What is the equation of the line that is perpendicular

Ответ:

gabriieldelacruz16

18.04.2020

The answer is B because 1/2 is the reciprocal of 2 and it has to be a negative slope in order for it to cross the other line perpendicularly

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Ответ:

morganluvsblueow9wtm

13.11.2022

Mathematics

$\begin{aligned} \dfrac{1}{4} \sin(4y) & = \sin(y) \cos^3(y) - \cos(y)\sin^3(y) \\ \end{aligned}$ we use the following identities $\sin(2y) = 2\sin y\cos y$ $\cos(2y) = \cos^2y - \sin^2y$ $\begin{aligned} \mathrm{rhs} & = \sin(y) \cos^3(y) - \cos(y)\sin^3(y) \\ & = \sin(y) \left[ \cos^3(y) - \cos(y)\sin^2(y)\right] & & \text{factoring out $\sin(y)$}\\ & = \sin(y) \cos(y) \left[ \cos^2(y) - \sin^2(y)\right] & & \text{factoring out $\cos(y)$}\\ & = \sin(y) \cos(y) \cos(2y) & & \text{use } \cos(2y) = \cos^2y - \sin^2y\\ \end{aligned}$ $\begin{aligned} \text{lhs} & = \dfrac{1}{4} \sin(4y) \\ & = \dfrac{1}{4} \sin\big(2(2y)\big) & & \text{manipulate to use $\sin(2y)$ identity}\\ & = \dfrac{1}{4} \cdot 2\sin(2y)\cos(2y) \\ & = \dfrac{1}{2}\sin(2y)\cos(2y) \\ & = \dfrac{1}{2}\left[2\sin(y)\cos(y)\right]\cos(2y) \\ & = \sin(y)\cos(y)\cos(2y) \\ & = \mathrm{rhs} \end{aligned}$

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What is the equation of the line that is perpendicular to y = 2x – 4 and has a y-intercept of –2?
a.
x + 2y = –4
b.
y=-1/2x+2
c.
y=-2x-2
d.
2x + y = 2

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What is the equation of the line that is perpendicular to y = 2x – 4 and has a y-intercept of –2? a. x + 2y = –4 b. y=-1/2x+2 c. y=-2x-2 d. 2x + y = 2

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What is the equation of the line that is perpendicular to y = 2x – 4 and has a y-intercept of –2?
a.
x + 2y = –4
b.
y=-1/2x+2
c.
y=-2x-2
d.
2x + y = 2