# Help! Part of an area questionWatch

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#1
Given that

2d^2 + 1/2 pid^2 = 1/2d^2 theta

show that theta is equal to: 1+1/4pi

thanks!
0
13 years ago
#2
What on earth does that equation mean? What are the d's, what's a "pid" (is that "pi*d"? Is it 1/(2d^2) or (1/2) d^2. You have to be more rigorous with your notation, dude.

Taking my best guess...if you cancel the d^2, then you get theta = pi + 4. But I have no idea what the question means.
0
13 years ago
#3
(Original post by Lawbutwhere?)
Given that

2d^2 + 1/2 pid^2 = 1/2d^2 theta

show that theta is equal to: 1+1/4pi

thanks!
2d²+0.5(pi)d² =0.5d²T
T= 2d²+0.5(pi)d² / 0.5d²
T= 4 + pi

Have you written the question out wrong again? Honestly you do this in all your threads make some effort!
0
#4
Sorry! I am not gd with computers, or writing out equations! Have solved it now! It was:

2d^2 + (1/2)(pi*d^2) = 2d^2theta

so you just divide by 2d^2, to get theta = 1+1/4pi.

Thanks.
0
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