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Semenov [28]
2 years ago
12

Subtract the rational expression 1/x^2-x-6 - x/x+2.

Mathematics
1 answer:
Aleksandr [31]2 years ago
4 0

Answer:

Step-by-step explanation:

x² - x -6 = x²  + 2x - 3x - 3*2

            =x(x + 2) - 3(x + 2)

            = (x + 2 )(x - 3)

\frac{1}{x^{2}-x-6}+\frac{x}{x+2}=\frac{1}{(x +2 )(x - 3)}-\frac{x}{x + 2}\\

                      = \frac{1}{(x+2)(x-3)}-\frac{x(x-3)}{(x+2)(x-3)}\\\\=\frac{1-x(x-3)}{(x+2)(x-3)}\\\\=\frac{1-x^{2}+3x}{(x+2)(x-3)}\\\\

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Complete question:

A police car is located 50 feet to the side of a straight road. A red car is driving along the road in the direction of the police car and is 130 feet up the road from the location of the police car. The police radar reads that the distance between the police car and the red car is decreasing at a rate of 75 feet per second. How fast is the red car actually traveling along the road.

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The red car is traveling along the road at 80.356 ft/s

Step-by-step explanation:

Given

Police car is 50 feet side off the road

Red car is 130 feet up the road

Distance between them is decreasing at the rate of 75 feet per sec

Let x be how far the police is off the road.

Let y be how far the red car is up the road.  

Let h be the distance between the police and the red car.

This forms a right triangle so we can use the Pythagorean theorem, to solve for h

h² = x² + y²

h² = 50² + 130²

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h = √19400

h = 139.284 ft

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Let dx/dt be how fast the police is traveling toward the road.

Let dy/dt  be how fast the red car is traveling along the road.

Let dh/dt be how fast the distance between the police and the car is decreasing.

Recall that, h² = x² + y² (now differentiate with respect to time, t)

2h(dh/dt) = 2x(dx/dt) + 2y(dy/dt)

divide through by 2

h(dh/dt) = x(dx/dt) + y(dy/dt)

since the police car is not, dx/dt = 0

and dy/dt is the how fast is the red car actually traveling along the road

139.284(75) = 50(0) + 130(dy/dt)

10446.3 = 0 + 130(dy/dt)

dy/dt = 10446.3 / 130

dy/dt = 80.356 ft/s

Therefore, the red car is traveling along the road at 80.356 ft/s

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