<p>forget the dice one
x^2 > 3/5 * y^2
PR =8
3a = 20
x + z = 7
24 combinations
y = x/ root 3
119
july had 8000 less
6000 computers difference
290 -locker question
radius of 5 semicircles = 1.2
12 for area of triangle
paralell lines was I and II only
x = 5/3
4500 dollars i think
15 paperback
1080 something
4 for height of cylinder</p>
<p>Nvm. I solved it. The equation was (x+a)(x+2b)=x^2+4abx+b^2
What I did was equate 2ab to b^2 which would mean 2a=b. Then it’s just algebra from there. Too bad I thought of this as I ran out of time. Oh well, I’m pretty sure that it was experimental, otherwise people would definitely be asking about it. At least my mind is at peace =] For anyone that wants to know the answer: a=0,5/8. But then I think the problem said a and b were integers so a=0</p>
<p>It was I and II for the lines. I was that two lines were perpendicular, which was true. II was that two lines were parallel, which was also true. III was that two lines were perpendicular, which was false since they were actually parallel.</p>
<p>forget the dice one
10 for area of rearranged rectangle</p>
<p>area of trinagle is 12 , in which sides are 5,5,6
x^2 > 3/5 * y^2
PR =8
3a = 20
x + z = 7
24 combinations
y = x/ root 3
119
july had 8000 less
6000 computers difference
290 -locker question
radius of 5 semicircles = 1.2
12 for area of triangle
paralell lines was I and II only
x = 5/3
4500 dollars i think
15 paperback
1080 something
4 for height of cylinder</p>