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lara [203]
3 years ago
10

Solve 2x2 − 8x = −7.

Mathematics
2 answers:
maks197457 [2]3 years ago
8 0

Answer:

D.2 plus or minus the square root of 2 end root over 2

Step-by-step explanation:

leva [86]3 years ago
7 0
\sf 2x^2 - 8x = - 7  \\  \\ Subtract \ -7 \ from \ both \ sides \ of \ the \ equation. \\  \\ 2x^2 - 8x - (-7 ) = - 7 - (-7)  \\  \\ 2x^2 - 8x + 7 = 0  \\  \\ Use \ quadratic \ formula \ a = 2, b = -8, c = 7 \\  \\ x =  \dfrac{- b \pm \sqrt{b^2- 4ac} }{2a}  \\  \\  \\ x =  \dfrac{-(-8) \pm \sqrt{(-8)^2 - 4 (2) (7) } }{2(2)}  \\  \\ x =  \dfrac{8 \pm  \sqrt{8} }{4}  \\  \\ x = 2 +  \frac{1}{2}\sqrt{2} \ or \ x = 2 +  \frac{-1}{2}  \sqrt{2}

Ur answer is the fourth one 



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Given the coordinate transformation f(x, y) = (- 2x, y) and the domain (0, 5), (8, - 1) and (- 6, 4) what is the range?
Mariana [72]

Answer:

{(0, 5), (-16, -1), (12, 4)}

Step-by-step explanation:

Given the coordinate transformation f(x, y) = (- 2x, y) and the domain (0, 5), (8, - 1) and (- 6, 4), to get the range, we will get f(x, y) for all the domain values

For the coordinate (0,5);

f(0, 5) = (-2(0), 5)

f(0, 5) = (0, 5)

For the coordinate (8, -1);

f(8, -1) = (-2(8), -1)

f(8, -1) = (-16, -1)

For the coordinate (-6, 4);

f(-6, 4) = (-2(-6), 4)

f(0, 5) = (12, 4)

Hence the range is {(0, 5), (-16, -1), (12, 4)}

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Answer:

14.9 min

Step-by-step explanation:

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