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GalinKa [24]
3 years ago
10

Line AB has an equation of a line y = 5x − 2. Which of the following could be an equation for a line that is parallel to line AB

? y = 5x + 3, y = 1 over 5x + 3, y = −5x + 3, y = −1 over 5x + 3

Mathematics
2 answers:
schepotkina [342]3 years ago
8 0
I think it should be y=5x+3
Ne4ueva [31]3 years ago
6 0

Answer:

y = 5x + 3

Step-by-step explanation:

Parallel lines have the same slope.

Your function is

y= 5x - 2

slope = 5

The line must have slope = 5.

(a) y = 5x + 3

slope = 5, ∴ Parallel.

(b) y = 1/(5x) + 3.

Not a straight line. ∴ Not parallel.

(c) y = -5x +3

slope = -5 ∴ Not parallel.

(d) y = -1/(5x )+ 3

Not a straight line. ∴ Not parallel.

The Figure below shows the graph of y = 5x -2 (black) and that of the parallel line y = 5x + 3 (green).

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Simplify the expression.<br><br> [(11 - 4)^3]^2 ÷ (4 + 3)^5
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Answer:

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Step-by-step explanation:

\left[\left(11\:-\:4\right)^3\right]^2\:\div \left(4\:+\:3\right)^5\\\\\frac{\left(\left(11-4\right)^3\right)^2}{\left(4+3\right)^5}\\\\\mathrm{Subtract\:the\:numbers:}\:11-4=7\\\\=\frac{\left(7^3\right)^2}{\left(4+3\right)^5}\\\\\mathrm{Add\:the\:numbers:}\:4+3=7\\\\=\frac{\left(7^3\right)^2}{7^5}\\\\\left(7^3\right)^2=7^6\\\\=\frac{7^6}{7^5}\\\\\mathrm{Apply\:exponent\:rule}:\quad \frac{x^a}{x^b}=x^{a-b}\\\\\frac{7^6}{7^5}=7^{6-5}\\\\\mathrm{Subtract\:the\:numbers:}\:6-5=1\\\\=7

5 0
3 years ago
Can someone please help me answer this??
rewona [7]
<h3>Answer:  x = (y-2)^2 + 5</h3>

In other words, y-2 goes in the first box and 5 goes in the second box.

=================================================

Work Shown:

y^2 - 4y - x + 9 = 0

y^2 - 4y + 9  = x

x = y^2 - 4y + 9

x = y^2 - 4y + 4 + 5 .... rewrite 9 as 4+5

x = (y^2-4y+4) + 5

x = (y-2)^2 + 5  .... apply the perfect square factoring rule

So we'll have y-2 go in the first box and 5 goes in the second box

note: One version of the perfect square factoring rule says (a-b)^2 = a^2-2ab+b^2.

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