#help-23
1 messages · Page 136 of 1
Would I half the 50 for the angle? So 25 degrees in the right triangle
Yes
Alright, so I have the radius and height but what’s the formula for the surface area of a cone? Like Google gives me a long thing with a square root but surely there’s an easier to remember formula or something?
It's just $\pi r^2 + \pi r l$ where $l = \sqrt{r^2 + h^2}$
A Lonely Bean
If google gave you this then no you can't simplify this
Ok that first one makes more sense- Google gave me which is the same thing but makes less sense at first glance
Yeah I prefer to write it this way since it's more intuitive
Since it's just base area + lateral area
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Where did the integral of 2 1(big f thingy)-1 change into 4 1(big f thingy)0?
even function
Since your function is even (symmetric around the y axis) the area from -1 to 1 is twice the area from 0 to 1
range?
bounds
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hello guys
can i multiply the question to make it easier to do the quadratical formula
yup
is it right
why not dont be scared bro its just those loser math laws
yeah I'm actually just trying to help my friend but i want to make sure im right
thanks ig
.close
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i need help on items 10-16
G8 laws of exponents
starting with number 10
x³y⁴
x²y⁷
When you divide, you subtract exponents with same base. When you multiply, you add the exponents with same base.
Sample: x^2 times x^3 = x^2+3 , x^3 divided by x^2 = x ^3-2
so all the variables become x?
Nope, it's only applicable to the ones with the same variable. In that case, you have to repeat the same process if theres another variable
so what will the answer of the fraction look like with two different variables
For example in number 10, you have 2 variables present, x and y
yea?
They'll just be side by side with each other.
If the other variable has a negative exponent, you have to put it in the denominator
Example y^-1 = 1/y
so it would be
xy⁷
xy⁹
?
x^-2 = 1/x^2
does that only apply for the negative exponent rule?
it's y^4-7
ohhh
Yesyes you put negative exponent in the denominator
il do that with both sides?
Yes
The first one, it's correct
Yess
To the other items, just take note of pemdas rule and order of operations
It's essentially the same process
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welcome🤗
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@upbeat peak Has your question been resolved?
<@&286206848099549185>
???
@upbeat peak Has your question been resolved?
Yes
Ok then lets call angle EBA x
What is EBC?
In terms of x
Are you here?
@upbeat peak
Yes sorry
I was busy
Ok now youre free?
1 min
Ok
90-x
90 90 x-90 and f
90-x not x-90
Oh yes my vas
What does that equal to
360=270-x
i think we can prove the angle DAF and ABE equal
Yes simplify a bit
We are doijg right now
then we can prove the triangle abe equal to triangle daf
Yeah
so angle PAB + ABP = 90
and DAF + PAB = 90
then PBA = DAF
What is angle AFC in terms of x then?
sry if my english is bad
let me think
we have: ADF = BAE = 90
AD = AB (ABCD is square)
DAF = ABE (has proved)
=> Triangle DAF = ABE
=> AF = BE
is that right? anything wrong?
mmmm
idk
90+x
that is what I thought
360=90+90+(90-x)+(90+x) so the x cancels out and we have 90+90+90+90=360
Now we have angle AFC what is angle AFD
I have a lesson rn sorry
Ok
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what does the double vertical line mean? In this example
I assume it is not simply a sum notation, but rather some kind of an operation on the vector w
is it a modulus of this vector?
so it is the same as a modulus? Based on what I can see it is the same term. Thank you!
yes
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can anyone tell me in easy way to remember these 3 terms
@wraith prism Has your question been resolved?
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!volunteers
Helpers are just people volunteering their time to help you. Be polite.
I'd be polite if I'd completed this stuff an hour ago
you only asked your question <20 minutes ago
also still not a good reason to be rude and entitled
buddy im here for help with my math
bro thinks I know how long an hour is
then wait and follow the rules
is that what you need help with? 1 hour = 60 minutes
60 minutes = 3600 seconds
,calc 60 * 60
Result:
3600
correct
1 hour = 1/24 days
@slate heart Has your question been resolved?
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not sure if i did either right
im p sure i got the graph right but i have no idea how to do the second one i just kind went for it :p
@teal gate Has your question been resolved?
<@&286206848099549185>
yes.
Hello, (x-4)(x-2) is not a function. It's part of the procedure when you find the intersections of the parabolas
yeah your graph is correct, so follow it. You have 3 different intervals here
state the intersections here so the intervals are clearer
each of these intervals have a top function and a bottom function
ohhhh so i solved for the first interval
kinda
just for the bounds
which would be 4, 2
ok those are the intersections of the parabolas
the first interval from the left would go from x=1 to ...
2? ish
yes
so those are the bounds for the integral
(first integral)
when calculating are, the integrand is top - bottom
do the same for the other regions
so the bounds for each three respectively are 2 and 1, 4 and 2, and 5 and 4
idk what the integrals would be though im stuck there
it's top function - bottom function
follow your shetch
so the first one would be int2,1 (x^2-6x+10)dx - int2,1 (-x^2+6x-6)dx
ELeonardo
?
yeah
ok
sorry idk how to do that
well you usually write it all in the same integral
so like int2,1 (x^2-6x+10) - (-x^2+6x-6)dx
yes
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how do I interpret this x|x≠2
maybe it's a set that has everything except 2
if it's {x|x≠2}
i guess it can;t be that without {} so no need to guess
that's the format I didn't put the curly brackets
{x|x≠2}
right, so it's a set that has no 2, but has everything else
U \ {2} in other words
so ≠ means the number can't be used?
no it measn not equals
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hey!
im pretty sure i found a, i have the integrals for b setup (hopefully) i just have no idea if i got a right or my setup for b right and any help would be great
ill send a pic of my work <3
(a) mostly correct, but don't forget the region in the third quadrant too
the question only asks for the first quadrant so thats all i did
ok hello discord decided to die
a is fine
for b these two should be switched
y=x makes a larger solid than y=x^3
also I don't think you need the third integral
but I could be wrong
@teal gate Has your question been resolved?
ohhhh
uhmmm
why wouldnt i need the third
or idk
thanks tho
wait yeah when u reflect over a point that isnt touching i think id need to have the r(3)
is there anyway someone else can look at it to see if the third integral is needed
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.reopen
✅
can someone look at it when u have a chance!
@teal gate Has your question been resolved?
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Hey I was hoping someone could help me with these concepts. They're from an economics textbook but this discord is more active than the econ one and it's pretty much all math.
I was hoping someone could help me with the difference between strong monotonicity monotonicity, and local non satiation
in principle I get the definitions
but, say we have a vector of two goods. if we assume strong monotonicity, then y>=x means y is preferred to x
this condition implies monotonicity, and local nonsatiation
so say we have two bundles (x_1, x_2) and (y_1, y_2)
let x = (1, 1) and y = (1, 1+e/2) for any e>0
as far as I understand the definitions, y would be preferred to x under strong monotonicity AND under local nonsatiation, but not under monotonicity
sorry I should mention that squiggly > is a binary relation
@nimble tartan Has your question been resolved?
@nimble tartan Has your question been resolved?
well what does >> mean
a vector is strictly larger to the other vector
based on what?
[1 , 1 , 1] >> [0, 0, 0] but [1,0,0] is not >> than [0, 0, 0]
@nimble tartan Has your question been resolved?
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x^x^3 = 36
$x^x^3 = 36$
Jatin
Compile Error! Click the
reaction for more information.
(You may edit your message to recompile.)
its x to the power x to power 3 or x to power 3x
The first one
write it as (x^x^3)^3 = 36^3
then try msking both sides of the form y^y for some y each side
@winter whale do you mean $(x^x)^3$ or $x^{(x^3)}$? these are not the same.
Ann
i was asking OP, not you.
wht?
i was asking OP what they meant. i was not asking you, @opaque bolt.
i alrdy asked whats the que he said me that only

@winter whale Has your question been resolved?
how are you allowed/expected to solve it? what's the syllabus that you're currently on?
its a algebra que ig
i have a soln but dont know it right or wrong
i do have an exact solution, but i want to know what he is expected to do
cross or GG
tell
we're not supposed to give the answer directly
dm ?
Here
u got it right?
that would be correct
This was an Olympiad question
the way i solved it initially would not be accepted on the olympiad :3
Well regardless yall ggs
u testing helpers

💀
what, you dont like being tested to keep sharp?
anyways if its done the .close?
.close
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ok so i need some help clarifying these wack ass words in logic:
A only if B means A => B
A if B means B => A
A if and only if B means A <=> B
A unless B means A => B
are these right?
mostly just the only if and unless parts
A only if B means B has to happen for A but B can happen without A
A if B means A has to happen for B but A can happen without B
A if and only if B means if one or the other condition is satisfied, the other condition must exist, so in that way it’s “bidirectional”
A unless B means A happens in all scenarios except those where B happens
but still if you think of them as A —> B and B—> A it makes more sense why if and only if could be an arrow pointing in both directions
So you want to translate the plain english to logic, correct?
The first three would be somewhat right (you'd use A -> B, B -> A and A <-> B in logic usually, the double line usually is used in maths because typically assumes that you already come from true statements)
The fourth one would be:
¬B -> ¬A
Isn't not B -> not A the same as A -> B
I think fourth is (not A only if B) i.e. (not A -> B)
Yeah you’re right
@dull sequoia Has your question been resolved?
so A unless B means B => A?
no wait that's not right
not B => A
you can't win unless you finish first
you didn't finish first => you cant win
hmm that's true
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I asked this question earlier but the anwser we came to as a conclusion was wrong
MelMetal was helping me i think
yea the anwser is 17496
omds
im so sorry
i literally came in to check the anwser
and i remembered it wrong
then i checked my work
because im driving
ahh
4 x 6 x9 x9 x9 is right
so sry
i didnt
its right
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Well in the above question the cylinder cut's the another one and my teacher told me that the sphere is obtained and the projection will be circle But I think that we will get cylinder as a cutted region what do you think about the answer
,rotate
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Ok sorry about that I was a little impatient
Well I know the projection is cylinder
But I am talking about the cutted region whether it is cylinder or sphere
But both the 3d shapes have same projection and that is circle
<@&286206848099549185>
now I can use this tag right
@elfin badge Has your question been resolved?
No
@elfin badge Has your question been resolved?
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is this correct
@red talon Has your question been resolved?
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I dont understand why this evaluates to b and not d?
show ur work
There was no work really. I just used theorem 1 here
those are not theorems.
also, use a trig sub here
x = (3sinu)/2
What would be the right thing to call them?
generalised results/formulas
there u have -1 before u , in ur task u have -4x
its not same
Ah, my bad
dopediscorduser
Like this?
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Hello. Calculus questoin, I got confused by this "prove" question.
HqppyFeet
there
I didn't fully understand how to proceed with this. I made it so that z=x+yi and w=a+bi
thereafter the conjugates of these two would be \bar{z}=x-yi and \bar{w}=a-bi... but what
How would you express z+w in (...)+i(...) form?
You could, but it's easier if you keep things in Cartesian form
can someone help lol
z + w = x + iy + a + ib = (x + a) + i(y + b)
bruh this is OCCUPIED, go to the Math Help AVAILABLE channels (they are actually caps, im not yelling)
If you take the complex conjugate of this, you'll find a familiar result
Oh this makes sense. Ah I should've known
thank you ❤️
No worries! It's easy to overthink proofs
.close
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hi i need help with math can i send it here?
yes
Start by calculating the side lengths in each right angled triangle starting from the biggest one the one with 36,39
but i don't understand how
Using pythagorean theorem
I don't think i know any other ways no
@faint hazel Has your question been resolved?
for this youre gonna need the Pythagorean theorem, which states:
c^2 = a^2 + b^2
c is the longest side of the triangle (the hypotenuse) and its opposite to the right angle
a & b are the two other sides which are next to the angle
careful we said c is the side opposite to the right angle
in this case we can see that the side opposite to the right angle (c) is 39, so you just have to solve for a
So I got the answer of the largest triangle wrong?
yeah you need to solve for a since you have c already
again c is the longest side and is opposite to the right angle
But
Is it right now?
nvm
it s not
yes!
you gotta repeat it a few more times but its the same thing just different numbers
now using the new a = 15 side you continue with the next smaller triangle as rosa said
and remember to check which side is the c side for the next triangle 😉
Ohhhh yeaaa
Sorry
I forgot
Okayy nice, Thank you very much. @spare forum and @desert pivot
But I have one more task that I don't understand
with u
I need to calculate the ABC of a triangle
a) Angle sizes
b) Edge lengths
c) And the inside of the triangle
But how do you do solve it?
Where is x?
x =13
I don't understand
How did you get the answers?
I need bc angle, ac angle and ba angle
what Ali or Alj im not sure is saying is that the three angles A B C when added are equal to 180 degrees, so
A + B + C = 180
then we substitute the equations for A B and C in there, solve for x and youre good to go from here :)
also for the sides you use the pythagorean formula again but this time solving for y
But how do you know that angle A is 53
so in the formula A + B + C = 180 substitute all the info of the picture in there and solve the equation for x
then plug in that x to the individual formulas for A B and C to find them
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I thought average velocity is slope
i tried the first one
where it'd be (2,28) and (3,54)
then did slope formula of
54-28 / 3-2
got 26, it wasnt right
It's asking for the 3s starting from t = 2
So you need to find the slope between t = 2 and t = 5
So find slope with (2,28) and (5,102)
ah okay got it
@stray socket could you explain this one please?
I thought it'd be (x+h) (or a+h) bc 1 = a
With that problem, it leads me confused onto this one also
@chrome berry Has your question been resolved?
sorry just seeing this
(x)^13?
hmm thats right but that makes no sense ot me
$\lim_{h\to 0}\frac{(x+h)^{13} - x^{13}}{h}$
Umbraleviathan
1
So what the function, and where is it evaluating its derivative at?
(x)^13
Same deal. 16 = 2^4
I dont understand
rather I get how theres 16
got it
okay thank you
.close
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how do i do 14
If the cosine of theta is root 3 / 2, then you can use the inverse cosine of root 3 / 2 to find theta
do you know your unit circle?
look up the unit circle
cos is x sin is y for the ordered pairs
can u rexplain
there's ordered pairs for specific angles
yes
yes
ouhh
so for 30 degrees
x
how did yk it was that one
it tells you
cos150 is also sqrt3/2 yes
😭
in your problem it says theta is from 0-90
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@lean otter Has your question been resolved?
don't you have it written correctly there?
Ik but it’s asking for a single surd
<@&286206848099549185>
buddy I think that's a single surd unless you're looking for sqrt(288)
Yes
That
Is the answerkey
How to get that
when you went from sqrt(4) * sqrt(3) = 2* sqrt(3)
you converted sqrt(4) to 2*sqrt(1)
sqrt(1) = 1
so to put something back under that root sign (forgetting what it's called right now)
you'd square it
i.e. 2 = sqrt(2^2) = sqrt(2 * 2) = sqrt(4)
so do you see what doing the same to 12 would look like?
@lean otter Has your question been resolved?
Wait i still@dont get it
sqrt(288) = sqrt (2 * 144) = sqrt(2) * sqrt(144) = sqrt(2) * sqrt(12 * 12) = sqrt(2) * sqrt(12^2) = sqrt(2) * 12
@lean otter
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What mistake did I do in tryna solve for x
I can't read your last line
how did you get from -5x + 42 = 12 to 5x=54
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Factor 4x^2-27x-40 = 0. Answer is (x-8) (4x+5) = 0 at least without fully solving it's the answer. My question is why isn't it (2x and (2x since 4 is factorable by 2 and 2
well you could write it as (2x - 16)(2x + 5/2) but that is a lot uglier
Can you explain why it's x and 4x instead of 2x and 2x
I had another question where 6x^2 was 6x and x instead of 2x and 3x
So I need to understand why
it's entirely a matter of convenience and desire for integer coefficients, nothing more..
i really don't know what else to say, sorry.
Ok well thanks for trying I still didn't understand but thanks
@cobalt acorn Has your question been resolved?
Oh okay thanks
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Hello I have put images of my working and the answeri
I would like to know where the sin(X) went
u-substitution while integrating sin(x)e^(-cos(x))dx
u = -cos(x) du = sin(x)dx --
y = (e^u)du = (e^u) + C = e^(-cos(x)) + C
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September 1 enjoyer
September 1 enjoyer
September 1 enjoyer
September 1 enjoyer
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,rotate
I have no idea what to do from here
You can cancel (s+t)
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I didn't understand the argument
They should use [] bracket?
well if both endpoints are included, it wouldn't be a function anymore
For example, with their first definition, if it were to be [-pi, pi], Arg(-1) could be pi or -pi
And then again, this wouldn't make Arg(z) a function at all
that's not what bijectivity is...
what I'm saying is that for z=-1, the function Arg would have two outpus if the range were [-pi, pi]
Bijectivity on the otherhand says that every element in the codomain can be mapped to
and to be clear, the definition of "function" that I'm talking about here is a single-valued function
Yes i was thinking that the function should get one value only
Okay so another question is (0,2π]
Why are we using two different terms for range?
two different terms for range?
Can you specify what these are? Since I don't understand what you mean by "different terms"
we actually do. Generally, the defintion for principal argument will output values in the interval of (-pi, pi]. But you can also use another definition that suits your preference.
But of course, be consistent with it
like they said
...there is no general convention about the definition of principal value...
I don't think I get your question
Are you asking if we can find what the range of f(theta) = e^(itheta) is or...
I meant can we not find a range of this function?
Pr interval for this function like period?
@wind stream
Period of it?
,w period of e^(i thita)
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Perhaps try rephrasing your question?
Click here to refine your query online
is this a file that you are trying to send?
In the file it is just euler formula
alright, so you are asking about the range of e^(itheta) = cos(theta) + isin(theta)?
it's not really something that has a "range"
it's not really a function that outputs anything, just an equality that is true
I see
if you are asking about the function f(theta) = e^(itheta) for real values of theta however. This function outputs complex numbers that has magnitude of 1
,w range of arctanx
Why are we using interval (-π,π] it should be (-π\2, π\2]?
I don't see why it should be (-pi/2, pi/2]
Is it not arctan (y/x)
no, generally Arg(z) != arctan(y/x) for z = x+iy
you will see why this is a problem for complex numebrs that lie in the second and third quadrant on the complex plane
and of course, evidently through your wfa search
I can't search for it
Can you please do it for me?
So that i can visualise
Thanks for that complicated part explanation
Iirc, there is a MathStackExchange for this discussion
Please provide that link
or you can search it...
oh wait nvm, it's not that
There was one where they had a derivation for it
oh well. Can't find it.
Anyway, the derivative is as simple as considering cases. The first obvious one is complex numbers that is in first and fourth quadrants. Meaning their real parts is positive. And these numbers have argument ranging from (-pi/2, pi/2)
Then, the second case if for numbers that are in second quadrant, these have arguments ranging (pi/2, pi], Hence you want to add pi, giving you arctan(y/x) + pi
And similarly, you have a case for complex numbers in the fourth quadrant
Then consider the cases where z is (pure) imaginary as well, since this is the case where the real part (i.e x) is 0
You actually don't need to specify that Arg(z) is undefined for x=y=0. This is already implied if you don't specify it
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how to simplify to simplest form
Dyssrupt
use the first one.
yeah
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Hello
If I were to use the integral formula for finding the length of a curve
To solve for y(x) = 2 + 5x^3/2, for 0 ≤ x ≤ 10
Would I just integrate the 5x^3/2 then apply the limits to find the solution
so for example 7.5x^1/2 into the formula
sqrt(1+(7.5x^1/2)^2))
sqrt(1+(7.5(10)^1/2)^2))-sqrt(1+(7.5(0)^1/2)^2))
You integrate the whole expression, not just the stated derivative.
So I would sub u for 1+(7.5x^1/2)^2)?
You are trying to solve $\int_{a}^{b}\sqrt{1+\left(\frac{15}{2}\sqrt{x}\right)^{2}}dx$
Chixen
I guess u sub would work
Try $\frac{p}{q}$ instead
Chixen
or $\sqrt{x}$
Chixen
Asahi
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Anyways, all I wanted to know was weather I could just sub the limits in at that point or weather u needed to be subbed in
Yeah, sorry if I phrased it badly
So I'm on the right track with it being u = 7.5x^1/2?
Yeah that for 1+ that.
oh?
There’s a square
u = (1+7.5x^1/2)^2?
No, where is the square in the integral?
From here?
2 + 5x^3/2 should equal 7.5x^1/2 when integrated no?
dropping the power of 3/2 infront of the number (5x) and then reducing the power by 1
What did you mean by this?
.close
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Sending this again cause I forgot to say "no" to the bot >_> Basically, new to complex analysis, cauchy residue theorem and whatnot. Tried this integral, and the result I get is not the same as wolfram's. So I need help understanding what I did wrong. Don't hesitate to ask me to clarify bits of the image as it's not very good :)
@tepid walrus Has your question been resolved?
You'd probably have better help in the advanced channels
Oh perhaps yeah
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how to do part D
what did you get for (c) ?
yes since B is diff so B is cont and (B(4)-B(1))/(4-1)=2.5 by MVT
so the function for chloe's velocity between 0 and 2 is te^(4-t^2)) and you can use MVT to get B for that interval
then you use an equality and solve for t in chloe's equation
and see if that t is within the range
dont i have to find if there is a t such that B(t)=C(t)
so can i use IVT on B(t)-C(t)=0
using the intermediate value theorem
yeah
if you plug in values of 0, 1, and 2 into C(t)
then you can see the values of B(t)-C(t) and see if they would go through 0
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How do you integrate
1/x?
But we solved this in reverse, like how are we supposed to know we start from ln f(x)
Consider how one function in the integrand is the derivative of the other
One of the first integration methods we learn is a great tool for that situation
When you do u-substitution
what do you do?
You solve for u'
then obtain du
then how do you plug that back in the integral?
in particular, what is du
$\int{\frac{f'(x)}{f(x)}}dx$, let $u=f(x)$ and $du=f'(x)dx&, $\implies \int{\frac{1}{u}}du$
lexitorius
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👆
Oh
Do you understand now?
Cool
Got it thanks
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Using interval notation, describe the set of all numbers x such that |x^2 + 8x| > 8
write it as two inequalities without the absolute value
if you’re unsure on how to do this, think about how you would do it for |x| > a
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speed = distance / time
Ok
I understood that part
This is supposed to be Solve inequalitys, and I’ve literally got no Idea what to do with this one question
well try to apply this
you are given the distance and the time
and I guess they want an inequality, so it would be speed >= distance / time
as it says at least 250 miles
Im sorry I have a smooth brain, Ima need a tiny bit more help
Everything, I only got 2 numbers to work with, and it’s telling me to find the average speed with only those 2 numbers
yeah
do you understand that the speed of something is equal to its distance divided by time?
like your car
drives at speeds in km per hour
(or miles per hour)
:/
so is this a no?
Its a No
what part do you not understand
Never mind im sorry for wasting your time
