I don't remember the midpoint formula exactly, but use the midpoint formula for this problem.
Answer:
Distributive Property
Step-by-step explanation:
Answer choice a is correct because you are distributing 2 to each term in the parentheses. When you multiply 2 to each factor in the parentheses, it simplifies to 2x-2, thus making the new equation 6x+2x-2=30
Answer:
The product is represented by Point A: -8.
Step-by-step explanation:
9.78(4.3x + 5.15)
Multiply everything inside the parentheses by 9.78
42.054x + 50.367
Answer:
1-i and -1+i
Step-by-step explanation:
We are to find the square roots of
. First, convert from Cartesian to polar form:
![r=\sqrt{a^2+b^2}\\r=\sqrt{0^2+(-2)^2}\\r=\sqrt{0+4}\\r=\sqrt{4}\\r=2](https://tex.z-dn.net/?f=r%3D%5Csqrt%7Ba%5E2%2Bb%5E2%7D%5C%5Cr%3D%5Csqrt%7B0%5E2%2B%28-2%29%5E2%7D%5C%5Cr%3D%5Csqrt%7B0%2B4%7D%5C%5Cr%3D%5Csqrt%7B4%7D%5C%5Cr%3D2)
![\theta=tan^{-1}(\frac{b}{a})\\ \theta=tan^{-1}(\frac{-2}{0})\\\theta=\frac{3\pi}{2}](https://tex.z-dn.net/?f=%5Ctheta%3Dtan%5E%7B-1%7D%28%5Cfrac%7Bb%7D%7Ba%7D%29%5C%5C%20%5Ctheta%3Dtan%5E%7B-1%7D%28%5Cfrac%7B-2%7D%7B0%7D%29%5C%5C%5Ctheta%3D%5Cfrac%7B3%5Cpi%7D%7B2%7D)
![z=2(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2})](https://tex.z-dn.net/?f=z%3D2%28%5Ccos%5Cfrac%7B3%5Cpi%7D%7B2%7D%2Bi%5Csin%5Cfrac%7B3%5Cpi%7D%7B2%7D%29)
Next, use the formula
where
to find the square roots:
<u>When k=1</u>
<u />![\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(1)}{2}\biggr)\biggr]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csqrt%5B2%5D%7B2%7D%5Cbiggr%5Bcis%5Cbiggr%28%5Cfrac%7B%5Cfrac%7B3%5Cpi%7D%7B2%7D%2B2%5Cpi%281%29%7D%7B2%7D%5Cbiggr%29%5Cbiggr%5D)
![\displaystyle \sqrt{2}\biggr[cis\biggr(\frac{3\pi}{4}+\pi\biggr)\biggr]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csqrt%7B2%7D%5Cbiggr%5Bcis%5Cbiggr%28%5Cfrac%7B3%5Cpi%7D%7B4%7D%2B%5Cpi%5Cbiggr%29%5Cbiggr%5D)
![\sqrt{2}\biggr(cis\frac{7\pi}{4}\biggr)](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D%5Cbiggr%28cis%5Cfrac%7B7%5Cpi%7D%7B4%7D%5Cbiggr%29)
![\sqrt{2}(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4})\\ \\\sqrt{2}(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)\\ \\1-i](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D%28%5Ccos%5Cfrac%7B7%5Cpi%7D%7B4%7D%2Bi%5Csin%5Cfrac%7B7%5Cpi%7D%7B4%7D%29%5C%5C%20%5C%5C%5Csqrt%7B2%7D%28%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D-%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7Di%29%5C%5C%20%5C%5C1-i)
<u>When k=0</u>
<u />![\displaystyle \sqrt[2]{2}\biggr[cis\biggr(\frac{\frac{3\pi}{2}+2\pi(0)}{2}\biggr)\biggr]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Csqrt%5B2%5D%7B2%7D%5Cbiggr%5Bcis%5Cbiggr%28%5Cfrac%7B%5Cfrac%7B3%5Cpi%7D%7B2%7D%2B2%5Cpi%280%29%7D%7B2%7D%5Cbiggr%29%5Cbiggr%5D)
![\sqrt{2}\biggr(cis\frac{3\pi}{4}\biggr)](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D%5Cbiggr%28cis%5Cfrac%7B3%5Cpi%7D%7B4%7D%5Cbiggr%29)
![\sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})\\ \\\sqrt{2}(-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)\\ \\-1+i](https://tex.z-dn.net/?f=%5Csqrt%7B2%7D%28%5Ccos%5Cfrac%7B3%5Cpi%7D%7B4%7D%2Bi%5Csin%5Cfrac%7B3%5Cpi%7D%7B4%7D%29%5C%5C%20%5C%5C%5Csqrt%7B2%7D%28-%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%2B%5Cfrac%7B%5Csqrt%7B2%7D%7D%7B2%7Di%29%5C%5C%20%5C%5C-1%2Bi)
Thus, the square roots of -2i are 1-i and -1+i