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MatroZZZ [7]
3 years ago
5

Find an equation for the line that passes through the points (1, -1) and (-3, 1).

Mathematics
1 answer:
saul85 [17]3 years ago
3 0

Answer:

<h3>y = - 1/2x - 1/2</h3>

Step-by-step explanation:

Equation of line is y = mx + c

where m is the slope

c is the y intercept

Slope of the line using points

(1, -1) and (-3, 1) is

m =  \frac{1 + 1}{ - 3 - 1}  =  -  \frac{2}{4}  =  -  \frac{1}{2}

Equation of the line using point

( 1 , - 1) and slope - 1/2 is

y + 1 =  -  \frac{1}{2} (x - 1) \\ y + 1 =  -  \frac{1}{2} x +  \frac{1}{2}  \\ y =   - \frac{ 1}{2} x +  \frac{1}{2}  - 1 \\ y =  -  \frac{1}{2} x -  \frac{1}{2}

Hope this helps you.

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A golfer hits a ball from a starting elevation of 2 feet with a velocity of 70 feet per second down to a green with an elevation
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Answer:

4.50 s

Step-by-step explanation:

The motion of the ball is represented by the following equation:

h(t) = 2+70t-16t^2 (1)

where

h(t) is the elevation at time t

2 ft is the initial elevation at t = 0

+70 ft/s is the initial vertical velocity

-32 ft/s^2 is the acceleration due to gravity

The green is located at an elevation of -7 feet, so the ball lands on the green when

h(t) = -7

Substituting into (1)

-7=2+70t-16t^2

And re-arranging we get

16t^2-70t-9=0

This is a second-order equation in the form

at^2+bt+c=0

which has solutions

t_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a} (2)

Here we have:

a = 16

b = -70

c = -9

Substituting into (2) we find the solutions:

t_{1,2}=\frac{70\pm \sqrt{(-70)^2-4(16)(-9)}}{2(16)}

The two solutions are:

t = 4.50 s

t = -0.125 s

The second solution is negative, and since negative time has no physical meaning, the only correct solution is

t = 4.50 s

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