10 C Exam 1 2008 VCAA

[lacoste]

Don't get 10 C?

The transposing of the composition and inverse.


thanks

[THem]

c)

[lacoste]

thanks THem

Im also stuck on question 9a can get the height of the prism?

[lacoste]

i dont get the second last step from your working can you explain?

[THem]

You have 1, but they require the answer in a certain form so I changed the 1 to have the same denominator as the other part. So 2x+1/2x+1 is still equal to 1. So in that case, the 1's cancel out , leaving -2x.

1  - (2x+1)
= 1-1 -2x

[lacoste]

nah i dont get the part of where it becomes 1/(2x+1) all of a sudden

[lacoste]

how did the become

[THem]

2 * 0.5 = 1
But there is a negative at the front, so I move it into the bracket, leaving it so the loge and e cancel out.
Since I moved the negative into the inside, it becomes (2x+1)^-1 which is 1/2x+1

aloge(x) = loge(x^a)

[dino]

e^log(x) = x

e to the power of log(x) = x

That's the theory behind it.

[lacoste]

thanks, dont fully get it but ill just remember that if theres a negative invert it to get it in the form of ax+b.
cheers!!
thank THem!! got it now

How do I do question 9A? seems easy but yeh kind of stuck

but damn questions 9

[Lighties]

It wouldn't matter if you don't put the modulus sign in, would it?

Volume of a prism = base times height, base being the triangle. Area of triangle is (base x height)/2, and to get the height I used algebra for a right-angled triangle with side lengths of height, x/2 and the hypotenuse being x.

so (sqrt(3)*x^2)/4 * y = 1000.

Sorry if that didn't make sense, I don't know how to use the thing that let's you write clearly or pictures. ^^;

[lacoste]

i got that but in vcaa sols it said its sqrt(3x)/2 why not sqrt(3x^2)/2?

[Lighties]

sqrt(3x)/2 is the height of the triangle. I put in '(sqrt(3)*x^2)/4' 'cause it was the area of the base.

(so sqrt(3x)/2 times x all divided by 2 for area)

[lookche]

an even better way to get the area of the triangle in Q9 is to note that its an equilateral triangle with equal side lengths, hence all angles are 60 degrees.  then you can go A = .5(x^2)sin60, which is (x^2 (root 3))/2 instead of using messy pythagoras to get the height.

[lacoste]

can someone put up the step by step solution to 9A?

i get the pythag stuff now, but not transposing? :/

I THINK i got it.

Anything divide by a sqr root number is wacked, eg. 4000/ root(3) = 4000*root(3) / 3