Secondary Maths questions, give me your maths questions

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magic_johnson! said:
The distance d travelled by a car when coming to rest from speed v is proportional to the square of the speed. A car travelling at a speed of 30 m/s along a residential street will stop in a distance of 15m. Find the stopping distance required for a car travelling at 60 m/s.
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Based on the way that question is worded, I think they want you to take a different approach. You know that the stopping distance is proportional to the sqaure of the speed. So if the speed doubles, what does that do to the stopping distance?
 

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PowerIStough said:
A few questions:

Find the asymptotes, x-intercept and y-intercepy ~ y = 2loge (x+1) - 1

Solve for x;

5 2x-4 (as a power) = 1. Solve for x.
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How about you do as much as you can and then we can see where you get stuck?
 
You need to memorise the exact values for sin/cos/tan of 0,30,60, and 90 degrees.

Inverse is effectively working backwards.

So if tan x = 1,

x = inverse tan of 1. If you know the exact values then you would know that x = 45 degrees.
 
It's not that bad, you only really have to remember sin(x). The pattern is sqrt(0)/2, sqrt(1)/2, sqrt(2)/2, sqrt(3)/2, sqrt(4)/2 for sin of 0,30,45,60,90. Cos is just that backwards, and tan is sin/cos.
 
Nasma said:
It's not that bad, you only really have to remember sin(x). The pattern is sqrt(0)/2, sqrt(1)/2, sqrt(2)/2, sqrt(3)/2, sqrt(4)/2 for sin of 0,30,45,60,90. Cos is just that backwards, and tan is sin/cos.
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I never thought of it like that back in the day. That is a much easier way to remember it.
 
Nasma said:
It's not that bad, you only really have to remember sin(x). The pattern is sqrt(0)/2, sqrt(1)/2, sqrt(2)/2, sqrt(3)/2, sqrt(4)/2 for sin of 0,30,45,60,90. Cos is just that backwards, and tan is sin/cos.
Click to expand...

Yeah i just went over my last semesters work and remembered this was how we learnt it. Funny how you forget stuff so quickly :p

I went well on the imaginary numbers test today though :thumbsu:
 

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Find the square root of i. I've done this question, but I thought some of you might like to have a go ;)
 
Ok i'll write the solution because I've got time haha :p

let x+iy = sqrt(i) where x,y are real numbers

therefore (x+iy)^2 = i

therefore: (expanding the bracket) x^2 - y^2 + 2xyi = i

therefore (x^2 - y^2) + i(2xy - 1) = 0

therefore (using thm if a+ib=0 then a and b= 0): x^2 - y^2 =0, 2xy-1=0

tf: x^2 - y^2=0 and y=1/2x

then you sub in (1/2x) into your first eqn

so: x^2 - (1/2x)^2=0

tf: x^2 - 1/(4x^2)=0

then you multiply by 4x^2 and let x^2 equal A

so you get: 4A^2 -1 =0

then facotorise the eqn so: (2A+1)(2A-1)=0

tf: 2A=-1 and 2A=1 which then tells us that x^2=-1/2 and x^2=1/2

but since x is a real number we can't use x^2=-1/2

so we have: x^2= 1/2 which gives us: x= + or - sqrt(1/2).

which simplifies to x= +(sqrt(2)/2) or -(sqrt(2)/2)

then to find y we go back and substitute x into the eqn y=1/2x

and we end up getting y=+(sqrt(2)/2) when x is +ve and y=-(sqrt(2)/2) when x is -ve.

Then you go back to your initial eqn which is that sqrt(i)=x+iy
sub in x and y and you get

sqrt(i)= (sqrt(2)/2) + (sqrt(2)/2)i

and also sqrt(i) = -(sqrt(2)/2) - (sqrt(2)/2)i

well... that probably wasn't worth it hahaha. Trying to do a maths question on the computer takes about 50 times longer:p
 
-TAZZA- said:
doesn't expanding the brackets give you...

x^2 + y^2 + 2xyi = i
Click to expand...
(x + iy)^2 = x^2 + (iy)^2 + 2ixy = x^2 + i^2*y^2 + 2ixy = x^2 + (-1)y^2 + 2ixy

However using i^2 = -1 in a proof to prove exactly that seems a little fishy to me.
 
The solution to the problem can be proposed using De Moivre's theorem for fractional powers of complex numbers:

Let z be any complex number then

z^(1/2)=r^(1/2){cos(phi/2)+i sin(phi/2)} ______ (1)

if z=i then r=1 and phi=pi/2 then substitution into (1) gives

z^(1/2)=cos(pi/4)+i sin(pi/4) =sqrt(2)/2+ i sqrt(2)/2

By the Fundamental theorem of algebra, the equation of degree n has n roots. The other root is given by

z^(1/2)=cos(pi/4+2pi/2)+i sin(pi/4+2pi/2)
=cos(5pi/4)+i sin(5pi/4)
=cos(-3pi/4)+i sin(-3pi/4)
=-sqrt(2)/2- i sqrt(2)/2
 
∫e^x(1-e^-x)/e^3x dx
 
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