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Verdich [7]
3 years ago
10

Find the equation of a line passing through the given point and perpendicular to the given equation (-5,-2) and y=-2x+2

Mathematics
2 answers:
liq [111]3 years ago
5 0

Answer:

y=1/2x+ 1/2


Step-by-step explanation:


ozzi3 years ago
5 0

Answer:

y = (1/2)x + 1/2

Step-by-step explanation:

An equation, y = -2x+2, is given; its slope is -2.  We are to find the equation of a line which is perpendicular to this given line; the slope of the perpendicular has a slope which is the negative reciprocal of that of the given line.  Thus, the slope of the perpendicular is                

                                                                        -1

                                                           m = -------------- = +1/2.

                                                                        -2

This perpendicular is to pass through (-5,-2).  Plugging these x- and y-coordinates and the slope m = 1/2 into the point-slope formula for a straight line, we get:

y-[-2] = (1/2)(x-[-5]), or y + 2 = (1/2)(x + 5).

This is the correct answer.  If, however, you want this equation in slope-intercept form, do the following:

Mult. all three terms of y + 2 = (1/2)(x + 5 by 2 to eliminate the fraction 1/2:

2y + 4 = x + 5.  Solve this for y:  Subtract 4 from both sides, obtaining:

2y = x + 1.  Finally, multiply all terms by 1/2:  y = (1/2)x + 1/2

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G(b) = 5b- 9 h(b) = (b - 1)^2 Evaluate. (hog)(-6) =
laiz [17]

Given:

The two functions are:

g(b)=5b-9

h(b)=(b-1)^2

To find:

The value of (h\circ g)(-6).

Solution:

We have,

g(b)=5b-9

h(b)=(b-1)^2

We know that,

(h\circ g)(b)=h(g(b))

(h\circ g)(b)=h(5b-9)

(h\circ g)(b)=[(5b-9)-1]^2

(h\circ g)(b)=[5b-10]^2

Putting b=-6, we get

(h\circ g)(-6)=[5(-6)-10]^2

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