Answer:
Hello,
Step-by-step explanation:
![I=\dfrac{Area}{4} =\int\limits^4_0 {\sqrt{16-x^2} } \, dx \\\\Let\ say\ x=4*sin(t),\ dx=4*cos(t) dt\\\\\displaystyle I=\int\limits^\frac{\pi }{2} _0 {4*\sqrt{1-sin^2(t)} }*4*cos(t) \, dt \\\\=16*\int\limits^\frac{\pi }{2} _0 {cos^2(t)} \, dt \\\\=16*\int\limits^\frac{\pi }{2} _0 {\frac{1-cos(2t)}{2}} \, dt \\\\=8*[t]^\frac{\pi }{2} _0-[\frac{sin(2t)}{2} ]^\frac{\pi }{2} _0\\\\=4\pi -0\\\\=4\pi\\\\\boxed{Area=4*I=16\pi}\\](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7BArea%7D%7B4%7D%20%3D%5Cint%5Climits%5E4_0%20%7B%5Csqrt%7B16-x%5E2%7D%20%7D%20%5C%2C%20dx%20%5C%5C%5C%5CLet%5C%20say%5C%20x%3D4%2Asin%28t%29%2C%5C%20dx%3D4%2Acos%28t%29%20dt%5C%5C%5C%5C%5Cdisplaystyle%20I%3D%5Cint%5Climits%5E%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20_0%20%7B4%2A%5Csqrt%7B1-sin%5E2%28t%29%7D%20%7D%2A4%2Acos%28t%29%20%5C%2C%20dt%20%5C%5C%5C%5C%3D16%2A%5Cint%5Climits%5E%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20_0%20%7Bcos%5E2%28t%29%7D%20%5C%2C%20dt%20%5C%5C%5C%5C%3D16%2A%5Cint%5Climits%5E%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20_0%20%7B%5Cfrac%7B1-cos%282t%29%7D%7B2%7D%7D%20%5C%2C%20dt%20%5C%5C%5C%5C%3D8%2A%5Bt%5D%5E%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20_0-%5B%5Cfrac%7Bsin%282t%29%7D%7B2%7D%20%5D%5E%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20_0%5C%5C%5C%5C%3D4%5Cpi%20-0%5C%5C%5C%5C%3D4%5Cpi%5C%5C%5C%5C%5Cboxed%7BArea%3D4%2AI%3D16%5Cpi%7D%5C%5C)
Answer:
C
Step-by-step explanation:
Answer:
£8400
Step-by-step explanation:
1st: Expand the bracket
3t-3=9+t
2nd: Add the 3 to get rid of it on the first side. Do this to both sides
3t=12+t
3rd: Take away the t on the left hand side and one on the right hand side to just be left with t's on one side and integers on the other side
2t=12
4th: Divide both sides by 2 to find 1t
t=6
Hope this helps!! :)
