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Anarel [89]
3 years ago
14

Which of the following pairs of equations represents lines that are parallel? 

Mathematics
1 answer:
vaieri [72.5K]3 years ago
3 0

Answer:

D

Step-by-step explanation:

The slope of the parallel lines have the same slope. So since D is the only answer choice with the same slope, the answer is D.

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Solve the equation:<img src="https://tex.z-dn.net/?f=%5Csqrt%7B%7D%20x%5E%7B2%7D%20%2B%204%20-%201%20%3D%202" id="TexFormula1" t
Georgia [21]

Answer:

D) x = √5

Step-by-step explanation:

\sqrt{x^{2}+4 }-1 = 2

\sqrt{x^{2}+4 } = 3

x^{2}+4 = 3^{2}

x^{2}+4 = 9

x^{2} = 9 -4

+x = \sqrt{5} ,  -x = \sqrt{5}

3 0
3 years ago
What is this graphs slope?
Ksivusya [100]

Answer:

slope = -4

Step-by-step explanation:

Given the following points on the graph: (-1, 4) (1, -4)

Let (x1, y1) = (-1, 4)

(x2, y2) =  (1, -4)

Use the following slope formula:

m = \frac{y2 - y1}{x2 - x1}

m = \frac{-4 - 4}{1 - (-1)} = \frac{-8}{1 + 1} = \frac{-8}{2} = -4

Therefore, the slope of the line is -4

3 0
3 years ago
Pleaseeeee helppppp
IRINA_888 [86]

Answer:

  (a)  (6, 2)

Step-by-step explanation:

The system of equations has one of them in y= form, so it lends itself to solution by substitution.

__

Using the equation for y to substitute into the first equation, we have ...

  2x -y = 10

  2x -(-1/2x +5) = 10 . . . . . substitute for y

  2x +1/2x -5 = 10 . . . . . eliminate parentheses

  5/2x = 15 . . . . . . . . . add 5, collect terms

  x = 6 . . . . . . . . . . . multiply by 2/5

Using the equation for y, we have ...

  y = -1/2(6) +5 = -3 +5

  y = 2

The solution is (x, y) = (6, 2).

3 0
2 years ago
What is the perimeter of the triangle?<br> O 22 units<br> O 30 units<br> O44 units<br> O 60 units
umka21 [38]

Answer:

22 units

Step-by-step explanation:

Add 12, 4 , and 6 and you will get 22

3 0
3 years ago
Read 2 more answers
Translate the sentence into an equation. Six more than the quotient of a number 5 and 9 equals . Use the variable y for the unkn
sashaice [31]
the answer is 6+5y=9
7 0
3 years ago
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