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3 years ago

13 is at least the difference of a number v and 1

Mathematics

1 answer:

Sphinxa [80]3 years ago

4 0

v - 1 ≥ 13

good luck!!

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Find the value of k so that 48x-ky=11 and (k+2)x+16y=-19 are perpendicular lines.

Rufina [12.5K]

Answer: k = -1 +/- √769

Step-by-step explanation:

48x - ky = 11

-48x -48x

-ky = -48x + 11

$\frac{-ky}{-k} = \frac{-48x}{-k} + \frac{11}{-k}$

$y =\frac{48x}{k} - \frac{11}{k}$

Slope: $\frac{48}{k}$

*************************************************************************

(k + 2)x + 16y = -19

- (k + 2)x -(k + 2)x

16y = -(k + 2)x - 19

$\frac{16y}{16} = -\frac{(k + 2)x}{16} - \frac{19}{16}$

$y = -\frac{(k + 2)x}{16} - \frac{19}{16}$

Slope: $-\frac{(k + 2)}{16}$

**********************************************************************************

$\frac{48}{k}$ and $-\frac{(k + 2)}{16}$ are perpendicular so they have opposite signs and are reciprocals of each other. When multiplied by its reciprocal, their product equals -1.

$-\frac{(k + 2)}{16}$ * $\frac{k}{48}$ = -1

$\frac{(k + 2)k}{16(48)}$ = 1

Cross multiply, then solve for the variable.

(k + 2)(k) = 16(48)

k² + 2k - 768 = 0

Use quadratic formula to solve:

k = -1 +/- √769

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