#help-39
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I am trying to save this linear diff eq
I put it in the correct form i think
But i am not sure this is correct because if i differentiate the right side I get an insanely long integral
Which typically means i did something wrong lol
@cedar coyote Has your question been resolved?
<@&286206848099549185>
i have asked for help on diff eq multiple time for past week and no one ever answers
is this server not for diff eq?
u mean solve?
yes
Secret
u wrote x^2
oh yea my bad
ok but i would still be left with a crazy integral though
because if we multiply everything by p(x)
yeah
then we would have 3 functions
and i basically fell into and endless loop of doing integration by parts
actually
there's a niche trick to do series of integration by parts
i think it's called DI method
Integration by parts by using the DI method! This is the easiest set up to do integration by parts for your calculus 2 integrals. We will also do 3 integrals to illustrate the 3 stops of the DI method.
Dear calculus teachers, please let students use the DI Method (& why it is really the same as integration by parts) 👉 https://youtu.be/8xPfNuXLS...
ive heard of it but the reason i didnt do it
is because it doesnt work with e
so i didnt learn it
hmm
riggt it doesn't work
i remember
u see the dillema
and if i do integration by parts normally
i just am stuck integrating by parts over and over again
and it doesnt help that online calculators or atleast the ones i use cant solve it
yes lmao
but did i set up the problem correctly?
because that was my concern
i feel like there might be an issue with the way i set it up
is it even a linear diff eq to begin with
maybe u can write it as
$$( \frac{x}{3} )^11 \cos x
$$\left( \frac{x}{3} \right)^11 \cos x$$
$$\left( \frac{x}{3} \right)^{11} \cos x$$
ok im just terrible lmao
that should be e^x
in the denominator
$$\left( \frac{x}{e^x} \right)^{11} \cos x$$
Secret
lmao i dint notice
but the P(x) is wrong
would it not be whatever is in front of the y
yea see i thought there was something incorrect with my setup
differentiate the final equation with respect to x and u will notice
$$ \frac{d}{dx} e^{-11x} y = \frac{dy}{dx} e^{-11x} - 11 y e^{-11x} $$
Secret
but u need a
$$ \frac{dy}{dx} - 11 \frac{y}{x}$$
Secret
hm
this is what im going based off of
here they derived p(x) using whatever was in front of the y
in this case it is -11
and if u integrate that
that would be e^-11x
lol there there
ok so
x^-11?
making sure i did that right
im kind of rust on the natural log stuff
when i did p(x)
i got e^(integral of -11/x)
and then got e^-11lnx
e and ln would cancel out and i think it becomes x^-11 iirc
yea?
<@&286206848099549185>
This is a similar problem i am doing
Question is
My answer is almost correct
Correct answer is what I got but without a denominator
please help me someone
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yes this looks right
.reopen
here when ur simplifying the LHS
u don't have -13
same here
u won't have a -11
no
$$ x^-13 \frac{dy}{dx} - \frac{13y}{x} x^{-13} = \frac{d}{dx} y x^{-13} $$
$$ x^-{13} \frac{dy}{dx} - \frac{13y}{x} x^{-13} = \frac{d}{dx} y x^{-13} $$
$$ x^{-13} \frac{dy}{dx} - 13 x^{-12} \frac{y} = \frac{d}{dx} y x^{-13} $$
$$ x^{-13} \frac{dy}{dx} - 13 x^{-12} y = \frac{d}{dx} y x^{-13} $$
Secret
so basically
it is just integrating factor*y
because if we differentiate -11yx^-13
we dont get this if we differentiate that
however
if we differentiate yx^-13
then we obtain this
which must mean
the shortcut for this prodcut rule
is just integrating factor*y(no matter what y has in front of it)?
yeah
ok ok
i think i understand now
i need to be careful on what what is my integrating factor which is what i messed up on earlier
and for product rule it is always just p(x)*y
i think those 2 things
was what i need to fix
thanks
i was struggling with this for a while so u saved my ass
hopefully exam goes well on wed
have a good day
no probs mate
happens
good luck
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I understand that k< 3/4, but how do u get 0 <or= k
!show
Show your work, and if possible, explain where you are stuck.
I see
When you divided by k, you assumed k>0
Otherwise you'd be dividing by zero or would need to flip the inequality sign
Ohhhh that makes sensee
So what should I do instead?
Think
You already proved k>0
And assumed k is nonzero
I'd go with factoring
4k^2-3k = k * (...)
Since that must be negative, the two numbers being multiplied must have opposite signs
@glossy hedge Has your question been resolved?
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Thanks
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can someone help me find lim x->-infinity f(x)
sin(x) osilicates between -1 and 1 i believe, and since the denominator is infinitely small, sin(x)/x becomes infinitely large?
so if 1 is infinitely large is it infinity?
but it also osilates to -1
so -infinity?
idk
The denominator gets larger, not smaller
Well I meant in absolute value.
wah
When you divide by something that's approaching -infinity or infinity, you get a number close to 0.
For example, 1/(-1000000000) is very close to 0
And so is 1/10000000000
so if the denominator is infinity it becomes infinitely small (0) and if the denominator is -infinity it becomes infinitely large...?
the -infinity is messing me up
it might be helpful to consider |f(x)| instead of f(x)
so the - doesn't matter?
or does it just make the infinitely small value (0) negative? but 0 can't be negative obv
the - doesn't matter in the special case where the result is 0
alr now that makes sense
basically f(x) -> 0 if and only if |f(x)| -> 0
oop oke ty
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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?
Could I just multiply by 3 to the entire thing to get rid of 1/3
But if I did that then the graph wouldn't be similar to the orginial graph
Or should I just convert this into vertex form?
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calculus help
i already have the instant rc formula written but im not sure where to go from hjere
@blissful briar Has your question been resolved?
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(g(1.001)-g(0.999))/(5.004-4.996)
how does that work?
@blissful briar Has your question been resolved?
I mean (g(1.001)-g(0.999))/(1.001-0.999)
i see
@blissful briar Has your question been resolved?
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I’m not sure how to do this do I do it by integration of parts and how do I start
Start with a u substitution
Oh wait
Integration by parts
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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
panica5193
And use fundamental theorem of calculus for integration
wait im not actually too sure about the limits part
how does that work?
thanks
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shouldnt the set up be 7.5+s=.25(75+s)
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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?
Do you know concept of one-sided limits in general? Or methods of solving them?
I tried substituting x with the number that it's trying to approach.
Like for this one.
In this case it won't work because it gives 0 in the bottom, right
I substituted x with -4, and got -8/0 as the answer.
But I know that we can't divide with 0.
but you can try numbers near to the -4 (from the left, so close to the -4 but less than it)
Okay, I'll try it with my calculator.
e.g. -4.1, -4.05 etc. and try to determine what does it approach to
or alternative method is to look at denom function behaviour
I got stumped at this step and didn't know how to proceed.
generally when dealing with one sided limits we often get -inf or inf as a result (especially for rational functions)
10.123 means nothing so far
so you should try with numbers closer to the -4
Oh, okay.
it's fine if you can use the calculator
-4.01 resulted with 100.125
okay
yes that's what I was about to say
the closer we are to -4, the output is bigger and bigger
this is because notice that for numbers close to -4 the denominator becomes less and less
dividing by really small number ---> huge result
Yes, I think I can see that.
Ohhh
Yeah, exactly
Nice, thank you.
How would I write my answer on paper though?
Do I substitute x with -4.01 or something?
mhm, you can it below, I'd rewrite only the result or
This was my answer earlier.
$$\lim_{x \to -4^-}\frac{2x}{16-x^2}=\Big[\frac{-8}{0^{-}}\Big]=+\infty$$
This is notation I used to use
How did you get 0+ as the denominator?
sorry, mb
Modus
0-?
Is it different from -0?
Wait, is it like the direction where x came from?
Since its 0^-, the 0 is approached from the left?
Did I get it right?
no, it's about sign, direction doesn't matter
Ah, okay.
If I were to write this on paper, do I write it directly straight?
I don't need to write it the middle of the solution?
This is just for example.
Ahh, I see.
So its that simple, I don't know why I'm trying to overcomplicate things.
in fact
if you know how graph of the denom looks like
(here if you can imagine the parabola 16 - x^2)
you can solve in a few seconds
without plugging anything
in fact
16 - x^2 = (4-x)(4+x)
zeros: x = -4, x = 4
Does the limit of a function depend on the denominator?
if we don't have zero in the numerator then yes
it's only about the bottom part
in fact
Are the zeroes 2, -2?
yeah, and a = 1 (leading coeff)
so you can easily decide if it's 0^+ or 0^-
in the denom
by leading coefficient, you mean the invisible 1 beside x^2? 1x^2?
Uh, wait.
if not, then try plugging -1.9, -1.95 etc.
from the right of -2, are positive number though?
Oh wait, they are still negative up to 0
okay, I'll try that.
-1.9 gave me 7.436, -1.95 gave me 14.937
so it's increasing right
why?
whole fraction increases
so if we are closer and closer to -2
the result is bigger and bigger
up to +inf
for infinitely close numbers
Ohhhh
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.
yes
to see that you can try same example with x approaches to -2 from the left
you should get negative infinity
Okay, I'll try.
-2.1 gave me -7.561, -2.01 gave me -75.062.
I can see that the numbers are decreasing the closer I get to -2.
Then, the limit should be -infinity right?
well
Uh, sorry I don't understand.
look at this
0^+ should be 0^- because dividing what by minus gives +?
Result:
-0.39
,calc (-1.95)^2 - 4
Result:
-0.1975
,calc (-1.99)^2 - 4
Result:
-0.0399
Ah, I think I see it.
Why does this give increasing numbers as it approaches -2?
I'm confused because of my calculator.
That's fine, don't confuse the whole fraction with only the denominator
OHHH
I forgot that I was only solving for the denominator!
The denominator is increasing as it approaches-2.
here what matters is sign
not monotonicity
thing is values are negative, period
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
it gives
Modus
In fact the first information is sufficient to get the result, so you can just rewrite +infty and that's it
Okay, I think I got it.
Thank you for all the help Modus, could I bother you with one more problem?
I don't mind
Let me guess, checking continuity?
Yes
Okay, so can you sketch its graph?
It's about drawing three independent graphs for given intervals
I probably can, it will take me sometime though because I'm learning it as I go.
Could you show me the graph for the first one?
Do you mean first part (interval)?
the x-4 if x<-4
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.
Its a linear function! I forgot lol
Let's take x = 0 and x = 4 for instance.
So the points are:
(0, -4) and (4, 0)
Plot them, connect, period.
Okay.
Ah I see.
How did that happen?
Line is infinite. We cut it at x = - 4 (we take the part for x's less than -4)
exactly
Hmm?
wait, where did these come from?
Ohhh
Did it come from here?
yes
This was meant for this
.
but you cannot plug x = -4 of x = 4 into that
is -4 < -4 or 4 < -4?
that's false
They're not, I see what you mean.
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 💀
@daring abyss Has your question been resolved?
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!status
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
1
The question you posted isn't even a question
it is a statement
is there any more context you can provide?
it does
it does not
it got the options too
Show us what you are looking at
send the options
can you show me a answer pic
no?
i want answer pic
no one is going to give you the answer
we are here to help guide you, not to give answers
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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.
@river aurora Has your question been resolved?
@river aurora Has your question been resolved?
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missed a day in class cause i got sick. help me with this please
Drawing a picture helps a lot on these.
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?

.close
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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
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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
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.

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Hello! I have 2 integrals that im having a bit of trouble with!
- Integral of x(1+x)^(1/2)
- Integral of 1/(1+4x^2)
i cant use partial integration for these
just u sub
I like u=1+x for 1.
i tried that let me post the result
(You'll need to use the fact that x=u-1 to change the x on the outside.)
Basically what I was thinking. The teacher in me is getting grumpy about a missing du and some missing +c's but you know.
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
Like, it looks a lot like arctan doesn't it?
yeah 1+u^2 maybe
how do i get tehre tho?
ooh
u
but in that case wouldnt dx =du/8w?
$\int \frac{1}{1+4x^2} , dx = \int \frac{1}{1+(2x)^2} , dx$
Oh I forgot I wrote code in my LaTeX to do \dx. Anyway.
Alti
ah and then i sub 2x for u?
That's what I'd do.

just because ive never done this
is u=2x
yes
nvm i got it
so ulitmately i get to
1/2 *arctan(2x) +C
Hax.
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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?
x1 + x2 + 3x3 = x1 + x2 + 3x3 + 0 x4
R^4, because thats a vector with four variables you are searching
not because there are 4 équations
It is because there are 4 variables
Yes but none of them have a non 0 of those
^
There are clearly no x5, or higher terms in any of the equations
i see
Yet there is a nonzero value of x4
yes okay
so you can add 0 values of the highest term presented
if that makes sense
in the terms missing
Yes
But it would be useless to add 0 values of other variables that aren’t presented
Sure
0 x2
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can some1 help me with 5
What do we know about angle <ABE already?
OH
ITS RIGHT ANGLE
See if you can use that to set up an equation in terms of X
would i have to add them all
because i don’t understand how to find the angles from that
What does <ABE being right angle tell you about <ABD?
it also equals to 90?
Yup
okay i think i figured it out
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!status
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
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
@prime bramble Has your question been resolved?
<@&286206848099549185>
From here and using 1) you have:
0 = (x1)^2-(x2)^2=(x1-x2)^2
since F is a field, you cannot have 0 divisors... ^^
I understand this, but how does this help me with the proof?
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
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Are the statements and proofs I gave here correct?
@kindred venture Has your question been resolved?
@kindred venture Has your question been resolved?
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if x is near -3, then |x| is always -x, right?
could you please elaborate
,tex .abs def
riemann
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@jolly basin Has your question been resolved?
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Help
-7
Show how you got -8 and -7
you factored wrong
How
plug in and see
(X-7)(x+8)
right so how do you get -7 here
-7
@spring crystal Has your question been resolved?
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@midnight haven Has your question been resolved?
@midnight haven Has your question been resolved?
If you have it in the form R*cos(x-a), then can't you just solve for when that is zero?
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why is this true? Could someone show me the steps
3l = -1 so l = -3^(-1) mod 5
I don't get how u can transition from 3l = -1 into -3^(-1) mod 5
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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How should i approach this question?
I don’t have a clue
Can someone guide me please?
So what does (x+1) giving remainder 0 tell you?
It is a factor
,rccw
Good (or you could say x = -1 satisfies the equation)
Now, do you know the Remainder Theorem?
If not, that’s fine
Yes, true.
so f(-1)=0
Good, yes. What can you say about f(2)?
Yep
So, with f(-1) = 0 and f(2) = 45, what can you do?
I don't know man. Sorrrry
Simultaneous equations is the way to go.
I dont know what that is
yeah sorry, it is the first time i have heard about it
Would you know how to solve, for instance,
x + 2y = 5
3x - 2y = -2
?
Maybe you use a different term
The principle is that we can add and subtract equations to get new equations also satisfying the original ones.
For example, these two added would give us
I see
Good. This gives us
4x = 3 so x = 3/4
That’s all right
yes
Then, we can use one of the original equations to find out y.
x + 2y = 5 so 3/4 + 2y = 5
2y = 17/4
y = 17/8
that's right
Good. Now, try substituting -1 into your polynomial and setting equal to 0
and do the same with 2, setting equal to 45
Are you sure about m - 1 and m + 2 ?
Oh i added the brackets now
But where do i put the remainder of 45 in the second equation
Ok
Now, please expand and simplify. What equations do you have?
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.
For instance, adding these guys allowed us to get rid of y.
How do i do that
Could you share your two separate equations that you added?
I would suggest you explicitly simplify each equation before adding
so the top one becomes - m - n = 15, right?
Can you explain please?🙏
Sure
Go term by term
What does m(-1) simplify to?
simply -m
What about 20(-1)^2 ?
I see
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
Yep
But check that number on the RHS
We should have
8m + 80 + 2n - 35 = 45
so 8m + 2n = 0
Well, that’s convenient.
Oh thanks
Do you see your errors?
in front of m?
Yeah isn’t it 8?
Oh it isnt -m+n=-15?
No!
M(-1)3 is not -m?
It is
But that gets you - m - n = 15
now if you multiply both sides by -1
m + n = -15
No
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
Oh right sorry
m + n = -15
4m + n = 0
-m-n=15
So these are our two equations
What do we do now? We can subtract them!
Observe: m - 4m + n - n = -15 - 0
Good, so m = 5
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
Can i check?
Go for it
🕺
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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
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
@winter rain Has your question been resolved?
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?
I think i got it im going to bed
@winter rain Has your question been resolved?
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helloo! pls help meeee
for 7. e, is there any other shorter way to do it??
Find the complement
what do you mean?
Complement set
how would i do that with this quesiton
Negate "at least 2 red cards"
Do you know what negate means?
to like remove?
2 red cards max
No
oh so just no restrictions?
Also no
Is 2 red cards part of your question or part of the negation?
part of the question
So adjust this
less than 2 red cards
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
Where did this come from
Why 2 red cards
oh wait ignore the number
so first total ways is 52C13 and then minus (ways with 0 reds + ways with 1 red)
Yup
@ivory ridge Has your question been resolved?
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?claim
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
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...
This is for completeing the square
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...
And thats for factoring
can i learn all dat in 30 mins
You could
@wanton thunder Has your question been resolved?
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