#help-27
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cos 3/5 cos 5/13 + sin 4/5 sin 12/13
and then i just put that into the calculate to get an answer
hm
Those fractions don't go in the trig functions
Those fractions are the values of the trig function
if cos(a) = 3/5
don't write cos(3/5)
just replace the whole cos(a) with 3/5
sin^2 x + cos^2x =1
you got the value of alpha?
because you have cos(a) = 3/5
not a=3/5
yeah
so in the future if its a = 3/5 i would use it how i did previously but if its cos(a)= 3/5 i can just put the raw fraction in
yes if you had a value for a, you could just plug it in for a
but in these types of problems, you're usually just playing around with the trig values
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reopen the channel
.reopen
✅
@restive river ^
You can ping the 'helpers' role after 15 min, but please don't ping individual helpers.

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Show your work
So ,
Perimeter of Circle = 2Πr
Perimeter of Semi Circle = 2Πr / 2
= Πr
Radius = Diameter / 2
= WX / 2
= 14 / 2
= 7 cm
Perimeter = Πr
= 22/7 * 7
= 22cm
Perimeter of square = 4a ( but here its 3a )
4 * 3
= 12
Now 22 + 12 = 36
But the answer is incorrect its B
wait .
How to find WT and XU
we need to add that also
so how?
Well you know WX, and you know TU since STUV is a square
@zenith lagoon Has your question been resolved?
<@&286206848099549185>
ohh
yes
alr so one side of the square is 4cm
and the diameter of the semicircle is 14 cm
so the radius is 7 cm
perimeter of the semi circle is 2pir/2
which is pir
so perimeter = 22/7 x 7
= 22cm
now since its on one side of the square we will subtract 4cm from the perimeter of the semicircle
so the perimeter now is 22-4 = 18cm
now add the remaining three sides with 18 cm
so 18 + 12 = 30 cm
there ya go
but 30 is not in the option
what are the options?
^
@zenith lagoon Has your question been resolved?
You know WX = 14 and TU = 4. Use this to find TW and UX
@zenith lagoon Has your question been resolved?
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Hi :) struggling to setup the integral here, I can't seem to find one without y^2 (need y on its own so they don't evaluate to 0, i think?)
because they're square I thought the integral $\int_{-2}^{2}y^2 dx$ works
Syrenate
but it just gives 0, obviously
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i do
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Oh ur right ty! Is everything else ok?
,w int 1/(sinx +1)
Ok tyy!
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hey guys
Integration doesnt work like that
This is correct.
My school called it reverse chain rule. But only works for linear inside.
Oh my god it does
@raven cape Has your question been resolved?
It's a useful trick sometimes. 🙂
You just have to beware answers in the back of the book might look nothing alike though.
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Would this be the correct answer?
#❓how-to-get-help delete your post
Seems good
Thankssss
Close the channel
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yeah it should be ab not b^2
weird
this is khan academy and hes like 10 minutes past this without correction
thanks
.close
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How to find shaded area
nothing is shaded
The pee color is apparently shading
everything looks pee colored to me tho
smh
Yeah
What though
Calculate the integral of y
there are only 2 things to integrate
one of them is very boring to integrate
and too easy
either you can subtract the integral pi/4 to 3pi/4 from the area of the rectangle
the other one is less easy
Between pi/4 and 3pi/4
yeah
or you could calculate the rectangle area and subtract integral pi/4 to 3pi/4
either way would work
why would you not?
I see that works too
Idk maybe he can't see the pee colored part
Nvm
pee on the question
2x+cotx?
You are right sorry
After integration
I am either an idiot for doing the integration wrong or substituting with the bounds wrong
Or both
looks right to me
After substitution is it not pi/2 + 1 - 3pi/2 +1
So -pi/2 + 2
Ah I reversed it oops
I am an idiot I got it thanks
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hey
RickyDicky
from -inf < x < inf
i have to find the value of K and the distribution function F_x(x)
do you know the defining property of a probability density function
well, one of
but the most important one
sure but thats not the one i meant here
no, i meant that $\int_{-\infty}^{\infty} f(x) \dd{x} = 1$
Ann
hmm
what is this exactly
like why is it = 1
im really new to probability so i just need a little bit more explanation if you could offer it
$\int_a^b f_X(x) \dd{x} = P(a < X < b)$
Ann
where X is your random variable and f_X is its density function
thats the point of the density function
but yes
RickyDicky
\implies
ah
$F_x(x) = \int_{-\infty}^{\infty} \frac{1}{\pi(1+x^2)}\dd{x} = \frac1{\pi}\cdot\left(\arctan(x) + \frac{\pi}{2}\right)$
RickyDicky
okay
and now what about finding the variance of this
var(x)
expected value and variance
your notation there was bad but i got too distracted to point out how exactly
$F_X(x) = \int_{-\infty}^x \frac{1}{\pi(1+t^2)}\dd{t} = \frac1\pi \left( \arctan(x) + \frac\pi2 \right)$
Ann
this is what it should have been
Right
thats what i did in my notebook
but went too quickly on writing the tex so i messed it up
from -inf to x
Ann
in your case the expectation ||does not exist||
how did u figure that out
i happen to have seen this particular problem before, but int x/(1+x^2) dx is not hard to calculate symbolically
$\frac1{2\pi}\left[\ln(1+x^2)\right]^{\infty}_{-\infty}$
RickyDicky
the integral does not converge
yes exactly
$Var(x)=E(x^2)-[E(x)]^2$
RickyDicky
when the expectation fails to exist, so too do any other higher moments.
RickyDicky
woops
yeah okay
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quick question, we have to like find x here: 2x^2-4x=0
i wrote:
2x(x-2)=0
2x=0 —> x=0/2=0 and x=2
is this correct
yeah
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yes mostly, please write "or" instead of "and", x cant be 2 values at the same time
yeah yeah i wrote or but i got confused on discord i thought saying or triggered like a channel so i just said and
thanks again :)
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Help
what are you asking?
To solve the question
the english in that is kinda, weird
It's still understandable though?
if you want i could speak in your native language if that is better
Do you know french
oui
French isn't my native but if you can understand the exercise in french and explain it to me in English that would be great
d'accord
envoyez-le en francais s'il vous plait
Looks like the straight line equation question which asks about y-intercept and slope and then asks for the equation itself
hehe okie
Alright here
f(1) = 2
Wow
Chief we don't give answers. We explain them
f(-2) = -4
oh ya i forgot lol
Thank u
for finding f(1) and f(-2), you need to look at the graph
I see, so how do I find it
where does the arrow hit the line @robust mason
3?
my bad, what i meant to say was where does the arrow intersect with the line
the line is f(x)
By the way sorry but where do we get the x from
we are asked to find f(1)
that's ok
So how did you find f(-2)
i will explain a bit better with this graph
this is the graph of the function f(x) = x+2
the axis which is horizontal or lying flat is called the x axis
the numbers on the x axis represent our x values
So x=2?
for f(1)?
Ah
Yes
ok well x doesnt equal 2
f(1) = 2 yes
but this is only when x =1
f(1) = 2
the thing inside the bracket is x
I see
look at when x = 2
f(2) is equal to 2+2 which is equal to 4
f(x) = x+2
f(2) = 2+2
f(2) = 4
look at the point which is right above 2
Damn so complicated
f(2) =4 right?
Yes
Oh yeah
look at when x =0
when x=0
f(x) = x+2
f(0)=0+2
f(0)=2
notice that the point above 0 is 2?
btw we can start calling "the point above" as y
or better, f(x) = y
y=x+2
do you see the pattern now?
the flat line is the x-axis
we call it the x-axis because it represents the x values
the line which is standing is called the y-axis, it represents the value of the function
Oh I see
what is f(1) in this case then
2
Damn my brains fried wait
ya bro just watch organic chemistry tutor
Whats that
you are laughing at yourself
Thanks for pointing it out
@robust mason Has your question been resolved?
@robust mason Has your question been resolved?
My question has not been solved
looks like your channel got trolled. can you close and open a new channel with the problem and your current status?
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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
@stark obsidian Has your question been resolved?
.close
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Ideas to show the sequence -1,1,1-1,-1,1,1,... and sequence -1,-1,-1,-1,1,1,1,1... without using mod?
Solved.
.close is the command u want
a_n = 1/3 - 4/3 * cos(2pin/3), starting at n = 0
Not any ideas yet for the second one
.close
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Thanks, this idea is better then mine
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Hi, can someone explain this to me?
This seems to be correct:
But using the quotient rule I get a slightly different result:
What am I doing wrong? I'm going crazy
would help if you showed what you tried to get there
,w diff e^(x/2) / (x^2+3)
for one: how did the 2 in the denominator came to be?
I just used the quotient rule, so the numerator being f(x), and the denominator g(x):
f'(x)g(x) - f(x)g'(x) / g(x)^2
you put the 2 you get in the denominator from differentiation e^x/2
But that only affects the first term, you need to get a common denominator with the second term befor eyou can do that
I differentiated e^(x/2) to get (1/2) e^(x/2), and just pulled down the 2
is that wrong?
yes
because now it affects the -2x * e^(x/2), which it shouldnt
until you get a common denominator there
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idk where to start
do you know the definition for a probability density function?
would be helpful to review your notes then
basically, for (a), we use the definition to equate the integral of f(x) to 1
(ie "the sum of the probability of all events is 1")
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I’m not sure where to go from here
make a substitution for 7^(x/2) and solve a quadratic
yeah
Hmmm how would I turn that 7^x/2 into a quadratic to power 2?
let u=7^(x/2)
what does this become when in terms of u?
That’s smart thinking that. The 7^x/2 would turn into u and -7^x would be -u^2
Hmm, interestingly I get two imaginary roots, which I know i dont want
hmm
would "stretch of half" perhaps mean that the first line should be y=7^(2x)-12?
I thought when you stretch something by b it meant that f(x) = (bx)
apparently its x/b
the point at x=1 of the old function needs to become x=2 of the new
,w roots x^2-x-12
roots make more sense too
The question asks for the intersection y = A
So which value do I sub in
you still have to take logs
Could you drop a hint on how I would continue on to do the logs for it
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if we have a loop that iterates i and j in it like
for (int j=0; j<10 j++)
WORK
then this is 10^2 of the work like this
**********
**********
**********
**********
**********
**********
**********
**********
**********
**********
but if the pattern is
for(int i=10; i>0; i--)
for (int j=0; j<i j++)
WORK
**********
*********
********
*******
******
*****
****
***
**
*```
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@proper thicket Has your question been resolved?
@proper thicket Has your question been resolved?
cos theorem?
@proper thicket Has your question been resolved?
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Can someone help me with this linear equation
And explain it to me
not rlly
i can
i think
um 0?
um
dontknow🥺
i dont know where the 10 is coming from
oooohh
10
is this connected to polynomials
yes
kinda
i know more than before
wait a min
is it 2?
oh i think its 8
oh wait
I think I got it
if you cant understand handwriting its x^2-16=x^2-8^2-(x-8)(x+8)
<@&286206848099549185>
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.close
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Is it the 2nd bullet?
The point is negative for the function but it’s increasing based on the first derivative
i think you're mixing up x and y here
Oh so it’s f > 0
yes
Alright and then f’ > 0
yes
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@prisma lava Has your question been resolved?
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idk how to solve this
<@&286206848099549185>
!help
Please read #❓how-to-get-help
<@&286206848099549185> ❤️
.
Please read #❓how-to-get-help
No. Actually read the link.
Ok
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As u know distance is speed x time
Just assume that he traveled x kilometres with bus
Then the distance he traveled with car would be 520-x
yeah
Let t = time, let d = distance.
t_bus + t_car = 6h.
d_bus + d_car = 520km.
The two equations for the distances traveled by the bus and the car would be:
d_bus = 80(t_bus)
d_car = 100(t_car)
Then you just have to do some substitution.
how
Ok I'm back
So the total travel time for bus will be x/80 , and for car it's (520-x)/100
You're given total time = 6 hours
So x/80 + (520-x)/100 = 6
Find x from this
Nah that's too much work
The question asks for the time, so it's much convenient to set the unknowns as the time
i remember you jay
Yeah I remember you too
so how do i do this the other guy keeps leaving after saying a couple things
So the question asks for the time spent on each the bus and the car, right?
yeah
If you say that the time on the bus is x and the time on the car is y
What would x+y be?
ok
Now you need to set the second equation using the total distance
Since the bus' speed is 80km/h, how much distance did it travel after x hours?
You need to express the distance the bus traveled using x
80(x) + 100(y) = 520?
Great
That's your second equation
Now all you need to do it solve x and y using the system of equations:
x+y=6
80x+100y=520
It's the same thing as we did last time
x=6-y
Great
and then substitute
80(6-y) + 100(y) = 520 right?
y =2
4
Great
So you're answer would be
Time spent on bus: 4hours
Time spent on car: 2hours
ty
Np
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how do i find common denominators of 1/2 + 3/4
Please don't occupy multiple help channels.
??
like factors of both of the 2 denominators?
no idk
ok
there are a lot of common denomaintors?
ok thx
i have another question
how do i represent these fractions like in that picture
a
1 a first
umm idk
it’s b?
nvm wait
i think it’s c?
idk.
i wrote 3/4 and 1/2 idk wym
yea same
it’s c right
help
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Is this mathematically solvable?
Actually these other 2 equations are also available if needed
Wait, x3+y3+z3=36 not 11 😅
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How is 3A not congruent
the SAS case of congruence requires that the angle be between the equal sides
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BCA and EFD are both acute, so they are congruent tho?
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I can send questions here right :>
you not only can but should lol
anyway
!status
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1 and 3
show the answer you got
the only four ways you can have girls not sitting adjacent is GBGBGB, GBGBBG, GBBGBG, BGBGBG
so 4 * 3!^2 surely
yep thats correct, but how did you know to ^2
oh nvmm!! I see it now
You're a star! Thank you!
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All the angles are equal in octagon. Find Area
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help please
how does that achieve its global max and min
between those 2 points?
<@&286206848099549185> 🙏
The function is only defined on the domain [a,b], so the global maximum is on that domain.
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we throw a die n times, let X be the number of times 5 is received, let Y be the number of times 4 is received
express using n the covariance of X and Y
lets see
[
cov(X,Y)=\sum_{x}\sum_{y}((x-\mathrm{E}(X))(y-\mathrm{E}(Y))P(X=x,Y=y)
]
metnal
im not even sure where to start with this
I think maybe break X and Y down into events representing individual dice rolls somehow
though I don't really understand the formula. I'm used to E(XY) - E(X)E(Y), is that one allowed?
no, I suppose I get it
yeah its allowed
In that case, calculate E(XY), E(X), and E(Y) for one trial
One dice roll
Then add up all the rolls using linearity of expectation
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How do I solve this
,tex \factoid original
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how do i convert the first equation to the second on
Given a right angled triangle. Determine the length of the median belonging to the hypotenuse,
if the legs are 12 cm and 16 cm long.
@tame ruin the other person got here first -- youll have to open a new one
oh you did alreadyt
ok
yeah
so its -9(....)?
-1
1
yes
i think after multiplying is just -9((-2/3)^n-1)
2x+6
-2x-6
wdym 2 common
2(-x-3)
But (-2/3) is still negative even after multiplying
How come?
so when im multiplying -1 by 9((-2/3)^n-1) im multiplying -1* 9 and -1(-2/3) and -1*-1?
im still a little confused
1
positive
But you said that the insides wasnt multiplied but was apparently left out
Thus, i said that 1 is positive which means that it was multiplied
if i multiply it inside the bracket it would be 1(2/3)^n)...
since -1* -2/3^n and -1 *-1
is it possible if you like write it out? i think thats easier for me to understand
ahhhhhhhhhhhhhh
okay nowi get it
wait why did u delete it
ok
yep got it
thx btw the arabic name guy
idk how to @ your name lol
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hi
ive tried doing that by subbing A and -9/40 into the tan(a-b) formula but I think thats wrong
any help?
@short tendon Has your question been resolved?
pls dont tag mods for math help
yeah
And do you know what tan(A/2+A/2) is?
Not really ;)
?
2tana/2?
wait no
its that formula no?
Which one?
..
Forget about everything else, and just focus on simplification of this
There is one very obvious thing you can do.
you could also ignore the tan
yhhhhhhhhhh
So
You are supposed to find tan(A/2)
You shouldn't use arctan I think
why not
You should us formula for tan(A+B)
Which is this thing
It would be too ez if you could just arctan it
when i put that into photomath i get 0.22
might be so
because its only part (ii) of a q
That's true but that's kind of cheating xD
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Please help
where are you stuck
!status
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6. None of the above
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(algebra, groups and rings)
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Am I on the right track here for finding the range or way off the mark?
I’m not sure what I need to do for the inverse equation.. isolate y on it’s own, yeah? So is making the inverse equation = 0 a bad idea?
you may have made your life somewhat more complicated here, but i'm not sure to what extent.
probably up to and including the penultimate line there's nothing you could've done significantly better
Yeah.. this is a long convoluted way to get to the range
wait, you're asked to get the range? just that...?
Yes
might you be better off finding the extrema by calculussy means
I thought I would just replace x with the smallest domain value (which is 2) but that’s not correct
Calculus is used to find range?
What I mean to say is, calculus is more effective than algebra for finding range?
for such mildly-complicated functions, yeah, i'd say so.
Finding the range can always be done both ways, however right? Via algebra or calculus
with algebra you may struggle a lot
Yes, which is where I am at
I’m looking at =0 and thinking it may give some false positives
I could divide and multiply the left hand side as much as I want, to remove terms, and it will still equal zero
But that may also be risky and could be eliminating answers, I’m not sure…
I keep hearing typically don’t wanna do a lot of this for equations that = 0.. not sure if that’s true
The right hand side will always be zero, so long as you are not dividing by a variable
I guess I have run into my “limit” with algebra. Time to learn calculus to help solve range
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Hello
Can anyone help?
Man find another channel
!help
Please read #❓how-to-get-help
and delete your message
what are the conditions on a,b,c?
They belong to R*
prove $a^2 + b^2 + c^2 \geq ab + bc + ca$
tushar
Ok and after that?
set a := a/b, b := b/c, c := c/a
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Can a log have a negative result?
^
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@uncut quarry chill man he closed
I did nothing man
You can add something if you want
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I am currently working in my math investigation for school. My inicial idea was to model access ramps and use optimization to find the best relationship between values (like length, cost, gradient etc) but after some drawing and starting my calculations I realized that all the ideas I had have a linear relationship. I am stuck on how to approach this investigation. I am open to suggestions or just your thoughts wether this topic is worth pursuing. thanks
Looks like you want a multi input function?
im not sure what that is
this is a pretty hard question to answer as we don't know what your school want from you in an "investigation"
its from the IB diploma programme it is for the Internal Assesment

do you have any math that you need help with?
I am struggling with ideas to then actually create a rough draft
Actually I want to ask
this is probably better for #discussion
How is your gradient thing a linear relationship
I’m pretty sure most escalators are only 30 deg
Going any higher pretty much makes it impossible to climb at all
And escalators have steps
what i found was that access ramps have a gradient between 1/12 and 1/16
and I couldn't equate it to another variable without the linear relationship
Also, isn’t length directly linked to the gradient
yeah
I think the main problem here is a not very well defined “best relationship between values”
Best based on what?
My idea was to use 3 factors as the judge of that which would be the length someone would have to walk to reach a specified height (I was thinking of climbing 3 meters in height) the cost of building it assuming it was just concrete and the gradient of the ramp

