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adoni [48]
3 years ago
9

What is the equation, in slope-intercept form, of the line that is perpendicular to the line y – 4 = –(x – 6) and passes through

the point (−2, −2)? y = –x – y = –x + y = x – 1 y = x + 1
Mathematics
1 answer:
Mkey [24]3 years ago
4 0
Let's convert the equation in point-slope form to slope-intercept form.

y - 4 = -(x - 6)
Multiply everything out.
y - 4 = -x + 6
Add 4 to both sides.
y = -x + 10

To be perpendicular, the slope of the new has to be the reciprocal of the other line. To slope in this equation is -1/1. To get the reciprocal, flip the numerator and denominator and change the sign too.

-1/1 becomes 1/-1, and changes to 1/+1, or 1.

y = x + b

Input the coordinate point and solve for b.

-2 = -2 + b
Add 2 to both sides.
0 = b

y = x
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a volley ball player servers the ball. The ball follows a path given by the equation y=-0.01x^2+0.5x+3 where x and y are measure
dolphi86 [110]

Answer:

(a)

Distance from player should be 13.82 feet or 36.2 feet

(b)

The ball will go over the net

Step-by-step explanation:

we are given

The ball follows a path given by the equation

y=-0.01x^2+0.5x+3

where

x and y are measured in feet and the origin is on the court directly below where the player hits the ball

(a)

net height is 8 ft

so, we can set y=8

and then we can solve for x

8=-0.01x^2+0.5x+3

8\cdot \:100=-0.01x^2\cdot \:100+0.5x\cdot \:100+3\cdot \:100

800=-x^2+50x+300

-x^2+50x-500=0

x^2-50x+500=0

we can use quadratic formula

x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

x=\frac{-50\pm \sqrt{50^2-4\left(-1\right)\left(-500\right)}}{2\left(-1\right)}

x=5\left(5-\sqrt{5}\right),\:x=5\left(5+\sqrt{5}\right)

x=13.82,x=36.2

So, distance from player should be 13.82 feet or 36.2 feet

(b)

we can plug x=30 and check whether y=8 ft

y=-0.01(30)^2+0.5(30)+3

y=9ft

we know that

height of net is 8 ft

so, the ball will go over the net


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Step-by-step explanation:

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