Secondary Maths questions, give me your maths questions

[Nasma]

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.

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?

 

[magic_johnson!]

OK, calculator exam tomorrow

Two circles of diameters 12cm and 8cm have their centres 8cm apart. Find the area of common region.

 

[PowerIStough]

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.

 

[SJ]

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.

How about you do as much as you can and then we can see where you get stuck?

 

[PowerIStough]

How about you do as much as you can and then we can see where you get stuck?

The first one I figured out just then

The other one, I don't know what to do if the the base is not a factor of 5.

 

[SJ]

The first one I figured out just then

The other one, I don't know what to do if the the base is not a factor of 5.

But it is.

5^2 = 25
5^1 = 5
5^0 =

 

[PowerIStough]

But it is.

5^2 = 25
5^1 = 5
5^0 =

Ohh yes. I now see it Thanks.

 

[magic_johnson!]

how do you work out the inverse of tan (1) without using the calculator?
or any other value for that matter...

 

[treefingers]

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.

 

[magic_johnson!]

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.

Sounds like fun Thanks

 

[Nasma]

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.

 

[SJ]

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.

I never thought of it like that back in the day. That is a much easier way to remember it.

 

[magic_johnson!]

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.

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

I went well on the imaginary numbers test today though

 

[Stan Van Gundy]

Find the square root of i. I've done this question, but I thought some of you might like to have a go

 

[magic_johnson!]

Find the square root of i. I've done this question, but I thought some of you might like to have a go

Ahh MJ give up

 

[Stan Van Gundy]

Ok i'll write the solution because I've got time haha

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

 

[treefingers]

therefore (x+iy)^2 = i

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

doesn't expanding the brackets give you...

x^2 + y^2 + 2xyi = i

 

[SJ]

doesn't expanding the brackets give you...

x^2 + y^2 + 2xyi = i

(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.

 

[magic_johnson!]

doesn't expanding the brackets give you...

x^2 + y^2 + 2xyi = i

It would be, but yi^2 = -y

 

[treefingers]

Yeah I forgot about the i part, just thinking it was a regular (a + b)^2 = a^2 +b^2 +2ab

 

[SJ]

Pretty sure you aren't allowed to use an identity in steps to prove the same identity though.

 

[jessCFC]

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

 

[Flawed Genius]

∫e^x(1-e^-x)/e^3x dx

 

[jessCFC]

∫e^x(1-e^-x)/e^3x dx

This is an easy problem. Using index laws for expansion and division you can reduce the equation to: ∫exp(-2x)-exp(-x)dx= -1/2exp(-2x)+exp(x) + c

 

[Pie Legend]

∫e^x(1-e^-x)/e^3x dx

I swear I had that exact question in a Methods SAC the other day..