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Soloha48 [4]
3 years ago
7

Which equation in point slope form contains the points (-1,3) and (2,-2)

Mathematics
1 answer:
kolezko [41]3 years ago
4 0

The point-slope form:

y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}

We have the points (-1, 3) and (2, -2). Substitute:

m=\dfrac{-2-3}{2-(-1)}=\dfrac{-5}{2+1}=-\dfrac{5}{3}\\\\y-3=-\dfrac{5}{3}(x-(-1))\\\\\boxed{y-3=-\dfrac{5}{3}(x+1)}

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Which expression is a fourth root of -1+isqrt3?
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          \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{\dfrac{-\pi}{3} +0}{4}+iSin  \ \dfrac{\dfrac{-\pi}{3}+0}{4}\right] \\\\\\z= \sqrt[4]{10} \left[Cos \ \dfrac{ -\pi  }{12}+iSin  \ \dfrac{-\pi}{12}\right]\\\\\\z = \sqrt[4]{10}\left[-Cos \ \dfrac{\pi}{12}-i \ Sin \ \dfrac{\pi}{12}\right]

For k =1,

         \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{5\pi}{12}+i \ Sin \ \dfrac{5\pi}{12}\right]

For k =2,

       z = \sqrt[4]{10}\left[Cos \ \dfrac{11\pi}{12}+i \ Sin \ \dfrac{11\pi}{12}\right]

For k = 3,

      \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{17\pi}{12}+i \ Sin \ \dfrac{17\pi}{12}\right]

For k = 4,

      \sf z =\sqrt[4]{10}\left[Cos \ \dfrac{23\pi}{12}+i \ Sin \ \dfrac{23\pi}{12}\right]

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