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larisa [96]
3 years ago
6

A company invests $94,000 for equipment to produce a new product. each unit of the product costs $11.20 and is sold for $16.98.

let x be the number of units produced and sold. (a) write the total cost c as a function of x. c(x) = incorrect: your answer is incorrect. (b) write the revenue r as a function of x. r(x) = incorrect: your answer is incorrect. (c) write the profit p as a function of x. p(x) =\
Mathematics
1 answer:
antoniya [11.8K]3 years ago
5 0
Number of units = x

(a) c(x)= $94000 + 11.2x

(b) r(x) = 16.98x

(c) p(x) = 16.98x - 11.2x - 94000
     p(x) = 5.78x - 94000
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The height of a stuntperson jumping off a building that is 20 m high is modeled by the equation h = 20 -57, where t is the time
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A stuntman jumping off a 20-m-high building is modeled by the equation h = 20 – 5t2, where t is the time in seconds. A high-speed camera is ready to film him between 15 m and 10 m above the ground. For which interval of time should the camera film him?

Answer:

1\leq t\geq \sqrt{2}

Step-by-step explanation:

Given:

A stuntman jumping off a 20-m-high building is modeled by the equation

h =20-5t^{2}-----------(1)

A high-speed camera is ready to making film between 15 m and 10 m above the ground

when the stuntman is 15m above the ground.

height h = 15m  

Put height value in equation 1

15 =20-5t^{2}

5t^{2} =20-15

5t^{2} =5

t^{2} =1

t =\pm1

We know that the time is always positive, therefore t=1

when the stuntman is 10m above the ground.

height h = 10m  

Put height value in equation 1

10 =20-5t^{2}

5t^{2} =20-10

5t^{2} =10

t^{2} =\frac{10}{5}

t^{2} =2

t=\pm\sqrt{2}

t=\sqrt{2}

Therefore ,time interval of camera film him is 1\leq t\geq \sqrt{2}

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