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Blababa [14]
2 years ago
5

The equation of line AB is y = 5x + 1. Write an equation of a line parallel to line AB in slope-intercept form that contains poi

nt (4, 5).
a) y = 5x - 15
b) y = 5x + 15
c) y = 1/5 + 21/5
d) y = 1/5 - 29/5
Mathematics
2 answers:
Ierofanga [76]2 years ago
6 0

Answer:

your answer is d. hope this helped and pls mark brainlest

Step-by-step explanation:

Zolol [24]2 years ago
4 0

Answer:

The equation of the line y= 5x+1. Line that is parallel to the function given should have the same slope as the equation which is 5. We use the point-slope form in order to find the second equation that contains the point (4,5).

y - 5 = 5(x-4)

y = 5x - 20 +5

y≈ 5x -15

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Endpoints of segment MN have coordinates (0, 0), (5, 1). The endpoints of segment AB have coordinates (1 1/22 , 2 1/4 ) and (−2
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Answer: k=18\dfrac{8}{11}.

Step-by-step explanation:

If a line passing through two points, then

Slope=\dfrac{y_2-y_1}{x_2-x_1}

Endpoints of segment MN have coordinates (0, 0) and (5, 1).

Slope of MN =\dfrac{1-0}{5-0}=\dfrac{1}{5}

The endpoints of segment AB have coordinates \left(1\dfrac{1}{22} , 2\dfrac{1}{4}\right) and  \left(-2\dfrac{1}{4} , k\right).

A=\left(1\dfrac{1}{22} , 2\dfrac{1}{4}\right)=\left(\dfrac{23}{22} ,\dfrac{9}{4}\right)

B=\left(-2\dfrac{1}{4} , k\right)=\left(-\dfrac{9}{4} , k\right).

Slope of AB =\dfrac{k-\frac{9}{4}}{-\frac{9}{4}-\dfrac{23}{22}}

=\dfrac{\frac{4k-9}{4}}{\frac{-99-46}{44}}

=\dfrac{4k-9}{4}\times \dfrac{44}{-145}

=4k-9\times \dfrac{11}{-145}

=\dfrac{44k-99}{-145}

Product of slopes of two perpendicular segments is -1.

Slope of MN × Slope of AB = -1

\dfrac{1}{5}\times \dfrac{44k-99}{-145}=-1

\dfrac{44k-99}{-725}=-1

44k-99=725

44k=725+99

k=\dfrac{824}{44}

k=\dfrac{206}{11}

k=18\dfrac{8}{11}

Therefore, the value of k is k=18\dfrac{8}{11}.

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