Recent content by FlopperJr

nevermind again lol. i just had to find the area of rectangle 3x2

however, now I'm stuck on the second part to that question.

nevermind i got it. it should be y=3 when x=2. i plugged it in wrong because i did x squared not cubed. calculation error

Homework Statement Question 2 http://www.xtremepapers.com/CIE/International%20A%20And%20AS%20Level/9709%20-%20Mathematics/9709_s11_qp_22.pdf Homework Equations A=1/2xh[(y1)+2(y2)+2(y3)...+(yn)] The Attempt at a Solution I know h=1 x=0 y=1 x=1 y=√2 x=2 y=√5 however when...

Okay, for negative square root: The inverse is -50.8 180-(-50.8). Is 230.8 which doesn't fit the range But if I subtract from 360 I get -129.2 So my my two answers for this is -50.8 and -129.2 giving me a total of four answers

Ohhh. So it would be sinx= +/- the sqr root 3/5. Which then the inverse would be 50.8. Then 180 minus 50.8 is 129.2. After that I would take those two answers and +/- 360 which doesn't work...

Aww. Alright. Then how would I do it if sin is squared? Sin^2=3/5

Ahhhh. I see now, that helps. Thank you!

Homework Statement Solve the equation 5sin2x-3=0, given values of x in the interval -180≤x≤180 correct to 1 decimal place. Homework Equations for sin: take inverse 180-θ ±360 The Attempt at a Solution I rearranged the equation to sin2=3/5 but I am not quite sure what to do...

Thanks so much for your help! I always have trouble with the domain and range.

Thank you so much for your time and help! It means a lot!

Okay thanks! I know that it is going to be oppsite of whatever the original function was. Now that I see the graph, it would be x≤4.

Oops i had my signs reversed but I got the answer now. x≤-8 x≥-2

I believe the domain and range of the inverse is x>4 and y≤0

Its a parabola. It opens down, with the max point on 4.