'm making a substitution.

this turns your limit into a different limit that's equal to the old (in the strong sense -- meaning that if the old limit didn't exist, neither will this one) but is hopefully easier to evaluate

[toxic cliff]

ohh

I'm still a bit confused lol. How did you get the t^2?

and the t^3

[toxic lichen]

$t = x^{1/6}$

[jolly parrot]

Ann

[toxic lichen]

so $\sqrt[3]{x} = x^{1/3} = x^{\frac{1}{6} \cdot 2}$ and likewise $\sqrt{x} = x^{1/2} = x^{\frac16 \cdot 3}$

[jolly parrot]

Ann

[pearl pond]

@toxic cliff Has your question been resolved?

[pearl pond]

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[neon meteor]

One circle is moving towards another circle with velocity v, calculate velocity with which the upper intersection moves give that the distance between the two centres is a and both have the same radius r.

[neon meteor]

Here's what I got

,tex $\frac{1}{4}\cdot\frac{av}{\sqrt{r^{2}-\frac{a^{2}}{4}}}$

[jolly parrot]

ℝage-EMILY

[neon meteor]

I did this using related rates but am not sure if this is correct or if I did it correctly.

[pearl pond]

@neon meteor Has your question been resolved?

[pearl pond]

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[radiant forge]

I am stuck on a question for functions, I just don't know how to do it

[nimble wing]

i would try substituting in the values

[regal herald]

starting with y=a(x-h)^2+k, the vertex tells you the values of h and k

you can then sub in point (-5,3) in order to determine a

[radiant forge]

okay I will try

[radiant forge]

I have to go eat so I will do it afther I finished

thank you for your help

[pearl pond]

@radiant forge Has your question been resolved?

[pearl pond]

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[pearl pond]

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[muted fable]

Bit confused by the first 2 questions, is it asking for the value at 1 on the x or y axis?

[rain mulch]

it's asking if you input x = 1 into g(x), what does g(x) equal

same for the second question but instead of x = 1 it's x = 0

[hybrid basin]

Just because there is a hole in the function, that doesn't mean its value is absent

[muted fable]

hold up im not gonna lie im lost

how do i solve for that without a given equation

OH nvm

so at x=1 the value is 3 and at x=0 its 1?

.close

[pearl pond]

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[unreal flare]

the problem is integral( (3x-1)/(x^2 -4x +29) )dx

[unreal flare]

I'm not sure what I was expecting when I tried to do the partial fractions

I'm just really confused rn

[pearl pond]

@unreal flare Has your question been resolved?

[hasty eagle]

still need help with this?

[unreal flare]

yes please

[hasty eagle]

ok

so you wanna get to a point where you can use a u-subsitution

i tried partial fractions on this didn't work

but u-sub did

[unreal flare]

oh interesting

[hasty eagle]

maybe i did the partial fractions wrong hahaha but i'm pretty confident with the u-sub

[unreal flare]

I don't have to use partial for this one so it's ok I think

[hasty eagle]

ok. then plain ol' u-sub then

[unreal flare]

for u sub maybe I could separate it like 3x/(x^2 -4x +29) - 1/(x^2 -4x +29)

[hasty eagle]

ah that's tempting

but how would you choose u

[unreal flare]

you're right maybe I should make u = 3x -1

[hasty eagle]

sure, but the bottom is x^2

[unreal flare]

true then u = (x^2 -4x +29)

[hasty eagle]

Exactly!

Now the problem is that -4x

how do you think you should deal with that

[unreal flare]

u = (x^2 -4x +29)
du = 2x -4
du +4 = 2x dx
(1/2) (du +4) = x dx ???

err

no

[hasty eagle]

can't do that

[unreal flare]

yeah

mmm

let me think

[hasty eagle]

sure thing

but you're on the right track. you want to use that choice of u

[unreal flare]

maybe if there is a way to make the numerator 2x -4

[hasty eagle]

Yes!

[unreal flare]

:O

[hasty eagle]

How would you do that?

[unreal flare]

no idea 💀

uhhh

a fraction

[hasty eagle]

Haha here's a hint

add 0

[unreal flare]

add 0?

[hasty eagle]

you don't want to add JUST 0

you want to write 0 in a fancy way

a - a is still 0

So you want to add something and subtract at the same time

Something that will that 2x-4 show up

[unreal flare]

I have a feeling

but I need a moment to think

[hasty eagle]

Yeah, no rush

[unreal flare]

ahh I'm not sure what to do

oh wait actually...

no..

I'm not sure

can I have another hint lol

[hasty eagle]

$3\frac{2}{2}x-1 + (5-5)$

[jolly parrot]

TooManyCooks

[unreal flare]

😨

[hasty eagle]

Hahaha too much ?

[unreal flare]

I'm just a bit confused how you would think of that 😭

[hasty eagle]

Like I said, you want to get 2x-4

[unreal flare]

yesyes

[hasty eagle]

I literally just forced it to show up by adding things

for example, that -1

is by itself

what would I add to -1 to get 4?

[unreal flare]

5

ohhh

5-5

[hasty eagle]

Right, so you add AND subtract 5

which is 0

classic math trick

[unreal flare]

and we want 2x, so you multiply by 2/2

[hasty eagle]

Yes! You got it!

[unreal flare]

mm I'm actually a bit confused by this one

I know I said it but I didn't get it

I got the 5

[hasty eagle]

No, that's fine

What's wrong

[unreal flare]

I'm not sure how that gets you 2x

[hasty eagle]

Well, I multiplied by 1

Adding 0 and multiplying by 1 doesn't change the value of the function

[unreal flare]

true 💀

[hasty eagle]

Since we have 3x, i inserted a 2/2

[unreal flare]

ok heard

[hasty eagle]

You can group things up, so you get (3/2) (2x)

[unreal flare]

that's what I was thinking sick then

I'm gonna look at this for a sec

[hasty eagle]

Sure, take your time

[unreal flare]

can you take out the 3/2 as a constant?

[hasty eagle]

Yes!

But only on the first term

I'm afraid you're stuck with the 5 on the other term

[unreal flare]

so I have ((3/2)(2x) -4 -5)/(x^2 -4x +29)

would I be able to u sub from here even tho the 3/2 is connected to the 2x?

[hasty eagle]

only on the first term

[unreal flare]

could you possibly explain what you mean by that

I'm v sorry

[hasty eagle]

the u-sub you're doing right now is only useful on the first term

the x-term

you still have the term with 5

that's a different problem

but we can deal with that later

[unreal flare]

OH

i think i get it

like this

shit it's sideways

[hasty eagle]

,rcw

[hasty eagle]

by the way, the substitution i gave you earlier (the trick with the +0 and x1) isn't the actual answer

that was just a hint 😅

just making sure we're clear

anyway that's right

[hasty eagle]

except for a sign

[unreal flare]

oh wait it's right tho

[hasty eagle]

https://tenor.com/view/rock-one-eyebrow-raised-rock-staring-the-rock-gif-22113367

[unreal flare]

oh?

was it the negative bc i put it on the outside

[hasty eagle]

The hint I gave you earlier is VERY close to the true one

you just need to make a slight adjustment

but that should be a +5

[unreal flare]

are you sure

(i know nothing)

[hasty eagle]

Let me see what you have so far. we can work through it

LIke how did you do the trick part

[unreal flare]

i'll send a photo i just p much like added and distributed

(sorry it's dark my roommate is asleep)

[hasty eagle]

https://tenor.com/view/calculation-math-hangover-allen-zach-galifianakis-gif-6219070

[unreal flare]

SORRY I KNOW ITS A MESS

i failed the last problem too pls ignore that

[hasty eagle]

You had a -1 + 5 - 5

and you somehow got -4 -5

[unreal flare]

OH

[hasty eagle]

yeah it's that one

[unreal flare]

MY BAD

[hasty eagle]

all good

[unreal flare]

okok

so it's the +5 got it all set

ok i'm gonna u sub this brb

[hasty eagle]

Yeah the proper grouping was -6 + 5. you got it

[unreal flare]

ok so the first integral is like (3/2)ln|x^2 -4x +29|

[hasty eagle]

That's right

On to the next one

[unreal flare]

for the second on i thought like take the 5 out as a constant for simplification and then maybe

idk if this works algebraically can you like separate the denominator that's not allowed right

i'm trying to think outside of the box rn 😭

[hasty eagle]

Here's a hint. Arctan

[unreal flare]

HEARD OK not me just remembering how to complete the square

ok small question

[hasty eagle]

Sure

[unreal flare]

i can't leave the 25 there but i don't know what to do about it

,rcw

[hasty eagle]

What do you think?

[unreal flare]

i feel like i could use that 5

but i've been staring at this problem for an hour so i've lost all concept of algebra

if i use the 5 uhhhh

[hasty eagle]

How about you factor 25 from the denominator?

[unreal flare]

😶

but there is no 25 in front of the u so should i multiply the top by 25 maybe possibly

[hasty eagle]

Right. If only there's a way to substitute things...

😉

[unreal flare]

PLS HAHAHA

[hasty eagle]

So you have x^2

25 = 5^2

Think you can exploit that?

[unreal flare]

oh! I think so

one moment

so it's like 1/(u^2 + 5^2) right so

there's a 5 on top

[hasty eagle]

Do you have an integral formula for that?

[unreal flare]

I mean there's 1/(x^2 +1)

[hasty eagle]

Ok so how can you get that 1?

[unreal flare]

exactly yes so

multiply the top by another 5 maybe

but

you can't just do that so

I know I'm missing something I'm sorry my bad

[hasty eagle]

No that's fine, that's why we're doing this

My suggestion is to factor out 25 from the denominator

That will give you 1

Now all you need to do is the z^2 part

[unreal flare]

but how can you do that when there is no 25 in front of the u?

[hasty eagle]

Multiply by 1 in a fancy way

The same trick from earlier

[unreal flare]

25/25

[hasty eagle]

Yes!

[unreal flare]

OK

like this right

[hasty eagle]

Not really the grouping I had in mind

How about this

Factor x from (1+x)

What do I multiply x with that gives me 1+x

[unreal flare]

1/1

wait

[hasty eagle]

1*x = 1+x?

[unreal flare]

nono

one sec I got this

pls don't give up on me lol

[hasty eagle]

I'm not

[unreal flare]

1 + 1/x maybe

x(1 + 1/x) = x +1

[hasty eagle]

Yes exactly

Use that same trick when I say factor 25 from the denominator

[unreal flare]

ok

25(((u^2)/25) + 1/25) = u^2 + 1

[hasty eagle]

Wait what. What's up with the right side

[unreal flare]

😨

u squared + 1

[hasty eagle]

I thought you had u^2 +25

[unreal flare]

I DO sorry ok one sec

[hasty eagle]

No worries

[hasty eagle]

Great

Now what

[unreal flare]

,rcw

[unreal flare]

i can take the 1/25 out as a constant

[hasty eagle]

What about the integral?

[unreal flare]

since d/dx arctanx = 1/(x^2 + 1) dx
i think the integral of 1/((1/25)u^2 +1) is

25arctanu maybe i think

[hasty eagle]

Do another sub

[unreal flare]

ok bet

[hasty eagle]

Maybe v this time 😂

[unreal flare]

LOL

so v = (u^2)/25

[hasty eagle]

Not quite

[unreal flare]

v = 1/25 ?

[hasty eagle]

That doesn't have u

[unreal flare]

i'm dying as we speak

[hasty eagle]

Hint: 25 = 5^2

[unreal flare]

withering away

[hasty eagle]

You have u^2 / 5^2

[unreal flare]

i do

u/5

[hasty eagle]

Ok good

[unreal flare]

oh

ohhhh this workkkkskssss

it's all coming together now...

i did it my keyboard just isn't working

thank you sm for your patience

[hasty eagle]

So what's the final answer?

[hasty eagle]

uh

that's not what u is

[unreal flare]

wait wrong u

i had 2 u's

my bad

[hasty eagle]

yes

[unreal flare]

x-2

[hasty eagle]

might wanna change that when you rewrite it

[hasty eagle]

Cool. Also, fun fact, the quadratic inside the log has no real roots

so it never touches 0

which means the abs value sign is not necessary

[unreal flare]

that's actually v cool

[hasty eagle]

Aight. Feel free to close when you're done

[unreal flare]

thank you

i love your username

.close

[pearl pond]

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• Show your work, and if possible, explain where you are stuck.
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[midnight haven]

I’m not sure how to solve this

Like I need someone to explain everything to me step by step in full detail, I’d appreciate it if you drew it out aswell 😭

[tacit mulch]

you're basically just filling in a blank. v is just a missing number

do you know what that 2 means?

above the v

[midnight haven]

Yes

That’s an exponent

So I’d multiply v x v

Oh my god

[tacit mulch]

yeah, what number would v be?

[midnight haven]

It would be 3 but

How would I

[tacit mulch]

there's a second solution too

[midnight haven]

Yea what’s the second solution

There’s only one answer yea?

[tacit mulch]

no, there are two

[midnight haven]

Wait

So it would maybe be the opposite

And I’d out

Put

Square root of 9?

[tacit mulch]

well, but that's just 3

[midnight haven]

Which equals 3^2

[midnight haven]

3 x 3 equals nine

[tacit mulch]

right, which is why 3 is the square root of 9

and why 3 is one of your solutions

but there's a different number too

remember (negative number) x (negative number) = (positive number)

[midnight haven]

OH

-3

[tacit mulch]

yep

[midnight haven]

IS ALSO AN OPTION

IDDINT REALIZE

[tacit mulch]

3^2 = 9
(-3)^2 = 9

[midnight haven]

yes bcs a negative x a negative equals a positive

[tacit mulch]

yep 👍

[midnight haven]

Can I add another question

I’m tryna get through this whole lesson

I’m 30 minutes

[tacit mulch]

sure

[midnight haven]

Whoops wrong one

[frosty bolt]

√x² = |x| and not just x

[midnight haven]

I’m confused

Ok so tgis

Tgis i dont really get

[tacit mulch]

so what if I said

I have two numbers

and when I multiply them, I get 0

what can you tell me about my two numbers?

[midnight haven]

That

Both of them multipled

Equals 0?..

[tacit mulch]

I think you can tell me a bit more than that

[midnight haven]

So you’d have to make it like 1 x 0

Or anything

X 0

[tacit mulch]

or 2x0, or 3x0, or 0x5

right, one of them always has to be 0

[midnight haven]

Let me write that down

How can I make it into a zero

Doesn’t the s need to be the same

[tacit mulch]

you've got (s+2) (s+8) = 0
That's two numbers (s+2) and (s+8), and when you multiply them you get 0

[tacit mulch]

sort of. There are two solutions

yes, s can only have one value at a time. but there are two different values you can give it that will make the equation equal 0

like how in the last one, v could be 3, or v could be -3

it's not both at the same time, but either one will make the equation true

[midnight haven]

So

Every equation like the last one will be a positive and a negative

For the two solutions

[tacit mulch]

I'm hesitant to say every time

x^2 = -1 doesn't have any real solution, for example

because x times x will always be positive, never -1

[midnight haven]

Oh yea

Because

It’s an exponent

Can’t make it a negative if the 1st number is a positive

But for this equation

Equation

[tacit mulch]

or even if the first number is negative

(positive) times (positive) = (positive)
(negative) times (negative) = (positive)

[midnight haven]

yes

[tacit mulch]

so if you have v^2 = (negative), that's not possible

[midnight haven]

yes bcs

• x + = +

How d wer find the answer to the equation

[frosty bolt]

Just break them into two equations

[tacit mulch]

for the (x+2)(x+8) one, you mean?

[midnight haven]

yes

[tacit mulch]

so, you agreed that if we multiply two things and get 0, one of them has to be 0, right?

[midnight haven]

yes

[tacit mulch]

so you're multiplying (s+2) (s+8) and getting 0, according to the equation

which means either

s+2 = 0
or
s+8 = 0

one of them has to be equal to 0 in order for their product to equal 0

[midnight haven]

but it’s adding

[tacit mulch]

no, (s+2)(s+8) means (s+2) x (s+8)

[midnight haven]

well yea

I’m saying

Until after

We’re don’t with ()

[tacit mulch]

well

let's take s+2

[midnight haven]

Okay

[tacit mulch]

what number would s have to be in order for s+2 to equal 0?

[midnight haven]

-2

[tacit mulch]

right

so look what happens when we let s = -2

[midnight haven]

It just makes the 8 go down to a 6

[tacit mulch]

(s + 2)(s + 8)
(-2 + 2)(-2 + 8)
(0)(6)

[midnight haven]

OH

THEN IT EQUALS 0

[tacit mulch]

right, only one or the other has to equal 0, in order for the whole thing to equal 0

that's why there are two solutions

[midnight haven]

And the other solution

Would be -8

[tacit mulch]

(s + 2)(s + 8)
(-8 + 2)(-8 + 8)
(-6)(0)

yep, that checks out

[midnight haven]

Woah

[tacit mulch]

yeah lol, it's cool, right?

[midnight haven]

Very

Can I send some more?

[tacit mulch]

sure, i'll still be here for a minute

[midnight haven]

Yay ok

[tacit mulch]

ah, okay, this is similar to the last one, but we need to factor it first

[midnight haven]

Oh okay

[tacit mulch]

since we only have two terms here, maybe there's something we can factor out?

do you know what I mean by that?

[midnight haven]

no

Can you rephrase

[tacit mulch]

hm

are you comfortable with the distributive property?

[midnight haven]

yes!

[tacit mulch]

it's like that, in reverse

h^2 - 23h

what do you notice those two terms have in common?

[midnight haven]

They both have the variable, h

[tacit mulch]

yeah, so, that means we should be able to "un"distribute that h

in other words

we can rewrite

h^2 - 23h

as

h ( blank - blank)

[midnight haven]

OH YEA

REVERSE

[tacit mulch]

so like, what would go in those two blanks, so that when you distribute the h, you get h^2 - 23h

[midnight haven]

h

And

Actually

I’m not sure

[tacit mulch]

h times h is h^2, yeah

h times what is 23h?

[midnight haven]

Oh yea

[midnight haven]

23

[tacit mulch]

so then

h^2 - 23h is the same as

h(h-23)

if you distribute that h, you get the same thing again

[midnight haven]

But it’s just in a looop

[tacit mulch]

right, don't distribute it again lol

I was just saying, it's equivalent

[midnight haven]

yes

[tacit mulch]

so now we're back to a situation like the last one

h (h-23) = 0

two things multiplied = 0

h and (h-23)

[midnight haven]

So

H would equal 0

Since 0 x anything. Is 0

[tacit mulch]

yes, that's one solution

[midnight haven]

Oh yay

[tacit mulch]

h(h-23)
0(0-23)
0(-23)
0

[midnight haven]

I cant think of another solution

[tacit mulch]

remember you have two things multiplied, and the product is 0

so, if either one is 0, the whole thing is 0

h (h-23) = 0

you made the first factor (h) equal 0

but how could you make the other factor (h-23) equal 0?

[midnight haven]

By multiplying the h by each number in the ()

[tacit mulch]

nah, don't distribute

h (h-23) = 0
means either

h must equal 0

or h-23 must equal 0

[midnight haven]

OH

SO

[midnight haven]

H HERE SOUKD BE 23

[tacit mulch]

yep

0 and 23 are the solutions

[midnight haven]

well I guess that’s all

You don’t know how much. I appreciate it

Thank you so much

[tacit mulch]

no problem 👍

hope that cleared some things up

[midnight haven]

It definitely did

Thank you!

[tacit mulch]

sure thing 👍

[midnight haven]

.close

[pearl pond]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
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• Be polite and have a nice day!

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[dusk rapids]

Find the vector function parametrized in term so of $t$ that represents the curve of intersection of the paraboloid $z = 2x^2 + 4y^2$ and the cylinder $y = 3x^2$

[jolly parrot]

o.O

[dusk rapids]

It tells me the first component of the vector function, $\vec r(t) = (t, ..., ...)$

[jolly parrot]

o.O

[dusk rapids]

The first step I took was to subsitute $3x^2$ for $y$ for the paraboloid, but I got stuck after that. $$z = 2x^2 + 4y^2 = 2x^2 + 4(3x^2)^2 = 2x^2 + 36x^4 = 2x^2(1 + 18x^2)$$

[jolly parrot]

o.O

[dusk rapids]

would I substitute $x = t$, creating $z = 2t^2 + 36t^4$ and $y = 3t^2$ which gives $\vec r(t) = (t, 3t^2, 2t^2 + 36t^4)$?

[jolly parrot]

o.O

[dusk rapids]

yes, that makes a lot of sense actually, i think i get it

.close

[pearl pond]

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[midnight haven]

help-

[pearl pond]

Please don't occupy multiple help channels.

[midnight haven]

;-;

i need help\

[copper kestrel]

just send your question lol

[midnight haven]

mk

[frosty bolt]

Close one channel

[midnight haven]

so

[slate eagle]

@I❤peepee whats the issue you are facing?

[fleet sky]

dude I recommend you change your status on your profile. It goes against #rules and will likely end with a ban.

[pearl pond]

@midnight haven Has your question been resolved?

[pearl pond]

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[pearl pond]

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[manic lake]

How would you parameterized x-y+z=1?

[tulip pewter]

Yes, you may first want to change the variables to incremental coordinates as you are now working with a parameterized equation

from there you can try and set x = u. y=v, z can be a function of both u and v so z= u,v

[pearl pond]

@manic lake Has your question been resolved?

[manic lake]

Ok

.close

[pearl pond]

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[pearl pond]

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[lost umbra]

is this right

[lost umbra]

im like dying with end behavior

can someone help explain it

[coarse dawn]

That should be enough explanation

[lost umbra]

my answer looks right

[coarse dawn]

Then there you go

[pearl pond]

@lost umbra Has your question been resolved?

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[river sun]

are these correct

[river sun]

these 3 questions