We want to get rid of the stuff in the denomenator
gcd of the denomenator is 2x
but wait, we can get rid of the deonmenator in the numerator by multiplying by 4x
muliplty the whole thing by 4x/4x
we get
Area of a quarter circle = 1/4 * pi * r^2
= 1/4 * pi * 4^2 = 12.6 m to 3 sig figs.
perimeter = 4 + 4 + 1/4 * 2 pi * 4 = 14.3 to 3 sig figs.
By de Moivre's theorem,
![\implies \sqrt[4]{(1 - i)^2} = \sqrt[4]{2}\,e^{i(2\pi k-\pi/2)/4} = \sqrt[4]{2}\,e^{i(4k-1)\pi/8}](https://tex.z-dn.net/?f=%5Cimplies%20%5Csqrt%5B4%5D%7B%281%20-%20i%29%5E2%7D%20%3D%20%5Csqrt%5B4%5D%7B2%7D%5C%2Ce%5E%7Bi%282%5Cpi%20k-%5Cpi%2F2%29%2F4%7D%20%3D%20%5Csqrt%5B4%5D%7B2%7D%5C%2Ce%5E%7Bi%284k-1%29%5Cpi%2F8%7D)
where 
. The fourth roots of 
are then
![k = 0 \implies \sqrt[4]{2}\,e^{-i\pi/8}](https://tex.z-dn.net/?f=k%20%3D%200%20%5Cimplies%20%5Csqrt%5B4%5D%7B2%7D%5C%2Ce%5E%7B-i%5Cpi%2F8%7D)
![k = 1 \implies \sqrt[4]{2}\,e^{i3\pi/8}](https://tex.z-dn.net/?f=k%20%3D%201%20%5Cimplies%20%5Csqrt%5B4%5D%7B2%7D%5C%2Ce%5E%7Bi3%5Cpi%2F8%7D)
![k = 2 \implies \sqrt[4]{2}\,e^{i7\pi/8}](https://tex.z-dn.net/?f=k%20%3D%202%20%5Cimplies%20%5Csqrt%5B4%5D%7B2%7D%5C%2Ce%5E%7Bi7%5Cpi%2F8%7D)
![k = 3 \implies \sqrt[4]{2}\,e^{i11\pi/8}](https://tex.z-dn.net/?f=k%20%3D%203%20%5Cimplies%20%5Csqrt%5B4%5D%7B2%7D%5C%2Ce%5E%7Bi11%5Cpi%2F8%7D)
or more simply
![\boxed{\pm\sqrt[4]{2}\,e^{-i\pi/8} \text{ and } \pm\sqrt[4]{2}\,e^{i3\pi/8}}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cpm%5Csqrt%5B4%5D%7B2%7D%5C%2Ce%5E%7B-i%5Cpi%2F8%7D%20%5Ctext%7B%20and%20%7D%20%5Cpm%5Csqrt%5B4%5D%7B2%7D%5C%2Ce%5E%7Bi3%5Cpi%2F8%7D%7D)
We can go on to put these in rectangular form. Recall
Then
and the roots are equivalently
![\boxed{\pm\sqrt[4]{2}\left(\sqrt{\dfrac12 + \dfrac1{2\sqrt2}} - i\sqrt{\dfrac12 - \dfrac1{2\sqrt2}}\right) \text{ and } \pm\sqrt[4]{2}\left(\sqrt{\dfrac12 + \dfrac1{2\sqrt2}} + i \sqrt{\dfrac12 - \dfrac1{2\sqrt2}}\right)}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cpm%5Csqrt%5B4%5D%7B2%7D%5Cleft%28%5Csqrt%7B%5Cdfrac12%20%2B%20%5Cdfrac1%7B2%5Csqrt2%7D%7D%20-%20i%5Csqrt%7B%5Cdfrac12%20-%20%5Cdfrac1%7B2%5Csqrt2%7D%7D%5Cright%29%20%5Ctext%7B%20and%20%7D%20%5Cpm%5Csqrt%5B4%5D%7B2%7D%5Cleft%28%5Csqrt%7B%5Cdfrac12%20%2B%20%5Cdfrac1%7B2%5Csqrt2%7D%7D%20%2B%20i%20%5Csqrt%7B%5Cdfrac12%20-%20%5Cdfrac1%7B2%5Csqrt2%7D%7D%5Cright%29%7D)
Answer:
f(7) = 2
Step-by-step explanation:
3f(7) + f(7)g(7) = 18
3f(7) + f(7)(6) = 18
f(7) (3 + 6) = 18
f(7) x (9) = 18
f(7) = 2
Answer:
Look in Step by Step Explanation
Step-by-step explanation:
3)SOHCAHTOA
tan=opposite/adjacent
tan(20)=x/83
83tan(20)=x
x=30.209
2) SOHCAHTOA
Tangent=opposite/adjacent
Tan(62)=x/8
8tan(62)=x
x=15.04
3)SOHCAHTOA
Sin=opposite/hypotonouse
sin(25)=30/x
x=30/sin(25)
x=70.986
