#help-39
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after finding g'(x) and plugging in the values into the power rule what do i do 😭
do you by chance know what the product rule or chain rule is
oh my bad i shouldve put g(0)
f’(0) f(0) although idk if thats g’(0) i just know its the derivative 😭😭😭
how to solve
plug in 0s
and the multiply,?
well yeah
its asking you for the slope of k(x) at x=0
aka the derivative of k(x) with respect to x when x is 0
so just plug in and solve for the slope of the tanget line at x=0 of k'(x)
am i misunderstanding the problem?
like that??
well yeah im not sure what else the question could be asking
just asking for the output of k'(0)
why are you confused tho
does it seem too easy or something
nono like i feel that wouldn’t be the answer but i could js be overthinking it
i mean it is the first question so its probably meant to be this easy
and if its telling you specifically that you should find k'(0) there is a good chance it is just a number
anyways goodluck
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I made an app that can
Square both sides, then move the resulting square root to one side by itself and square again
Try assuming x+1 = y² , that might help
dm me if u want to use it everyone in this sever gets premium free for life
Please do not advertise your stuff in here.
btw this has no solutions
Is it free?
if we still care about solving the problem
Bro is not slick with the advert💀
sqrt(2-2) + sqrt(2+1)
sqrt(0) + sqrt(3) definitely is not 1
same thing if you plug in 10
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not quite sure where i went wrong here
It's not clear whether a = AB or AC
,w calc (10/8) * Sin(64°)
would a not be opposide of A?
Try taking a and c as different sides
This triangle isn't possible
Where a is the side opposite to a
,w calc (8/10) * Sin(64°)
,w calc Sin^-1((8/10) * Sin(64°))
so am i just allowed to switch up the sides like that?
Yup
oh what
Unless they specify
okay let me try this out then
,w calc Sin^-1((8/10) * Sin(64°)) *180/π
Which is approximately 46°
is this correct?
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can someone explain number two to me
i dont understand how f(x) equals the sqrt of x
sqrt(x) is our "outer" function here
if you let $g(x) = \ln(x)$, then $\sqrt{\ln(x)} = \sqrt{g(x)}$
higher!
now you need to write $\sqrt{g(x)}$ as $f(g(x))$
higher!
the only choice you have is $f(x) = \sqrt{x}$
higher!
the reason we want to write it in this way is because that's what the chain rule applies to
functions of the form f(g(x))
so we need to find what f(x) is, and this is how you can tell
this part was really helpful
i think i get it nonow
tysm
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Axiom A5:
Let there exist two distinct additive inverses, $a;b$ of $v$. So $a+v=0$ and $b+v=0$. So $a+v=b+v$. We know that $-v$ is one of the many ( If there are many) additive inverses. so $a+v-v = b+v-v$. From which we can conclude that $a=b$. Thus the additive inverse is unique!
Veni, vidi, perii is not f(wai)
yes that works
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Here's my proof
Base case :- $a \cdot v_1= av_1$
Assuming this holds true up until $v_n$, we get $a(v_1+v_2 \dots + v_n) = av_1+av_2+av_3 \dots av_n$
Now we add $av_{n+1}$ to both sides, thus gettin $a(v_1+v_2 \dots +v_n)+av_{n+1} = av_1+av_2 + \dots +av_{n+1}$. Which is in turn equal to $a((v_1+v_2 \dots + v_n)+v_{n+1}) = a(v_1+v_2 + \dots + v_{n+1})$. Thus concluding our proof
Veni, vidi, perii is not f(wai)
The second proof follows similarly. The base case is $(a_1)v =a_1v$. We then assume $(a_1+a_2 + \dots + a_n)v =a_1v+a_2v+ \dots +a_nv$. We now add $a_{n+1}v$ to both sides to obtain $(a_1 + a_2 + \dots +a_{n})v = a_1v+a_2v + \dots + a_nv+a_{n+1}v$. We now apply the distributive law of vector fields to obtain$(a_1+a_2 + \dots + a_{n+1})v= a_1v+a_2v + \dots +a_{n+1}v$
Veni, vidi, perii is not f(wai)
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Could someone explain to me how/why the midnight formula / abc formula works? I feel like I’m missing some mathematical basics to understand it.
quadratic equation?
The quadratic formula gives you the solutions for ANY quadratic equation. I explain how to find it, why it is also called Midnight formula, and do some examples.
they use completing the square
ive never heard it referred to as "midnight formula"
"If I wake you at midnight you have to be able to say the formula"
bacc the sigma😔🤞
You simply copy and paste the coefficients a,b and c and compute it.
This here seems like a derivation of the formula, which can be derived by completing the square of the equation ax²+bx+c=0
Muss wie aus der Pistole rausgeschossen kommen 🗣️
@midnight haven Has your question been resolved?
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If 3(T-R)=10 which of the following correctly expresses T in terms of R?
do you have a photo sending or something?
where to begin
you can start by dividing both sides by 3
multiply 3 with the parenthesis ?
that's one way to do that
now can you go ahead and apply the same step to the question 3(T-R)=10
im getting t = 10+3R/3
Did you forget to multiply T with 3?
now divide by 3 on both the sides
im getting t=10+r
where does this come from
you can seperately divide a number to each term if it is common in denominator
oh
got it?
yea ty
@quaint crag Has your question been resolved?
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your lim resolves to -1/2 * (sinx + tanx)/x
=> -1/2 * (sinx/x + tanx/x)
=> -1/2 (1+1)
a^t-a^-t/2t=lna?
wait why a^2t-1/2t=lna?
try LH
d/dx x = 1
d/dx a^x - 1 = a^x ln a
lim x-> 0 a^x ln a / 1 = ln a
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hi
Yes
Yupsies
$sin^{3}(x)$ is the same as $sin(x) * sin(x) * sin(x)$, so if you were to write them out like that, visualizing how they cancel is more clear
m. frost
yes
how do u type like this
ok hthanks
It's a markup language called LaTeX that the bot @jolly parrot can render
oh, dont know how to use it, new here
$You type markup code between US dollar signs$ Here is an introductory guide to using it
.close
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,rotate
Need help with 13
Yeah
Use that lol
I tried and prob messed up somewhere I’ll try it again
Send your calculations then
Yeahh
Imma plug it in and check my answer
Lmao
btw dont write in such way in boards tho
You Indian too? @red hedge
How’d you know 😭
Was it my finger slightly showing in the picture that gave it away

You in which class?
12
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@red hedge f(-2) = 16 and f(2) = 6?
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Show your steps too
okay
Oh No
I just made huge mistake Brb
a) 2 m/s
b) 4 m/s
c) 1 m/s
new answers
Why are your ans in m/s lmao
mol/L/sec imo
Give me the corrected calculations lol
i gave already
a) 2 m/s
b) 4m/s
c) 1m/s
@finite stump Has your question been resolved?
I think there are correct
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what is the second question bruh
i don't understand it
Which?
Question 2
@urban sun Has your question been resolved?
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<@&286206848099549185>
@twin fox Has your question been resolved?
@twin fox Has your question been resolved?
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<@&286206848099549185>
Where that 2 and 3 came from?
Hello
🖐️
Where that 2 and 3 came from
Yes
Kinda
Thanks man
👍
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hi guys sorry to be a pain but could someone check these answers for me please? i dont want to ask teacher in fear of being wrong for simple qs
,w simplify 5x + 3xy - x -xy
,w simplify x(1-y) - 8xy + 5(x+y)
,w simplify 4x^2 -4x+7y^2 - 3y + 2x^2 - y^2 +xy
Thats a usefull command! thanks for introducing it to me :)
yes wolfram alpha is a good website
for the last question
which is a preferred way of presentation? 6X2 -4x +6Y2 -3y + xy or the one with brackets?
,w
,w simplify 4(x/(2xy)) +(6y/2y)-((x^2 y^2 /(xy))
not sure what i did wrong here
Are we not supposed to multiply 4 by the initial bracket?
due to 2xy being inside another set we cant touch it?
It's 4 multiplied by (x/2xy)
Multiplying means you only have to multiply the numerator
which is x
The (2xy) doesn't mean it's untouchable, it's there just to specify that 2xy is the entire denominator
You have an x in the denominator and numerator so they cancel
What's your concern exactly?
my answer doesnt match
so i think i have done something wrong? idk
sorry to be a pain
Huh? This isn't even the same equation as the one you sent
They're 2 completely different equations
You only care about this, no?
oops, sorry deleted the picture from the bot when asking for more information before
,w simplify 4(x/(2xy)) +(6y/2y)-((x^2 y^2 /(xy))
yeah :(
You're really confusing me here, if we're just talking about the question you sent here, then we've already gone through it together just now
Are you mixing up that question to one of the previous 3?
oh, sorry, i think i might be!
thanks for pointing out my mistake, i am less stressed now
so sorry about the confusion
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i don't know how to solve this
,rotate
try setting each quadratic to 0, and solving for x in each case
ok
one of them is going to have roots not equal to 6sqrt(2) and -6sqrt(2)
you mean making f(x) into zero so 2X squared would equal 144?
yeah
that would be once instance
though I recommend factoring instead of moving terms around
how would you factor
i got the anser by moving the terms but how do you solve it by factoring
ah, you can't exactly factor this one 
you can apply the quadratic formula though
that would suffice
@dawn folio Has your question been resolved?
ok ty
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How do this
do you know how to find the absolute value of a complex number?
$|x+iy| = \sqrt{x^2 + y^2}$
riemann
for any real numbers x, y
no
if not then its root 225 plus 64 right?
y^2 = (-8)^2
,calc sqrt(15^2 + (-8)^2)
Result:
17
yes
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are you sure? substitute i root 5 into the equations and see what you get
How did you conclude that none of them have a negative root or i
If you just rearrange B, C and D for x^2 it should make sense
it’s C
You aren't meant to just directly give the answer but a'ight
ik its c i dont how to get ther tho
you just move the constant to the other side
x^2 = -5
It should make sense that a square can never be negative unless it's not real
oh so x =iroot5
Yes
so you just solve for x?
ye plus or mines thank you @urban swift @twilit lion
np
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what am i doing wrong? im not getting the answer in the mark scheme
@supple citrus Has your question been resolved?
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<@&268886789983436800>
isn't it tan^-1(x)
yeah it is
can i just use a u substitution for now or something and come back to it later in the expression
I would give the cos(a+b) formula a try
yeah i was gonna use that usig u substitution
i saw that angle sum difference identity
with pi/2 and u
yeah i got
okay -x/sqrt(x^2+1)
so x = -x/sqrt(x^2+1)
and solving for x gives x^4 = 0
and fourth root of 0 is 0
so 0
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what'd you do
is the key wrong then?
is i on the inside or outside of the sqrt
It can be outside
Because i = sqrt(-1)
so it doesn't matter
if you rewrite i to sqrt(-1)
you can plug it back into the other square root
and take it out too
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it's a 2-degree equation
?
is there another way cause i dont know it by head
you can try factorisation
wdym
to put it in form like a(x-b)(x-c) = 0, for some a, b, c real
then x = b and x = c will be solutions
so -b/ac
what?
one sec
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hey guys im really confused on how to do part (iii)
i like dont even know where to start
if you have X as a vertex, how many ways can you choose two other vertices?
omg sorry
👍
ok lemme think for a sec
ok so if you already have x
you have to pick one from both lines
so (n-1) x (n-1)
Yo
oHHH
i think u need to use a new channel
I did no one is responding
too bad
💀
Me rn
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Dr p can you help me now then
uhh i can try
Logic

Proof tables
Cooked
😭
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what is the graphical technique from section 3.1
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how precise does it have to be
getting the general shape is easy
approximating the slope is egregious
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like do we really need to scale the axes and have coordinates
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💀
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well for the max you can look at the middle section
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where it has a point of inflection
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think of it like critical point for f’
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what
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no it more so looks like a straight line
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notice the maximum here
it changes from concave up to concave down
yea
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if you have the solution what’s confusing you
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cant you learn from looking at the solution
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you just desperately wanted to talk to me
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i get it
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well it should be obvious that it starts ~0 and ends ~ 0
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yes
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hopefully you can tell it’s suymmetrical
just by looking at it before and after the point of inflection
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yea about the point of inflection
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the shape should be obvious
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need i say more about the shape?
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let’s look at the slope near point of inflection
it’s approximately linear
so we can just good old slope formula
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little more
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the coordinates
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yuh
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notice they used the 20 for scale but it doesn’t actually go that high
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keep it simple
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15
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huh
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no idea where 1.6 came from
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the peak here is about halfway between 10 and 20
so either works
that was me making fun of you
👍🏻
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yea
perse
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yes way
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Closed by @buoyant garnet
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whom
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ah thanks ;)

Yeah ive been trying to come back to mathcord and help more
literally me

you need the green my man
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green grind starts today
Yep
yes
wrt time
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Isnt there supposed to be a negation
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yeah
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brother
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derivate of f(g(x)) is f’(g(x)*g’(x)
yep
no
in this case g(x) is 30-t or whatever
from -t
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😭
no
no
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derivative of e^(x^2) is 2xe^(x^2)
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^
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🤔
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yo kenzo we have
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what is f(x) in this case
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yes but chain rule is arguably the most important differentiating rule
and youll need it eventually
if not today in 2 weeks
this isn’t the place to learn it though
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@buoyant garnet
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ah alr
did you read the damn book
Then expand it
by mcmullen
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libgen 
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^
we can’t promote this though
i’ll dm you
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Yes
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Is derivative of 2x the same as x
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Exactly
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..what
brother
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d/dx cf(x) = cf’(x)
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constant multiple
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give your brother back the controller
okay look studying while sleep deprived is not
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You can derive constant multiple from product rule
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anyway
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whats the right one
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oh wait
The question might be kind of cheeky
wait nvm
Also the answer cannot be a positive number
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fuck it put it in wolframalpha
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,w derivative of 200(30-t)^2 at t=10
see
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Wait what are the steps
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Closed by @buoyant garnet
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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My main question is about v-w. When graphing, do you graph the resulting matrix or rather just v connecting v?
to me, this is what I see.
What I am unsure about is whether v-w would just be [-7, 9] or rather what I described earlier. Because the vector would be totally different.
u would join vector v and w, with the arrow pointing towards the tip of vector v
What would technically be the resulting vector then?
