This means it’s 6(3)^2. The PEMDAS process will help in this situation. Since exponents (E) come before multiplication (M), you would raise 3 to the 2nd power first. 3^2 is the same as 3 times 3, which equals 9. Now, you can multiple 6 by 9, which would result with 54. 54 is your answer!
From the given above, the point that still needs to be incurred by the group is calculated by subtracting 424 points from 1000. This as shown below,
n = 1000 - 424 = 576
If we are to determine the number of points each game, we divide 576 by 9. That is,
x = 576 / 9 = 64
<em>ANSWER: 64 points/ game</em>
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:
(w + 11)(w + 7)
Step-by-step explanation:
Consider the factors of the constant term ( + 77) which sum to give the coefficient of the w- term ( + 18)
The factors are + 11 and + 7, since
11 × 7 = 77 and + 11 + 7 = + 18
w² + 18w + 77 = (w + 11)(w + 7)
It is the last choice, 12
seven times two is 14
74-14=60
60/5=12
Hope this helps!!
