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svlad2 [7]
2 years ago
11

Which equation represents a line which is parallel to the line 2x + 7y = -56?

Mathematics
1 answer:
kicyunya [14]2 years ago
6 0

Answer:

last option:  y = -2/7x + 4

Step-by-step explanation:

original equation, in slope-intercept form, is:  y = -2/7x -8

Parallel lines have equal slopes

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Please answer correctly !!!!!!!!!!!!!!!!!!!! Will mark brainliest !!!!!!!!!!!!!!!!!
masha68 [24]

Answer:

The answer is 114 degrees.

Step-by-step explanation:

Since the angles are opposite from each other, it will be the same degrees.

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2 years ago
What is another way to name ∠MNO ?
katrin [286]

Alright, lets get started.

Please take a look to the diagram I have attached.

The base side is ON

The terminal side is MN.

We could say angle MNO as angle N.

We could day it as angle ONM also.

So, the answer is ∠ N and ∠ONM.   :  Answer

Hope it will help :)


7 0
3 years ago
What’s the correct answer
n200080 [17]
The third one would be the correct answer. I separately multiplied each number in the first section by each number in the second section. For instance, 0 x 2 is 0, 4 x 5 is 20, 5 x 3 is 15, and 2 x 0 is 0. All of these numbers are in the third answer.
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2 years ago
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d1i1m1o1n [39]
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5 0
2 years ago
X^2+4x+y^2-10y+20=30 find the center of the circle by completing the square
swat32

Answer:

a). Center of the circle = (-2, 5)

b). Equation of the line ⇒ y = -\frac{4}{5}x+\frac{58}{5}

Step-by-step explanation:

Equation of the circle is,

x² + 4x + y²- 10y + 20 = 30

a). [x² + 2(2)x + 4 - 4] + [y²- 2(5)y + 25] - 25 + 20 = 30

   [x² + 2(2)x + 4] - 4 + [y² - 2(5)y + 25] - 25 + 20 = 30

   (x + 2)² + (y - 5)²- 29 + 20 = 30

   (x + 2)² + (y - 5)²- 9 = 30

   (x + 2)² + (y - 5)² = 39

By comparing this equation with the standard equation of a circle,

    Center of the circle is (-2, 5).

b). A point (2, 10) lies on this circle.

    Slope of the line joining this point to the center (-2, 5),

    m_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}

          = \frac{10-5}{2+2}

          = \frac{5}{4}

    Let the slope of the tangent which is perpendicular to this line is 'm_{2}'

    Then by the property of perpendicular lines,

          m_{1}\times m_{2}=-1

          \frac{5}{4}\times m_{2}=-1

                 m_{2}=-\frac{4}{5}

   Now the equation of the line passing though (2, 10) having slope m_{2}=-\frac{4}{5}

           y - y' = m_{2}(x-x')

           y - 10 = -\frac{4}{5}(x-2)

           y - 10 = -\frac{4}{5}x+\frac{8}{5}

                  y = -\frac{4}{5}x+\frac{8}{5}+10

                  y = -\frac{4}{5}x+\frac{58}{5}

Therefore, equation of the line will be, y = -\frac{4}{5}x+\frac{58}{5}

7 0
3 years ago
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