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kenny6666 [7]
3 years ago
13

Determine whether the given linear equations are parallel, perpendicular, or neither. 5x - 3y = - 6 3x + 5y = - 20

Mathematics
1 answer:
qwelly [4]3 years ago
3 0
We need to put these equations into y=mx+b format.
5x-3y=-6
-3y=-5x-6. (take 5x from both sides)
y=5/3 x +2 ( divide both sides by -3)
The slope is 5/3

3x+5y=-20
5y= -3x -20 (subtract 3x from both sides)
y=-3/5x -4 (divide both sides by 5)
The slope is -3/5

The slopes of perpendicular lines are the opposite (+/-) reciprocal. We have 5/3 and -3/5. This shows they are perpendicular.
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Answer:

\large\boxed{1.\ (4)^{-3x^2}=\left(\dfrac{1}{4}\right)^{3x^2}}

\large\boxed{2.\ ab^{-3x}=a\left(\dfrac{1}{b}\right)^{3x}=a\left[\left(\dfrac{1}{b}\right)^3\right]^x}

Step-by-step explanation:

Use:\ a^{-n}=\left(\dfrac{1}{a}\right)^n\\------------\\\\(4)^{-3x^2}=\left[(4)^{-1}\right]^{3x^2}=\left(\dfrac{1}{4}\right)^{3x^2}

Use:\ a^{-n}=\left(\dfrac{1}{a}\right)^n\ and\ (a^n)^m=a^{nm}\\--------------------\\\\ab^{-3x}=a\cdot b^{-3x}=a\left[(b)^{-1}\right]^{3x}=a\left(\dfrac{1}{b}\right)^{3x}\\\\ab^{-3x}=a\left(\dfrac{1}{b}\right)^{3x}=a\left[\left(\dfrac{1}{b}\right)^3\right]^x

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