#help-39

1 messages · Page 38 of 1

midnight haven
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Alright ill just use the formulas instead

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thanks

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.close

pearl pondBOT
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cedar coyote
pearl pondBOT
cedar coyote
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I am trying to save this linear diff eq

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I put it in the correct form i think

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But i am not sure this is correct because if i differentiate the right side I get an insanely long integral

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Which typically means i did something wrong lol

pearl pondBOT
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@cedar coyote Has your question been resolved?

cedar coyote
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<@&286206848099549185>

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i have asked for help on diff eq multiple time for past week and no one ever answers

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is this server not for diff eq?

sudden charm
cedar coyote
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yes

sudden charm
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line 3

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$x^{12} / x = x^{11}$

jolly parrotBOT
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Secret

sudden charm
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u wrote x^2

cedar coyote
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oh yea my bad

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ok but i would still be left with a crazy integral though

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because if we multiply everything by p(x)

cedar coyote
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then we would have 3 functions

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and i basically fell into and endless loop of doing integration by parts

sudden charm
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i think it's called DI method

cedar coyote
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ive heard of it but the reason i didnt do it

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is because it doesnt work with e

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so i didnt learn it

sudden charm
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hmm

cedar coyote
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there were some eweird restrictions

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with this trick

sudden charm
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riggt it doesn't work

cedar coyote
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i remember

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u see the dillema

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and if i do integration by parts normally

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i just am stuck integrating by parts over and over again

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and it doesnt help that online calculators or atleast the ones i use cant solve it

sudden charm
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yes lmao

cedar coyote
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but did i set up the problem correctly?

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because that was my concern

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i feel like there might be an issue with the way i set it up

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is it even a linear diff eq to begin with

sudden charm
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maybe u can write it as
$$( \frac{x}{3} )^11 \cos x

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$$\left( \frac{x}{3} \right)^11 \cos x$$

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$$\left( \frac{x}{3} \right)^{11} \cos x$$

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ok im just terrible lmao

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that should be e^x

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in the denominator

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$$\left( \frac{x}{e^x} \right)^{11} \cos x$$

jolly parrotBOT
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Secret

cedar coyote
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oh like

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ok so you are saying

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wait

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i dont understand the e^x

sudden charm
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lmao i dint notice
but the P(x) is wrong

cedar coyote
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would it not be whatever is in front of the y

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yea see i thought there was something incorrect with my setup

sudden charm
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differentiate the final equation with respect to x and u will notice

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$$ \frac{d}{dx} e^{-11x} y = \frac{dy}{dx} e^{-11x} - 11 y e^{-11x} $$

jolly parrotBOT
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Secret

sudden charm
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but u need a
$$ \frac{dy}{dx} - 11 \frac{y}{x}$$

jolly parrotBOT
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Secret

cedar coyote
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hm

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this is what im going based off of

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here they derived p(x) using whatever was in front of the y

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in this case it is -11

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and if u integrate that

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that would be e^-11x

sudden charm
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no

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in this case it's -11/x

cedar coyote
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ahhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

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ok ok

sudden charm
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lol there there

cedar coyote
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ok so

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x^-11?

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making sure i did that right

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im kind of rust on the natural log stuff

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when i did p(x)

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i got e^(integral of -11/x)

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and then got e^-11lnx

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e and ln would cancel out and i think it becomes x^-11 iirc

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yea?

cedar coyote
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<@&286206848099549185>

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This is a similar problem i am doing

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Question is

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My answer is almost correct

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Correct answer is what I got but without a denominator

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please help me someone

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pearl pondBOT
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sudden charm
pearl pondBOT
sudden charm
#

.reopen

sudden charm
sudden charm
cedar coyote
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Why?

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Because of product rule

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Is it not -13y

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Or -11 y

sudden charm
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no

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$$ x^-13 \frac{dy}{dx} - \frac{13y}{x} x^{-13} = \frac{d}{dx} y x^{-13} $$

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$$ x^-{13} \frac{dy}{dx} - \frac{13y}{x} x^{-13} = \frac{d}{dx} y x^{-13} $$

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$$ x^{-13} \frac{dy}{dx} - 13 x^{-12} \frac{y} = \frac{d}{dx} y x^{-13} $$

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$$ x^{-13} \frac{dy}{dx} - 13 x^{-12} y = \frac{d}{dx} y x^{-13} $$

jolly parrotBOT
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Secret

cedar coyote
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so basically

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it is just integrating factor*y

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because if we differentiate -11yx^-13

cedar coyote
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however

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if we differentiate yx^-13

cedar coyote
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which must mean

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the shortcut for this prodcut rule

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is just integrating factor*y(no matter what y has in front of it)?

sudden charm
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yeah

cedar coyote
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ok ok

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i think i understand now

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i need to be careful on what what is my integrating factor which is what i messed up on earlier

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and for product rule it is always just p(x)*y

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i think those 2 things

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was what i need to fix

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thanks

sudden charm
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yeah

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no probs

cedar coyote
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i was struggling with this for a while so u saved my ass

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hopefully exam goes well on wed

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have a good day

sudden charm
sudden charm
sudden charm
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night tho

cedar coyote
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oh ok

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good night then

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its 1pm here for me

sudden charm
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its around 1am lmao

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glossy hedge
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I understand that k< 3/4, but how do u get 0 <or= k

pearl pondBOT
glossy hedge
plush bramble
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I see

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When you divided by k, you assumed k>0

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Otherwise you'd be dividing by zero or would need to flip the inequality sign

glossy hedge
#

Ohhhh that makes sensee

glossy hedge
remote roost
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Think

plush bramble
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And assumed k is nonzero

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I'd go with factoring

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4k^2-3k = k * (...)

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Since that must be negative, the two numbers being multiplied must have opposite signs

pearl pondBOT
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@glossy hedge Has your question been resolved?

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glossy hedge
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Thanks

pearl pondBOT
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shy ginkgo
pearl pondBOT
shy ginkgo
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can someone help me find lim x->-infinity f(x)

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sin(x) osilicates between -1 and 1 i believe, and since the denominator is infinitely small, sin(x)/x becomes infinitely large?

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so if 1 is infinitely large is it infinity?

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but it also osilates to -1

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so -infinity?

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idk

autumn narwhal
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The denominator gets larger, not smaller

shy ginkgo
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oh

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wait but its -infinity on the denominator

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-infinity is infitely small

autumn narwhal
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Well I meant in absolute value.

shy ginkgo
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wah

autumn narwhal
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When you divide by something that's approaching -infinity or infinity, you get a number close to 0.

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For example, 1/(-1000000000) is very close to 0

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And so is 1/10000000000

shy ginkgo
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so if the denominator is infinity it becomes infinitely small (0) and if the denominator is -infinity it becomes infinitely large...?

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the -infinity is messing me up

west sapphire
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it might be helpful to consider |f(x)| instead of f(x)

shy ginkgo
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so the - doesn't matter?

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or does it just make the infinitely small value (0) negative? but 0 can't be negative obv

west sapphire
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the - doesn't matter in the special case where the result is 0

shy ginkgo
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alr now that makes sense

west sapphire
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basically f(x) -> 0 if and only if |f(x)| -> 0

shy ginkgo
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oop oke ty

west sapphire
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more generally:
f(x) -> L if and only if |f(x) - L| -> 0

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here L=0

shy ginkgo
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yeh thats how it says it in the textbook

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oke ty

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analog topaz
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Whats a trick to graphing 1/3x^2 +8x +36. I want to find a way to get rid of the fraction in front?

analog topaz
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But if I did that then the graph wouldn't be similar to the orginial graph

analog topaz
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.close

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blissful briar
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calculus help

pearl pondBOT
blissful briar
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i already have the instant rc formula written but im not sure where to go from hjere

pearl pondBOT
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@blissful briar Has your question been resolved?

blissful briar
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<@&286206848099549185>

dusty patio
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(g(1.001)-g(0.999))/(5.004-4.996)

blissful briar
#

how does that work?

pearl pondBOT
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@blissful briar Has your question been resolved?

blissful briar
#

<@&286206848099549185>

dusty patio
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I mean (g(1.001)-g(0.999))/(1.001-0.999)

blissful briar
#

i see

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rustic hawk
#

I’m not sure how to do this do I do it by integration of parts and how do I start

worldly glacier
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Oh wait

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Integration by parts

pearl pondBOT
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@rustic hawk Has your question been resolved?

pearl pondBOT
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@rustic hawk Has your question been resolved?

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rancid laurel
#

I'm not doing a maths question, but I just wanted to have some help in understanding: Why do the integration and differentiation formulae work? I've watched the 3b1b videos and I understand the theory of differntiation and integration, I understand the rate of change and the accumulated change, and adding up infinitesimals, I just don't understand the formulae that are used

rancid laurel
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the formulae:

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$$ Differentiation: x*n^{x-1}, Integration: \frac{n^{x+1}}{x+1}$$

jolly parrotBOT
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panica5193

plush bramble
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And use fundamental theorem of calculus for integration

rancid laurel
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how does that work?

plush bramble
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...

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You should learn limits first

rancid laurel
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thanks

pearl pondBOT
#

@rancid laurel Has your question been resolved?

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gusty prism
pearl pondBOT
gusty prism
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shouldnt the set up be 7.5+s=.25(75+s)

pearl pondBOT
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daring abyss
#

Could anybody online help me with finding the limit of the function (2x)/(16-x^2) as x approaches -4 from the left. Also, the limit of (x-1)/(x^2-4) ax x approaches -2 from the right?

buoyant panther
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Do you know concept of one-sided limits in general? Or methods of solving them?

daring abyss
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I tried substituting x with the number that it's trying to approach.

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Like for this one.

buoyant panther
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In this case it won't work because it gives 0 in the bottom, right

daring abyss
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I substituted x with -4, and got -8/0 as the answer.

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But I know that we can't divide with 0.

buoyant panther
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but you can try numbers near to the -4 (from the left, so close to the -4 but less than it)

daring abyss
#

Okay, I'll try it with my calculator.

buoyant panther
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e.g. -4.1, -4.05 etc. and try to determine what does it approach to

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or alternative method is to look at denom function behaviour

daring abyss
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I inputed -4.1 and got 10.123 as the result.

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What does 10.123 mean for my case?

daring abyss
buoyant panther
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generally when dealing with one sided limits we often get -inf or inf as a result (especially for rational functions)

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10.123 means nothing so far

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so you should try with numbers closer to the -4

daring abyss
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Oh, okay.

buoyant panther
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it's fine if you can use the calculator

daring abyss
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-4.01 resulted with 100.125

buoyant panther
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okay

daring abyss
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The numbers are increasing the closer I get to -4?

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Why is that?

buoyant panther
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the closer we are to -4, the output is bigger and bigger

buoyant panther
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dividing by really small number ---> huge result

daring abyss
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Yes, I think I can see that.

buoyant panther
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and now

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thing is

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we can deduce the result is +inf

daring abyss
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Ohhh

buoyant panther
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Yeah, exactly

daring abyss
#

Nice, thank you.

buoyant panther
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I'll paste the graph

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to visualize

daring abyss
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How would I write my answer on paper though?

buoyant panther
daring abyss
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Do I substitute x with -4.01 or something?

buoyant panther
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mhm, you can it below, I'd rewrite only the result or

daring abyss
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This was my answer earlier.

buoyant panther
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$$\lim_{x \to -4^-}\frac{2x}{16-x^2}=\Big[\frac{-8}{0^{-}}\Big]=+\infty$$

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This is notation I used to use

daring abyss
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How did you get 0+ as the denominator?

buoyant panther
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sorry, mb

jolly parrotBOT
daring abyss
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0-?

buoyant panther
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0^- means denom is negative

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when plugging numbers closer to -4

daring abyss
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Is it different from -0?

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Wait, is it like the direction where x came from?

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Since its 0^-, the 0 is approached from the left?

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Did I get it right?

buoyant panther
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no, it's about sign, direction doesn't matter

daring abyss
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Ah, okay.

daring abyss
buoyant panther
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I'd do that

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and that's it basically

daring abyss
#

I don't need to write it the middle of the solution?

daring abyss
daring abyss
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So its that simple, I don't know why I'm trying to overcomplicate things.

buoyant panther
#

in fact

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if you know how graph of the denom looks like

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(here if you can imagine the parabola 16 - x^2)

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you can solve in a few seconds

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without plugging anything

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in fact

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16 - x^2 = (4-x)(4+x)

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zeros: x = -4, x = 4

daring abyss
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Does the limit of a function depend on the denominator?

buoyant panther
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it's only about the bottom part

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in fact

daring abyss
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Thank you.

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Then this one should be the same process then?

buoyant panther
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yeah

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exactly same process

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but from the right

daring abyss
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Are the zeroes 2, -2?

buoyant panther
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yeah, and a = 1 (leading coeff)

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so you can easily decide if it's 0^+ or 0^-

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in the denom

daring abyss
buoyant panther
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ye

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look at -2

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from the right

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values are negative, right?

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you see that?

daring abyss
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Uh, wait.

buoyant panther
#

if not, then try plugging -1.9, -1.95 etc.

daring abyss
#

from the right of -2, are positive number though?

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Oh wait, they are still negative up to 0

daring abyss
daring abyss
buoyant panther
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so it's increasing right

daring abyss
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yes

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Then, the limit is -infinity right?

buoyant panther
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why?

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whole fraction increases

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so if we are closer and closer to -2

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the result is bigger and bigger

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up to +inf

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for infinitely close numbers

daring abyss
#

Ohhhh

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if the numbers get bigger and bigger, the limit is positive infinity. And if the numbers become smaller and smaller, the limit is negative infinity.

buoyant panther
#

yes

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to see that you can try same example with x approaches to -2 from the left

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you should get negative infinity

daring abyss
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Okay, I'll try.

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-2.1 gave me -7.561, -2.01 gave me -75.062.

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I can see that the numbers are decreasing the closer I get to -2.

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Then, the limit should be -infinity right?

buoyant panther
#

well

daring abyss
buoyant panther
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-3/0^+ -> -inf

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it should be -3/0^-

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because minus divded by minus gives +

daring abyss
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Uh, sorry I don't understand.

buoyant panther
#

look at this

daring abyss
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0^+ should be 0^- because dividing what by minus gives +?

buoyant panther
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nope, because for -1.9, -1.95 etc. x^2 - 4 is negative

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,calc (-1.9)^2 - 4

jolly parrotBOT
#

Result:

-0.39
buoyant panther
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,calc (-1.95)^2 - 4

jolly parrotBOT
#

Result:

-0.1975
buoyant panther
#

,calc (-1.99)^2 - 4

jolly parrotBOT
#

Result:

-0.0399
buoyant panther
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etc.

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this is why it's 0^-

daring abyss
#

Ah, I think I see it.

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Why does this give increasing numbers as it approaches -2?

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I'm confused because of my calculator.

buoyant panther
#

That's fine, don't confuse the whole fraction with only the denominator

daring abyss
#

OHHH

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I forgot that I was only solving for the denominator!

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The denominator is increasing as it approaches-2.

buoyant panther
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here what matters is sign

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not monotonicity

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thing is values are negative, period

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all in all, values of the whole fraction increases when x goes to -2
values of the denominator are negative when x goes to -2

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it gives

jolly parrotBOT
buoyant panther
#

In fact the first information is sufficient to get the result, so you can just rewrite +infty and that's it

daring abyss
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Okay, I think I got it.

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Thank you for all the help Modus, could I bother you with one more problem?

buoyant panther
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I don't mind

daring abyss
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Alright, thank you.

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This is the problem

buoyant panther
#

Let me guess, checking continuity?

daring abyss
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yes

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This is a piecewise function right?

buoyant panther
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Yes

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Okay, so can you sketch its graph?

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It's about drawing three independent graphs for given intervals

daring abyss
#

I probably can, it will take me sometime though because I'm learning it as I go.

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Could you show me the graph for the first one?

buoyant panther
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Do you mean first part (interval)?

daring abyss
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the x-4 if x<-4

buoyant panther
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How to draw it:
It's a line. What do you need to draw a line? Only two points. Because there is exactly one line which passes through two different points.

daring abyss
#

Its a linear function! I forgot lol

buoyant panther
#

Let's take x = 0 and x = 4 for instance.
So the points are:
(0, -4) and (4, 0)

#

Plot them, connect, period.

daring abyss
#

Okay.

buoyant panther
daring abyss
#

Ah I see.

buoyant panther
#

Now for the restricted domain it would be

daring abyss
buoyant panther
#

Line is infinite. We cut it at x = - 4 (we take the part for x's less than -4)

daring abyss
#

Ahh

#

Okay, I got it.

#

because x<-4 right?

buoyant panther
#

exactly

daring abyss
#

Okay, I'll try drawing it.

#

at x=-4, I got -8, and at x=4, I got 0

buoyant panther
#

should be 0

#

at both

daring abyss
#

Hmm?

buoyant panther
#

(-4)^2 = 4^2 = 16

#

16 - 16 = 0

#

sqrt(0) = 0

daring abyss
#

Ohhh

#

Did it come from here?

buoyant panther
#

yes

daring abyss
#

This was meant for this

daring abyss
buoyant panther
#

but you cannot plug x = -4 of x = 4 into that

#

is -4 < -4 or 4 < -4?

#

that's false

daring abyss
#

Modus, thanks for all the help. I need to go to sleep right now because I have a class at 8 am and the current time is 5 am 💀

pearl pondBOT
#

@daring abyss Has your question been resolved?

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lapis flower
pearl pondBOT
#

Please don't occupy multiple help channels.

lapis flower
#

help me

#

.

last summit
pearl pondBOT
# lapis flower
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
lapis flower
#

1

last summit
#

The question you posted isn't even a question

#

it is a statement

#

is there any more context you can provide?

lapis flower
#

it asked the value of a

#

?

#

tell me please

midnight haven
#

no it doesnt ask for the value of a

#

reread

lapis flower
#

it does

midnight haven
#

it does not

lapis flower
#

it got the options too

last summit
#

Show us what you are looking at

midnight haven
#

send the options

lapis flower
#

see

#

now ?

midnight haven
#

you could try moving ax-2 to the right

#

and compare the coefficients

lapis flower
#

can you show me a answer pic

midnight haven
#

no?

lapis flower
#

i want answer pic

last summit
#

no one is going to give you the answer

#

we are here to help guide you, not to give answers

lapis flower
#

ok

#

thanks

pearl pondBOT
#

@lapis flower Has your question been resolved?

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#
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river aurora
#

Could someone double check my work? This was kind of a mess of a question, so I'm not sure if I got all the parts right.

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#

@river aurora Has your question been resolved?

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#

@river aurora Has your question been resolved?

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west venture
#

missed a day in class cause i got sick. help me with this please

shy thistle
#

Drawing a picture helps a lot on these.

west venture
#

way ahead of ya

#

still dont get it

shy thistle
#

If you label the picture you've got 2x+10 combined with x-4, the total of which is 21. Can you write an equation for that?

west venture
#

ohhhh

#

okay

#

i see

#

thank you sir

shy thistle
west venture
#

.close

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frozen estuary
#

first blank would be 1 as pure nickel is 100%

#

2nd blank would be total weight which would be (6+x) mg

#

no that is incorrect. think about the filled and unfilled spaces in your equation

#

0.45 is purity multiplied by x which is weight

#

yes

pearl pondBOT
#
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obtuse dagger
#

Hey! Um just wondering if anyone could help me finding the height of the cliff? Here’s the question and the information it gave me

shy thistle
#

For what it's worth, the horizontal speed doesn't matter for finding the height.

#

Since you can break the problem up into the vertical bit and the horizontal bit.

obtuse dagger
#

ohh

#

K thanks, I think I got it from here!

shy thistle
obtuse dagger
#

.close

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somber magnet
#

Hello! I have 2 integrals that im having a bit of trouble with!

somber magnet
#
  1. Integral of x(1+x)^(1/2)
#
  1. Integral of 1/(1+4x^2)
#

i cant use partial integration for these

#

just u sub

shy thistle
#

I like u=1+x for 1.

somber magnet
#

i tried that let me post the result

shy thistle
#

(You'll need to use the fact that x=u-1 to change the x on the outside.)

somber magnet
#

How does this look

shy thistle
#

Basically what I was thinking. The teacher in me is getting grumpy about a missing du and some missing +c's but you know.

somber magnet
#

yeah

#

i was rushing ill put them in

#

for the second integral i dont really know where to start

#

in my mind i feel like ln has to be there but given there would be nothing to solve the chain rule bit

shy thistle
#

Like, it looks a lot like arctan doesn't it?

somber magnet
#

yeah 1+u^2 maybe

#

how do i get tehre tho?

#

ooh

#

u

#

but in that case wouldnt dx =du/8w?

shy thistle
#

$\int \frac{1}{1+4x^2} , dx = \int \frac{1}{1+(2x)^2} , dx$

#

Oh I forgot I wrote code in my LaTeX to do \dx. Anyway.

jolly parrotBOT
somber magnet
#

ah and then i sub 2x for u?

shy thistle
#

That's what I'd do.

somber magnet
#

i always forget i can do these algabraic tricks

#

hahaha

#

tyvm

shy thistle
somber magnet
#

just because ive never done this

#

is u=2x

#

yes

#

nvm i got it

#

so ulitmately i get to

#

1/2 *arctan(2x) +C

shy thistle
#

Yeah that looks right.

#

,w integral of 1/(1+4x^2) dx

shy thistle
#

Hax.

somber magnet
#

yay!

#

ty for the help!

#

.close

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#
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midnight haven
pearl pondBOT
midnight haven
#

if i want to define a set

#

do i use r^4 og r^3

#

i use r^4 because there r 4 equations or?

foggy mason
#

x1 + x2 + 3x3 = x1 + x2 + 3x3 + 0 x4

midnight haven
#

im asking

#

is is r^4

foggy mason
midnight haven
#

since there r 4 equations

#

yes thank you

foggy mason
#

not because there are 4 équations

last summit
#

It is because there are 4 variables

midnight haven
#

ok

#

but then can i not just add another

#

+0 5_x

#
  • 0 6_x
last summit
#

Yes but none of them have a non 0 of those

foggy mason
#

^

midnight haven
#

explain further

#

quite new to this ))

last summit
#

There are clearly no x5, or higher terms in any of the equations

midnight haven
#

i see

last summit
#

Yet there is a nonzero value of x4

midnight haven
#

yes okay

#

so you can add 0 values of the highest term presented

#

if that makes sense

#

in the terms missing

last summit
#

Yes

#

But it would be useless to add 0 values of other variables that aren’t presented

midnight haven
#

yes okay

#

would it be possible to do so

last summit
#

Sure

midnight haven
#

even if it'll only make it more difficult to solve

#

i'd suppose

last summit
#

Right

#

So don’t do that

midnight haven
#

i wont

#

ty

#

also

#

in the second term is goes x1 - x3 - x4

#

can i then add a x2

foggy mason
#

0 x2

midnight haven
#

a 0x_2 term

#

nice

pearl pondBOT
#

@midnight haven Has your question been resolved?

pearl pondBOT
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devout plinth
#

can some1 help me with 5

pearl pondBOT
sick adder
#

What do we know about angle <ABE already?

devout plinth
#

OH

devout plinth
sick adder
#

See if you can use that to set up an equation in terms of X

devout plinth
#

because i don’t understand how to find the angles from that

sick adder
#

What does <ABE being right angle tell you about <ABD?

devout plinth
sick adder
#

Yup

devout plinth
#

okay i think i figured it out

pearl pondBOT
#

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prime bramble
pearl pondBOT
prime bramble
#

!status

pearl pondBOT
#
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
prime bramble
#

for b), 2 and for c), 1

#

I know that if h(x) is going to be injective then we assume that h(x1) = h(x2)

#

so that (x1)^2 = (x2)^2

#

but I've tried for a bit to get this to x1 = x2 which is what I want

#

and I cannot do it

pearl pondBOT
#

@prime bramble Has your question been resolved?

prime bramble
#

<@&286206848099549185>

acoustic spear
#

since F is a field, you cannot have 0 divisors... ^^

prime bramble
acoustic spear
#

because if (x1-x2) != 0 then is a zero divisor, since (x1-x2)^2 is 0

#

so the only possible answer is that x1-x2 = 0 UwU

prime bramble
#

I see

#

thank you

#

.close

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kindred venture
pearl pondBOT
kindred venture
#

Are the statements and proofs I gave here correct?

pearl pondBOT
#

@kindred venture Has your question been resolved?

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#

@kindred venture Has your question been resolved?

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lime salmon
pearl pondBOT
lime salmon
#

here is my work so far

#

how would i evaluate the top part of the piecewise?

west sapphire
#

if x is near -3, then |x| is always -x, right?

lime salmon
#

could you please elaborate

plush bramble
jolly parrotBOT
#

riemann

lime salmon
#

oh shoot i got it now

#

.close

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jolly basin
pearl pondBOT
#

@jolly basin Has your question been resolved?

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@jolly basin Has your question been resolved?

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spring crystal
pearl pondBOT
spring crystal
#

Help

plush bramble
#

You have -7 and 7

#

Which one is it

spring crystal
#

-7

plush bramble
#

Show how you got -8 and -7

spring crystal
#

X=7 and -8

plush bramble
spring crystal
#

How

plush bramble
spring crystal
#

(X-7)(x+8)

plush bramble
spring crystal
#

Idk

#

So which part is wrong

plush bramble
spring crystal
#

Oh so I remove -7 to 7 and the answer is right?

#

@plush bramble

pearl pondBOT
#

@spring crystal Has your question been resolved?

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midnight haven
pearl pondBOT
midnight haven
#

How do i do part 3

#

Heres th first 2 parts:

pearl pondBOT
#

@midnight haven Has your question been resolved?

pearl pondBOT
#

@midnight haven Has your question been resolved?

midnight haven
#

Can someone help?

#

<@&286206848099549185>

frosty creek
#

If you have it in the form R*cos(x-a), then can't you just solve for when that is zero?

midnight haven
#

im not sure what you mean. How would i go about doing that?

#

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finite rover
#

why is this true? Could someone show me the steps

remote roost
#

3l = -1 so l = -3^(-1) mod 5

finite rover
#

I don't get how u can transition from 3l = -1 into -3^(-1) mod 5

remote roost
#

Multiply both sides by 3^(-1)

#

Basically "dividing" both sides by 3

finite rover
#

oh ok, I remembered that you couldn't just straight up divide on both sides

#

so multiplying by the inverse is fine

#

.close

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nimble coral
pearl pondBOT
nimble coral
#

How should i approach this question?

#

I don’t have a clue

#

Can someone guide me please?

normal sigil
#

So what does (x+1) giving remainder 0 tell you?

nimble coral
#

It is a factor

dusty patio
#

,rccw

jolly parrotBOT
normal sigil
#

Good (or you could say x = -1 satisfies the equation)

#

Now, do you know the Remainder Theorem?

#

If not, that’s fine

nimble coral
#

I am reviewing lol

#

I should have known

normal sigil
#

Yes, true.

nimble coral
#

so f(-1)=0

normal sigil
#

Good, yes. What can you say about f(2)?

nimble coral
#

it equals to 45?

#

45 is what you get when you divide the function by (x-2)

normal sigil
#

So, with f(-1) = 0 and f(2) = 45, what can you do?

nimble coral
#

I don't know. I am slow on this

#

I could have a table?

#

of the function

normal sigil
#

Hint- f(-1) = 0 and f(2) = 45 lead to two equations

#

and you have 2 unknowns

nimble coral
#

I don't know man. Sorrrry

normal sigil
#

Simultaneous equations is the way to go.

nimble coral
#

I dont know what that is

normal sigil
#

Oh!

#

Unexpected

nimble coral
#

yeah sorry, it is the first time i have heard about it

normal sigil
#

Would you know how to solve, for instance,

x + 2y = 5
3x - 2y = -2
?

#

Maybe you use a different term

nimble coral
#

no, i didnt learn about it

#

could you please teach me?

normal sigil
#

Hmm… that’s interesting.

#

Sure

nimble coral
#

it is probably in the last year's lesson

#

it looks like simple algebra

normal sigil
#

For example, these two added would give us

nimble coral
#

I see

normal sigil
#

x + 2y + 3x - 2y = 5 - 2

#

Do you see what has happened?

nimble coral
#

Yup

#

I should have known it, but i skipped a lot of lessons

normal sigil
#

Good. This gives us
4x = 3 so x = 3/4

normal sigil
nimble coral
normal sigil
#

x + 2y = 5 so 3/4 + 2y = 5

#

2y = 17/4

#

y = 17/8

nimble coral
#

that's right

normal sigil
#

Good. Now, try substituting -1 into your polynomial and setting equal to 0

#

and do the same with 2, setting equal to 45

nimble coral
#

This?

normal sigil
#

Are you sure about m - 1 and m + 2 ?

nimble coral
#

Oh i added the brackets now

#

But where do i put the remainder of 45 in the second equation

normal sigil
#

equals 45

#

and the first one equals zero

nimble coral
#

Ok

normal sigil
#

Now, please expand and simplify. What equations do you have?

nimble coral
#

@normal sigil

#

Is this right?

normal sigil
#

No need to keep the (1)

#

let me just check

#

Oh, you’ve added them already

#

Now, something about the adding

#

You need to add them in such a way that you get rid of a variable by adding them.

normal sigil
nimble coral
#

How do i do that

normal sigil
#

Could you share your two separate equations that you added?

nimble coral
normal sigil
#

I would suggest you explicitly simplify each equation before adding

#

so the top one becomes - m - n = 15, right?

nimble coral
#

Can you explain please?🙏

normal sigil
#

Sure

#

Go term by term

#

What does m(-1) simplify to?

#

simply -m

#

What about 20(-1)^2 ?

nimble coral
#

Is this what you mean?

normal sigil
#

Oh good

#

Yes

nimble coral
#

I see

normal sigil
#

Move m and n over and we get m + n = -15

#

Please do the same thing for the second equation and see what you get

nimble coral
normal sigil
#

Not quite

#

Remember it is mx^3

nimble coral
#

Oh right

#

So it is 8m?

normal sigil
#

Yep

#

But check that number on the RHS

#

We should have

#

8m + 80 + 2n - 35 = 45

#

so 8m + 2n = 0

#

Well, that’s convenient.

nimble coral
#

Oh thanks

normal sigil
nimble coral
#

I don’t

#

Could you please tell me?

normal sigil
#

You forgot to cube the 2 in front of m

#

You also forgot the -35 entirely

nimble coral
#

in front of m?

normal sigil
#

Yes, it was mx^3

#

so you forgot to cube the 2

nimble coral
#

Yeah isn’t it 8?

normal sigil
#

Yep

#

Just saying beforehand

#

you forgot

nimble coral
#

Oh ok

#

What is next

normal sigil
#

Now we have two Wes

#

eqs

nimble coral
#

8m+2n=0

#

-m+20-n-35=0

normal sigil
#

Good, and the other one is m + n = -15

#

(rearranged a bit)

nimble coral
#

Oh it isnt -m+n=-15?

normal sigil
#

No!

nimble coral
#

M(-1)3 is not -m?

normal sigil
#

It is

#

But that gets you - m - n = 15

#

now if you multiply both sides by -1

#

m + n = -15

nimble coral
#

Oh i see

#

-m-n=-15

normal sigil
#

No

nimble coral
#

Oh sorry

#

I mean m-n=-15

normal sigil
#

Still no

#

Why do you think n has a different sign to m?

#

Remember that m(-1)^3 = -m and n(-1) = -n

#

so they have the same sign

nimble coral
#

Oh right sorry

normal sigil
#

m + n = -15
4m + n = 0

nimble coral
#

-m-n=15

normal sigil
nimble coral
#

And you switch the signs

#

Yes

normal sigil
#

Observe: m - 4m + n - n = -15 - 0

nimble coral
normal sigil
#

Good, so m = 5

nimble coral
normal sigil
#

Why is 5 + n = 0?

#

Use the equations.

#

m + n = -15

#

5 + n = -15

nimble coral
#

Oh

#

N=-10

#

Let me check if it fits

normal sigil
#

Minus 5 from both sides

nimble coral
#

Ohhhhh

#

-20

#

Aghhhh

#

Idk why i make so many mistakes

normal sigil
#

Practice makes perfect

#

Right, I recommend you try some simultaneous equation problems online

#

That can be quite key so it’s really important you have that skill

nimble coral
#

Yeah

#

I need a lot more practices

#

It is just lately i have picked up math again

normal sigil
#

🕺 It will pay off

#

Are we all done?

nimble coral
#

Can i check?

normal sigil
#

Go for it

nimble coral
#

👍 yup it is right

#

Thank you very much

normal sigil
#

🕺

nimble coral
#

I need a lot more practices

#

Thank you🥹 bye

#

.close

pearl pondBOT
#
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pearl pondBOT
#
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winter rain
#

Hi, I need help with part A and B of this question. How do I show that a = 2 and b = -4? I'm not sure where to start about this

winter rain
#

Bad picture but this is what he showed for part a in class, can't seem to wrap my head around how he answers it

pearl pondBOT
#

@winter rain Has your question been resolved?

winter rain
#

For A, i see the factors in the denominator, is that all that's necessary to answer the question? Given the vertical asymptotes and the hole/removable discontinuity?

#

For B, i was not sure how to come up with proper work to show how i got my answer

my teacher is really keen on showing our work, making sure it's very sound, so this is as far as I can get (B): since there's a r.d. at x=-2, (x+2) must be at the top of the numerator. then I just tried what values i could multiply (x+2) with to get -5x^2+cx. But how would I write this on paper??? and is this valid?

winter rain
#

I think i got it im going to bed

pearl pondBOT
#

@winter rain Has your question been resolved?

pearl pondBOT
#
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#
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Remember:
Ask your math question in a clear, concise manner.
Show your work, and if possible, explain where you are stuck.
After 15 minutes, feel free to ping <@&286206848099549185>.
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ivory ridge
#

helloo! pls help meeee

pearl pondBOT
ivory ridge
#

for 7. e, is there any other shorter way to do it??

plush bramble
ivory ridge
#

what do you mean?

plush bramble
#

Complement set

ivory ridge
#

how would i do that with this quesiton

plush bramble
#

Negate "at least 2 red cards"

ivory ridge
#

uhm

#

sorry can you explain it in steps?

plush bramble
ivory ridge
#

to like remove?

plush bramble
#

No

#

What's the opposite of "at least two red cards" in words?

ivory ridge
#

2 red cards max

plush bramble
#

No

ivory ridge
#

oh so just no restrictions?

plush bramble
ivory ridge
#

idk im rly confused

#

2 black cards?

plush bramble
#

Is 2 red cards part of your question or part of the negation?

ivory ridge
#

part of the question

plush bramble
ivory ridge
#

less than 2 red cards

plush bramble
#

Yea find that

#

Then subtract that from the total number of hands

ivory ridge
#

lemme try

#

so i would do

#

10400600

#

minus the ways it can be for 2 red cards and 1 red card?

#

wait

#

is the answer 6.3x10^11?

#

@plush bramble

plush bramble
plush bramble
plush bramble
#

"less than 2" means 0 or 1

ivory ridge
#

oh wait ignore the number

#

so first total ways is 52C13 and then minus (ways with 0 reds + ways with 1 red)

plush bramble
#

Yup

ivory ridge
#

yEUH

#

so it would be 6.3 x10^11 right?

pearl pondBOT
#

@ivory ridge Has your question been resolved?

#
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#
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After 15 minutes, feel free to ping <@&286206848099549185>.
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wanton thunder
#

?claim

pearl pondBOT
wanton thunder
#

hey

#

cna someone send a guide

#

a youtube video or a book pcitures

#

on how to solve quadratic equations by ,completing the squares, extracting the squares, factoring, using quadratic formula

#

pleasee ty

analog topaz
#

Here use this for quadratic formula

#

This algebra 2 video tutorial shows you how to complete the square to solve quadratic equations. This video is for high school students taking algebra 1 & 2 and college students taking introductory or college algebra. This video contains plenty of examples and practice problems with solutions / answers that can help you with your next workshee...

▶ Play video
analog topaz
#

This algebra 2 video tutorial explains how to factor by grouping. It contains examples of factoring polynomials with 4 terms and factoring trinomials with 3 terms. This video contains plenty of practice problems for you to finish your next worksheet assignment.

Algebra Review: https://www.video-tutor.net/algebra-for-beginners-bas...

▶ Play video
wanton thunder
#

can i learn all dat in 30 mins

analog topaz
pearl pondBOT
#

@wanton thunder Has your question been resolved?

pearl pondBOT
#
Channel closed

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Ask your math question in a clear, concise manner.
Show your work, and if possible, explain where you are stuck.
After 15 minutes, feel free to ping <@&286206848099549185>.
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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

oblique reef