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julia-pushkina [17]
3 years ago
15

Find the solution(s) for x in the equation below.

Mathematics
2 answers:
liraira [26]3 years ago
7 0

Answer:c

Step-by-step explanation:

Maru [420]3 years ago
6 0

The answer is C.) x = 7; x = -6

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Towns X, Y, and Z are on the same straight road. Town Y is between Town X and Town Z. Town X is 57 miles from Town Z, and Town Y
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The distance between town X and Town Y is equal to the difference between the distance between Town X and Town Z and the distance from Town Y and Town Z. Mathematically
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Let u = (1, k) and v = (2, 1). Find k such that The distance between u and v is 1 u and v are orthogonal The angle between u and
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Answer:k=1,k=-2,k=8\pm 5\sqrt{3}

Step-by-step explanation:

Given two vectors

u=1\hat{i}+k\hat{j}

v=2\hat{i}+1\hat{j}

\left ( i\right )Distance between them is given by

|u-v|=\sqrt{\left ( 2-1\right )^2+\left ( 1-k\right )^2}=1

squaring both side

1^{2}+\left ( 1-k\right )^2=1

k^2-2k+1=0

\left ( k-1\right )^2=0

k=1

\left ( ii\right )

angle between u and v is 90 i.e. orthogonal

u\dot v=0

\left ( 1\hat{i}+k\hat{j}\right )\dot \left ( 2\hat{i}+1\hat{j}\right )=0

2+k=0

k=-2

\left ( iii\right )

angle between u & v is \frac{\pi }{3}

u\dot v=|u||v|cos\left (\frac{\pi }{3}\right )

|u|=\sqrt{1^2+k^2}

|v|=\sqrt{2^2+1^2}

2+k=\left ( \sqrt{1+k^2}\right )\left ( \sqrt{5}\right )cos\left ( \frac{\pi }{3}\right )

\left ( 4+2\right )^2=\left ( 1+k^2\right )5

k^2-16k-11=0

k=8\pm 5\sqrt{3}

7 0
3 years ago
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