2 years ago

Help! I don't understand this

Mathematics

1 answer:

olchik [2.2K]2 years ago

3 0

PQ=5i-4j 3QR+PQ-PQ=11i-4j-(5i-4j)=6i

3QR=6i QR=2i

R=Q+2i R=(4, 1)

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Has a slope of -2/5 and goes through (-10,-1)

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Answer:

Below in bold.

Step-by-step explanation:

Using the point-slope form of a straight line equation:

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3 years ago

Read 2 more answers

Explain why S = {(5, 4, -2), (-15, -12, 6), (10, 8, -4)} is NOT a basis for R^3 (1 point Sis linearly dependent and spans R3 S i

earnstyle [38]

$S$ would be a basis for $\mathbb R^3$ if

(1) the vectors in $S$ are independent, and

(2) the vectors span $\mathbb R^3$ .

Linear independence requires that $c_1=c_2=c_3=0$ is the only solution to

$c_1(5,4,-2)+c_2(-15,-12,6)+c_3(10,8,-4)=(0,0,0)$

These vectors are not linearly independent because if $c_1=3$ , $c_2=1$ , and $c_3=0$ , we have

$3(5,4,-2)+(-15,-12,6)=(15-15,12-12,-6+6)=(0,0,0)$

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Help! I don't understand this​

Help! I don't understand this