Recent content by Yazan975

I am not able to understand what the question asks of me in Q75, part a)

In this question, I tried this: sin^2(180-x) cosec(270+x) + cos^2(360-x) sec(180-x), where cosec(x) = 1/sin(x) and sec(x) = 1/cos(x) -sin^2(180-x) = sin^2(x) and cos^2(x) = cos^2(x) -The sin^2 and the 1/sin(x) cancle out along with the cos^2 and the 1/cos(x) Therefore, I am left with...

I do not know how to start thinking about part a and then got even more confused when I saw the answer be: P<x^2-5x. I ask for your guidance please.

This is where I got so far. I can't figure out how to prove AH = DK in order to prove the HL property of congruency

Yeah, here's the formula: x^2-40x-400=0(after dividing the equation by 40) Hence, x=20?? Thing is my answer sheet says y= 30 and x=50

I see where it comes from but problem is I got stuck at 40x = 1600+10y I couldn't get an equation in terms of x - - - Updated - - - I see where it comes from but problem is I got stuck at 40x = 1600+10y I couldn't get an equation in terms of x - - - Updated - - - I see where it comes from...

xy/2 cm^2 = 800 xy = 1600 Is that right?

xy/2 cm^2

1600cm^2? For 3200 I multiplied Base and Height but didn't divide by 2

3200cm^2

What I have done: I changed all fractions to common denom and that gave me 5y-5x=1 (1) *I numbered the fractions 5y+2x=5 (2) Then: 5y=5-2x Substitute into equation 1 (5-2x)-5x=1 5-7x=1 x=4/7 Thing is my answer says I should be getting x=0 Any hints?

It is supposed to test us