#help-13
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<@&286206848099549185>
@lean sail Has your question been resolved?
ok so, you have a right triangle
you want to find the rate of change of the hypotenuse
one leg is the vertical distance between the radar and the bird (5km)
the other leg is the horizontal distance between the radar and the bird. After 1/6th of an hour, that's 7km
$x^2+y^2=z^2$
Scythe
x is the horizontal, y is the vertical, z is the hypotenuse/distance from radar to bird
oh
with me so far?
we don't want a distance, we want a rate
so we differentiate this!
$2xdx+2ydy=2zdz$
Scythe
this is still unhelpful, we don't want a change in distance, we want a change in distance over time. so divide by dt (and divide by 2)
$x\frac{dx}{dt}+y\frac{dy}{dt}=z\frac{z}{dt}$
ok so, now we have 6 variables
Scythe
oh
we know x: it's 7km after 1/6th of an hour /10mins.
we know dx/dt: it's 42km/hr
we know y: it's 5km
we know dy/dt: it's 0km/hr (constant)
we know z: it's the hypotenuse of a triangle with sides 5km and 7km (so sqrt74 km)
$742+50=\sqrt{74}\frac{dz}{dt}$
Scythe
34.1768 km/hr
thank u so much
glad I could help
thjis is the final answer right
how do I do b
since you sent the image sideways
issok i think ur right
well you do the same thing
you get 0*42+5*0=5*dz/dt, thus dz/dt = 0
since it's directly over for that moment its rate of change of distance is going from negative to positive, as it approaches and then leaves
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Im having an issue with a statistics question. I dont know how to attempt it
what is the question exactly?
its this
Im not sure how to do the square regression line
<@&286206848099549185> hi can someone please assist me with this question? :))
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@craggy pagoda Has your question been resolved?
@craggy pagoda Has your question been resolved?
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@smoky grove Sorry for ping, Ive had to use another channel but this is where it came from
this is from my theses and its a trigonometric series
@oblique lynx Has your question been resolved?
@oblique lynx Has your question been resolved?
@oblique lynx Has your question been resolved?
btw the question is what does the R(C) means
@oblique lynx Has your question been resolved?
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Can anyone help please?
v(t) =t²/2 -8t +c
As v(0)=3 at t=0 therefore c=3
v(t) =t²/2 -8t +3
S(t)=t³/6 -8t²/2 +3t +c
At t=0 ,s(0)=5. c=5
Therefore s(t) =t³/6 -4t²+3t+5
Just keep integrating with taking c into consideration
how’d you figure that out😲
There was a reason why v(0) and s(0) was give that's how
(B) part is same keep integrating keeping c in consideration
ohh i see. Can i ask how you got t^2/2
oh i see
Integration of t is t²/2
You may also say it as anti derivative
Yep
ok!
i understand!! thank u so much!
Could u help with B in 5 min? ill write A out really quickly
but it should be the same right
except its just cos(x)
Ye
Yep
You should
how would it look?
i know the integral sign but idk what’d be the top and bottom number
Just in case your teacher give marks on formulas you should
0 for both?
What
Kk
Yep
cool!
so then for b. I start eith cos(x)
this is the first one how does it look @cinder garnet
then i do the others?
F''(x)=sinx+c
F''(0)=sin0 +c=a
c=a
F''(x)=sinx +a
F'(x)=-cosx+ax+c
F'(0)=-cos(0)+c=b
F'(0)=c=b+1
F'(x)= -cosx+ax+b+1
F(x)=-sinx+ax²/2+bx+x+c
F(0)=0+c=f
F(x)=-sinx+ax²/2+bx+x+c
Check yourself I gotta go bye
@tough nacelle Has your question been resolved?
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Is there a difference between multivariable functions and parametric curves? If so, what is it?
parametric curves express coordinates as functions of 1 variable, x(t), y(t), etc. multivariable functions have multiple variable inputs eg f(x,y)
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how does one go about proving:\ $2k^2 > (k+1)^2 \ \forall \ k\geq 3$ \ Context: this was a proof by induction question but i've got to here
i laik turtles
only use the lhs
i know it might be something to do with writing lhs as
k^2 + k^2 then changing one of the k's to 2 or 1 so that you can write another inequality expression
but its not working
@crisp jay Has your question been resolved?
@crisp jay Has your question been resolved?
sorry, i forgot to mention all k's are integers.
and it was an induction question, initially being 2^(n+1) > n^2, for n≥3
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hi i was wondering how i should go about to solve this following question: determine the derivative function to the function f. this is my first question of this type and im completely lost
i assume you’re using the limit definition of the derivative?
wait
i don’t think you can do this using the limit definition
since it’s a fractional power
sorry if my english is off, but yes i am if that is the step i do since the exponent isnt an even number i need to find x>0. but now that the equasion it self isnt even im a bit lost
quantum
no i can send you an example of a previous question
sorry if its a bit unclear but it should be clear to understand the point of it . in the book it tells me i need to solve the question in this style.
so you need to use derivative rules
yes
well you can pull out the constant
and multiply that back in afterwards
$\dv{x}x^n = nx^{n-1}$
quantum
okay thank you
if you’re done, do .close
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currently working it through again. I have expanded and removed the parentheses
rn I have sin^2(t)=10/23
to remove ^2 I squared both sides?
so now I have sin(t)= +/- rad10/23
That would make 4th powers
sry I mean square root
@misty sedge Has your question been resolved?
So you have $\sin(t) = \pm \sqrt{\frac{10}{23}}$ right ?
Twentycents
Well there is no "simple" answer besides the one that uses arcsinus function
like this
theta=arcsin rad10/23 ?
Yep
ok simple enough, so I got 41.3
The other solution being not of interest here since t>0
Then I guess you can enter that number 👍
Okay let me try, I think I already have and it was rejected
nope got it wrong, I have to include 41.3, 138.7, 221.3, 318.7 etc
ugh, I fucking hate this prof
thanks for your help
Oooh ok
in other problems, he'll use the same exact working but we get it wrong if we dont put it like this "41.3deg+360degk" etc
he fucks us
Yeah I guess that would be from that step $\sin(t) = \pm \sqrt{\frac{10}{23}}$ to the next step
literally no diff in the wording
Twentycents
You should have included those 360degk indeed
My bad I haven't read the beginning of the question (he asked for solution_s_)
Yeah 😅 well next time you'll think about those details 👍
I did. I even solved it correctly with all the angles before I even posted here. His terrible attention to detail and lack of english is the problem.
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I'm stuck at the first part
@slow shuttle channel occupied
Any thoughts on the question so far?
Not yet
That doesn't make any sense, it needs parenthesis
AND has greater precedence than OR, no?
the braces are implied
If that is the case, then you are done
..no? My expression does not fulfil the requirements of CNF
Or, sorry it's backwards
To be clear, what I got is $\lnot p \lor (q \land p) \lor (\lnot q \land r) \lor r$
Tāhā
This is in DNF isn't it
You can convert from one to the other using a double DeMorgan's application
Been a while since I've done it, haha. I suggest Google to get that fully
let me try
@bronze swan Has your question been resolved?
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How do I solve this problem? I legit have no ideas.
wow it took me a hot second to decipher this
use multiple time thales
it's just thales and thales and thales all the way
you know when you got a triangle and a line in it (here DE) if DE||AC then you have basic formula
like AC/DE = AE/DB =CB/EB
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Hi folks. If 2^x = 5, then what is 2^x+2?
I can't seem to get around to solving this. I'm getting a little confused.
If 2^x = 5 then if we add +2 on both sides we get 2^x + 2 = 5 + 2 = 7 meaning we have 2^x + 2 = 7 ?
$$ 2^x=5 \iff 2^x + 2 = 5 + 2 \iff 2^x + 2 = 7 $$
mrbrown
Not that.
Let me write the question rq.
No problem, I was able to get the solution with my calculator but I need to know how I should solve it manually
You'd say that like 2^(x+2) 
first time writing maths equations on a keyboard :P
one of the first anyway
my bad
hint: $a^b \times a^c=a^{b+c}$
Kanga Gang ¬Sam
@graceful karma why bring up logs?
logs

so i've got 2^(2x+2)

okay let me try the hint differently? 
isn't there a specific rule where if the base = base then power = power?
wait no
alright i'm lost
today's not a day for my maths brain 
This should be the only rule you need
repost (no, that isnt it.)
wait what
wait no
why did i say +
idk

Kanga Gang ¬Sam
so do i multiply 2^x with 2^(x+2)?
$2^{b+c} =2^b \times 2^c$
PapaBread
Indeed
that got confusing for me
I think it was just because the formula was """backwards"""
You were trying to apply it left to right
Instead of right to left
maybe
alright, thank you guys a lot!
i actually wonder what you feel when you see some people struggling at a maths question that you probably consider easy :P
thanks again :)
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Hiii
I have a question related to the D.I. method
So i have $\int \qty(\frac{e^{-kx}}{x}+\frac{e^{-kx}}{kx^2})dx$
Insert nerdy name
And if you try to integrate with the traditional integration by parts the uv - int u'v one
Its easily solvable
Because int u'v of the first integrand is the same as the second so they subtract each other out
But if I try to do that with the D.I. method its not really obvious at first
I'd like to know how to spot conditions like that if im not clear with my question then pls ping me
<@&286206848099549185>
@crimson sedge Has your question been resolved?
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i need help with calc
it says Write Eq (3) in terms of A and B to derive Rule (3)
this is eq:3 cos(a+b)+cos(a-b)=2cosacosb
this
A=a+b and B=a-b.
thats it?
yes.
wait there is no working out?
it said terms of A and B
wait, is there a way for me to get the solving, not just the answer?
Upper
The solving is just look at what you need the sub to be to turn eqn 3 into rule 3.
what does that mean-
a and A are different variables.
i dont know what a sub is
can I upload a document here? maybe itll make more sense
I dont understand my teacher
Never understood any of your teachers?
I understand my algebra teacher
but calculus is hard fsr
subbing stuff is an algebra concept
Yeah part b is literally what I said.
I did b, i had help from someone
so c is obvious.
i just need to figure out the rest
it is?
apply what you just derived..
im gonna be dead honest right now
i dont understand ANYTHING youre saying
from sub to derive
god i hate this
how do i close the channel?
i dont want to solve this anymore
.close
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Bruh
Exactly bruh
I was here dude
Issok
@fierce egret I'd start by summarizing the two fractions on the right hand side into one fraction
You need to solve for the one on the right
yeah
oh right
first get the expression 1/de = .................
and then summarize the two fractions
and then you can take the reciproce value
Trying to summarize them is the hardest part
Don't I have to get common denominators to subtract them
yes exactly! 🙂
Basically it becomes this
ofc I can factor these
But then I don't know how to actually subtract them becuase theres a fraction over a fraction
you can use this $$ \frac{1}{a} = \frac{1}{b}- \frac{1}{c} \iff \frac{1}{a} = \frac{c}{bc} - \frac{b}{bc} \iff \frac{1}{a} = \frac{c-b}{bc} \iff a = \frac{bc}{c-b} $$
mrbrown
yoo how do u use texit? do i have to learn how to write in it?
wait a second 😅 we are working on Voximity's problem atm here
im relative new here, sorry if its dumb question and not the theme
you can use one of the 'free" help channels
yesss okok
🙂
@fierce egret I will assume you already got to this part, right? $$ \frac{1}{d_e} = \frac{1}{f} - \frac{1}{d_i} $$
mrbrown
let's summarize the two fractions on the right handside
$$ \frac{1}{f} - \frac{1}{d_i} = \frac{d_i}{d_i f} - \frac{f}{fd_i} = \frac{d_i - f}{fd_i} $$
mrbrown
now let's insert the bad bois 😄
$$ \frac{1}{d_e} = \frac{\frac{x^2+2x-3}{2x^2 + 10}- \frac{x^2-9}{x^2+x-2}}{\frac{x^2-9}{x^2+x-2} \frac{x^2+2x-3}{2x^2 + 10}} $$
mrbrown
surely it is possible to simplify this further
i am not sure yet but we can try to factorize things, hopefully some terms will cancel out each other
no, actually it doens'T look like much will cancel each other out
we can try to rewrite it
$$ \frac{1}{d_e} = \frac{(x^2+2x-3)(x^2+x-2) - (x^2-9)(2x^2+10)}{(x^2-9)(x^2+2x-3)} $$
mrbrown
looks good!
yeah it doesn't look like it can be simplified much further
Thanks for your help man @crisp jackal
Been struggling on this for a good 2hours
😂
no problem :D:D
i have similar issues haha, the day before yesterday i was sitting at a problem for 5 hours lol
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This is a Desmos graphic calculator question. My z_step function contains an action V->V+tA where V is a list, t is a number, and A is a list. This action does not have an error, but it doesn't execute, even if I put it on its own line. I think this is because A is defined as such:
Is there a way to fix this? Link for reference: https://www.desmos.com/calculator/aybjlzytr5
If there is another way to calculate acceleration in a way that desmos allows to be used in an action, that would be fine
A list that is simply a set of numbers CAN be used in an action, but A is still a list so it should work as far as I know
@wet gorge Has your question been resolved?
what was the question ?
how do I fix the z_step function to have the velocity action work
@wet gorge Has your question been resolved?
I'm assuming I don't ping helpers again?
@wet gorge Has your question been resolved?
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how to get from this to that
i dont know what they did'
ok but where does 6x come from
,tex $(x^2+3)^2=x^4+6x^2+9$
Brun。
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So does anyone know how to find the area and volume of a three sided prisms using the diagonal of the side?
I got 9√3/4 x 6√2. And i do not know how to solve that
@crimson sedge Has your question been resolved?
<@&286206848099549185>
yeah.. cause your question is vague as hell.
No prism with 3 sides exists
yes, a prism with 2 bases and 3 sides
Ah well i have been waiting for a response
so the diagonal of the side is 9 cm
like a base side is 3 cm
I need to calculate volume and area
<@&286206848099549185>
Of a triangular prism
@crimson sedge Has your question been resolved?
@crimson sedge Has your question been resolved?
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Is this true for all A and B if the angles can form a closed triangle?
Is this just implied to be true due to the construction of the triangle?
there's not enough information (in some cases)
let theta 1 be 50 degrees, A be 1, B be 1.5
good to know, and thanks for the response, @bright surge
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welp-
for your case, there are no triangles like that(here) (also @teal thorn do you understand why i'm drawing these?)
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I get its not a solid triangle.
Original problem is an offset crank, @bright surge
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Hello, I am given the basis (1,1),(1,0)
(or any basis where this is the matrix )
( a a)
(b +1 b )
and I am asked to find a similar basis such as (2,1)’s first coordinate is 0. How do I do this?
The correction states that this is just the equivalent of saying that (2,1) is a scalar multiplication of (b+1, b)
But it’s like ???
@hollow basin Has your question been resolved?
<@&286206848099549185>
@hollow basin Has your question been resolved?
The trick would be to notice that (2,1) is of the form (x+1, x) for x=1, and that this can be expressed as (0,0) + (1+1,1) = 0.A + 1.B (with A,B the basis vectors).
Otherwise you can just setup the system: x = 2 = a + b + 1 and y = 1 = a + b, from which the solutions are clear ( (0,1) is solution)
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<@&268886789983436800>
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I never got how the square root of 4x^2 is 2x. Why isnt it 4x
square root of 4 is what?
try doing (4x)^2
do you get 4x^2?
No
so should 4x^2's square root be 4x?
Nah
also, you can use this very handy thing:
$\sqrt{ab}=\sqrt{a}\sqrt{b}$
Kanga Gang ¬Sam
(both are the positive square root of ab, if you don't believe me, just try squaring them)
anyways, for this problem, you must find $\sqrt{4x^2} = \sqrt{4}\sqrt{x^2}$
Kanga Gang ¬Sam
i hope you can take it from here
@mental tangle Has your question been resolved?
2x • 2x = 4x^2
But 4x • 4x = 16x^2
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Help
Do you have a formula for sin(a + b)?
@distant gale Has your question been resolved?
This channel is occupied
ask your question somewhere else in the available channels
@heavy igloo
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Hey! Anyone able to help me with these 3 questions, been practicing and don't understand these questions:
lol i also need help but i think i can do this
lets say 5x+40 + 2x = 180 cus abc is also 5x (corres. angles) and they ar3 linear pair now, x = 20
5:
2x+1 = 85 (alternate interior angles)
solve for x
then solve for y since u know 2x+1 is 85 degrees, 180 - 85 = 3y + 5
6:
2m = 4m - 120 (corresponding angles)
once u get M, whatever 4m-120 is equals 5n cus vertical angle
@high flare
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I think he just made common denominators
really?
Yeah
ok i see it now
lol it seemed so foreign
it must be because of the brackets
these square brackets dont mean anything else do they?
Nope, you can view them as regular brackets
Long story
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Help?
with what exactly
I don't know how to solve it
are you solving for y
Both x and y
huh
System of equations
oh
Solve for one variable in one equation in terms of the other
u have 2 variables in 2 equations
Plug this expression into the other equation for the variable you solved for
this should be doable
This should yields an equation with just one variable.
You can then solve for this variable.
so maybe try to rearrange one of the equations in terms of 1 variable, then substitute that for the same variable in the OTHER equation
Once you get the value of this variable, you can then plug its value into either one of the original equations to solve for the other variable
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B
Exponential:
Use u-substitution where u = -x. All this will do is negate the non-constant term
$\int{2^{-x}}dx \implies u=-x, du=-dx \implies -\int{2^u}du = -\bigg[\frac{2^u}{\ln(2)}+C\bigg] \implies -\bigg[\frac{2^{-x}}{\ln(2)}+C\bigg] \implies -\frac{2^{-x}}{\ln(2)}+C$
Fang
dont just give answers.
i was expanding on what disorganized said because they just gave an answer
Theirs is less of an answer, you posted the entire solution.
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Let $f(x) = Ax + B$ and $g(x) = Bx + A$, where $A \neq B$. If $f(g(x)) - g(f(x)) = B - A$, what is $A + B$?
crabbo
i got $abx^2+a^2x+b=b-a$
crabbo
how do i continue
I guess I'm confused as to how you got an x^2 term. When I plug g(x) info f I get f(g(x)) = A(Bx + A) + B. This won't end up with an x^2 term. Similarly for g(f(x)).
@crimson sedge Has your question been resolved?
A(Bx+A)+B - B(Ax+B)+A
f(g(x)) = Ax(Bx + A) + B
yes.
so ABx+A^2+B=B-A
how did you get this?
oh right
A(Bx+a)-B(Ax+B)=0
solve time
this means that
Ax=-B
and Bx=-A
or
A=0
or B=0
i need help on how to continue
where's all that coming from
from this #help-13 message
oh
math error
A(Bx+a)-B(Ax+B)=-2A
steps were skipped but
A(Bx+a)-B(Ax+B)=0
looks ok
what did you do after that
where's all that coming from
where's Ax=-B
and everything that follows coming from
oh
i see
either
Ax+B=Bx+A
or A = 0
or B = 0
that's what you're saying but what's leading you to those conclusions
i.e where's Ax + B = Bx + A coming from
It's like saying if m+n=0 then m=0, n=0. That's the wrong conclusion. You can't assume anything like this.
where's
orA=0
or B=0
coming from
show all work and/or reasoning leading to your statements
i dont know how to solve it
Nothing stopping you from expanding the brackets.
A(Bx+A)-B(Ax+B)=0
ABx+A^2-BAx+B^2=0
A^2+B^2=0
is the answer 0
A+B=0?
The last term is -B² and not +B²
oh
so
(A+B)(A-B)=0
either A-B=0
or A+B=0
why not?
read the question
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hello I am struggling with this question
i mean, have you tried each of them?
yup
d is -1 so not true
indeed
e is not true it is 1
yes
yeah, i was for a second as well
and i think this is just the question being overly finnicky with the definition of "the limit exists"
because technically, if it tends to infinity it diverges so the implied finite limit doesn't exist
but usually you would disambiguate this by saying "converges towards a finite limit"
or something
i think it was b because e^x exists for all values so there is a limit of 0
yep, the others are simply false, and for this one i can see how you could play with the definitions to make it work
but if someone actually expected someone to answer b) without any hesitation i'd slap them
figure of speech, but still
Things going to ±∞ are in general "kinda convergent". Like yeah it doesn't tend towards any particular real value, but c'mon it's regular, it's not like sin(n) or anything
so why is the answer for b not exist because as that function approaches 0 from the right it gets closer to negative infinity
The only way i see the question working is if you say "-∞ isn't a number so the limit doesn't exist"
because limits are often understood to be finite values
and you're right
if you ask 100 mathematicians what this limit is, 99 will answer "-∞"
im just confused because as it approaches 0 from the right it goes toward negative infinity, it never really hits 0
Just like when x approaches -∞ in e^-x, it never really hits -∞
Limits don't deal in "what happens at the value", you have a wonderful tool for that called a calculator
Limits deal in "what happens if i get arbitrarily close to this value"
so yeah, you never reach 0, but that wasn't the point
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how many event possible in random experience
what do you mean it depends on the question
the question ask if the number of possible events could be greater than the fondamental set
issued results
Post the question verbatum
is the numbers of events associed to a random exprience greater than number of results in the fundamental set of that experience.
Fundamental set is something from ODEs... it sounds like you're trying to talk about probability
yes
So... what's fundamental set in this context...?
there is non
???
just a question
well it is not me who posing the question it is the teacher
No, it's getting lost in translation...
Given your broken English, you're translating the question
yes indeed
Hassan, um what language is the original question in?
french
@dusty stone Has your question been resolved?
mdr un fr
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can someone help me prove this statement ?
@rugged tusk Has your question been resolved?
@rugged tusk Has your question been resolved?
Define set non-empty A such that the statement a is true for only numbers in the A set and similarly define non-empty B such that the statement b is true for only numbers in the B set
This is basically what we conclude from (∃x a(x) ∨ ∃x b(x))
Then ∃x (a(x) ∨ b(x)) is clearly true, because we can choose x to be in the set $A\cup{B}$
Wait how do I write union in LaTeX
Touch Our Beans
Oh thanks
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I know this is wrong, but where is it wrong?
@clever jewel Has your question been resolved?
View $\frac{x - 1}{x + 1}$ as $1 - \frac{2}{x + 1}$
Touch Our Beans
but like, where is it wrong in this image?
$\lim_{x\to\infty}(\frac{1 - \frac{2}{x} + \frac{1}{x^2}}{1 - \frac{1}{x^2}})^x \ne 1$
Touch Our Beans
Only this part
can't I simply take each of those fractions as a separate limit and evaluate them separately?
and then plug in the values
No, I can give you an example where such conclusion fails
$e = \lim_{n\to\infty}(1 + \frac{1}{n})^n$
Touch Our Beans
hmm
yeah, I know it's e
but like, what's stopping me from breaking that limit into smaller limits?
can't the limit of some addition be broken into an addition of 2 limits?
Yes (1 + 1/n)^n is the same as (1 + 1/n) being multiplied n times
But the limit should be the value that the expression approaches as n gets arbitrarily large
We can't even guarantee that it's a natural number in the first place
To break (1 + 1/n)^n into (1 + 1/n) being taken n times
It's almost the same mistake as writing $\sum_{k=0}^N{k} = k\sum_{k=0}^N{1}$
Touch Our Beans
I hope it's clear now
Did you try this yet?
yeah, it works
what I was about to say
was that I thought I could do something like this:
but it only works if g(x) is a constant
I think it works as long as limits for f and g exist generally
oh
g doesn't have to necessarily be constant function
honestly, I still don't understand why this wouldn't work
if the limit can be broken into separate ones
cause you ignored the power entirely
so I can't just evaluate the function inside the paranthesis and then worry about the power?
like this
yes
so this doesn't work at all?
That I'm aware no, however those type of limits (ex. e's definition) require log tricks
$L:=\lim_{x\to a}(f^g)(x) \ \ln(L)=\lim_{x\to a}g(x)\ln(f(x))$
Mosh
yeah but you never have that
you have like 1.001^(something not 1 WLOG)
So you were in fact saying this
ohhh
I see
alright, it's crystal clear now
can't separately evaluate the paranthesis and then the exponent, they both affect each other
I see
well, thanks a lot @south tundra, @dense wing and @elfin hemlock ! :P
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so i gotta minimise (a-1+1/a)(2-a+1/(3-a)) where 0<=a<=2
there's a symmetry that makes (a-1+1/a) and (2-a+1/(3-a)) the same thing
and the minimum will just be the center of the symmetry
hmmm I'm not entirely sure
something about the limit at a=0 and a=3 being +infinity, and the degree of the polynomial meaning it can only turn once
shift it to the origin and apply the definition of symmetry
it does make sense, just that idk how to convince someone that the minimum will be at the middle
so like x'=x+1.5?
oh oops, its symmetric so at the middle point it must have an extrema
yea thanks got it
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I forgot how to this, Am i suppose to give the highest exponent or give the sum of degrees
Sum.
Alright thanks
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i need help with this <@&286206848099549185>
don’t immediately ping helpers
When he was walking 800m NE, you could conclude that he walked 800/sqrt(2) due to N (and 800/sqrt(2) due to eE)
So since he's walked 800/sqrt(2) due to N already and he wants to have walked 1000m
He now would need to walk 1000 - 800/sqrt(2) in the direction of North
(Feel free to rationalize the denominator in 800/sqrt(2))
ok
@boreal mist Has your question been resolved?
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I need to map a function onto a torus
You want to map a function?
Let $T$ be the torus defined with $[0,1]\times[0,1]$ (for example)
Define a function $f : \R^n \to T$
One such way to do this is to make the function 1-periodic
kirby
Hi, yes I do
I want to use stereographic projection to map the x and y values of a function onto a torus
The circle going around the top being the y-axis, and the circle going around the sides being the x-axis
So you can sorta see the whole infinite function in a finite space
@ionic peak Has your question been resolved?
@ionic peak Has your question been resolved?
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Define the function $g(x)=3x+2$. If $g(x)=2f^{-1}(x)$ and $f^{-1}(x)$ is the inverse of the function $f(x)=ax+b$, find $\dfrac{a+b}{2}$.
crabbo
x not y
and that is equal to....
ah right
crabbo
Compile Error! Click the
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that is the inverse of f(x)
and it is equal to...
and then use the information given about g(x)
$\frac{3x+2}{2}=\frac{x-b}{a}$
crabbo
yep!
now i know what $f^{-1}$ is equal to
crabbo
6x-2?
hmmm
no
equate coefficients and constants to determine the values of a and b
2x+2?
forget about finding f(x)
equate coefficients and constants to determine the values of a and b
how are you getting that?
im equating coefficients and constants
^^^^^^
show all work
ig its wrong
how exactly do you want me to do that?
which expressions should i work with
by actually equating the coefficients of x and the constant terms
perhaps it would be easier to see if you split your fractions
to explicitly identify the coefficients and constants
just because a is the denominator on one side and there is a 2 in the denominator on the other side, that doesn't necessarily mean that a is 2
nor is that in anyway equating the coefficients and/or constants
you mean split my fraction, by splitting the numerator and the denominator
no
im confused
i just dont understand what you want me to do
$\frac{3x+2}{2}$
ℝamonov
split that fraction and express that as a sum of your x term and a constant
there are no like terms
or
who said anything about like terms