A car sharing service offers a membership plan with a $50 per month fee that includes 10 hours of driving each month and charges $9 for each additional hour
(A) Write a piecewise definition of the cost F(x) for a month in which a member uses a car for x hours
(B) Graph F(x) for 0 < x s 15
(C) Find lim x--->10 F(x), lim x--->10F(x), and lim x--->10F(x), which- ever exist.

Answers

Answer 1

Answer and Step-by-step explanation:

Given:

Charges includes 10 hour per month = $50

Charges for additional hour = $9

Piecewise definition of cost F(x) for a month in which a member use a car for  hours.

Find limits?

Find lim x--->10 F(x), lim x--->10F(x), and lim x--->10F(x), which- ever exist.

Lim x → 10- F(x) = 50$

Lim x → 10 [ 9x – 70] = 9( 10) – 70  

                                = 90 – 70

                                = 20 > 10

Limit does not exist.

Lim x → 10+ F(x) = 50$

Lim x → 10+  F(x) = Lim x → 10+   [90x – 70]

                         = 90 – 70

                          = 20 > 10

Limit does not exist.

Lim x → 10 F(x) = 50$

An integer or an decimal.

The limit does not exist.

Answer 2

The piece wise definition of the cost [tex]f(x)[/tex] for a month in which a member uses a car for [tex]x[/tex] hours is given by the equation,[tex]f(x)= 50+9(x-10)[/tex].

Given,

Membership plan is [tex]\$50[/tex] per month.

We have to write a piece wise definition of the cost [tex]f(x)[/tex] for a month in which a member uses a car for [tex]x[/tex] hours.

Here, [tex]\$[/tex][tex]$50[/tex] per month fees includes 10 hours of driving each month, so the cost for a month is given as,

[tex]f(x)= 50+9(x-10)[/tex]

Here [tex]x[/tex] is the no. of additional hours for the month.

Further,

[tex]f(x)= 9x-40[/tex].

Now at,

[tex]lim _{x \to 10^+}f(x) = 9\times 10-40\\[/tex]

Or,

[tex]f(x)= 90-40\\f(x)=50[/tex]

Similarly, at left limit,

[tex]lim _{x \to 10^-}f(x) = 9\times (-10)-40[/tex]

Or,

[tex]f(x)=-90-40[/tex]

[tex]f(x)=-130[/tex]

Finally at [tex]x=10[/tex], [tex]f(x)=9\times10-40[/tex]

[tex]f(x)=90-40\\f(x)=50[/tex]

Since [tex]lim{x \to}10^+[/tex] [tex]\neq limx\to 10^-\neq f(x) (at x= 10)[/tex]

So the required function [tex]f(x)[/tex] is discontinuous at [tex]x=10[/tex]

Hence the piece wise definition of the cost [tex]f(x)[/tex] for a month in which a member uses a car for [tex]x[/tex] hours is given by the equation,[tex]f(x)= 50+9(x-10)[/tex].

For more details on limit follow the link below:

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Step-by-step explanation:

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Answers

Answer:

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espera y termino y te ayudo vale que me falta poco vale

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Answer:

1. Its about 14.90697674418605. I’m not sure if u want it rounded up if u want it rounded up to the nearest whole it would be 15 if I wanted it rounded up to the nearest tenth then it would be 14.9

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A ball is thrown from the top row of seats in a stadium. The function h(t) = −16t2 + 64t + 80 gives the height, h, in feet, of the ball t seconds after it is thrown. How long will it be before the ball hits the ground?

Answers

Answer:

Step-by-step explanation:

Given the height of the ball expressed as:

h(t) = −16t^2 + 64t + 80

The ball hits the ground the when h(t) = 0

Substitute h(t) = 0 and find t as shown:

h(t) = −16t^2 + 64t + 80

0 = −16t^2 + 64t + 80

−16t^2 + 64t + 80 = 0

Divide throigh by -16

t^2-4t-5 = 0

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t(t-5)+1(t-5) = 0

(t+1)(t-5)= 0

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time can not be negative:

t = 5seconds

Hence it will take 5 secs before the ball his the ground

Solving a quadratic equation, it is found that the ball hits the ground after 5 seconds.

The height of a ball after t seconds is given by:

[tex]h(t) = -16t^2 + 64t + 80[/tex]

It hits the ground at the values of t for which:

[tex]h(t) = 0[/tex]

Then

[tex]-16t^2 + 64t + 80 = 0[/tex]

Simplifying by -16:

[tex]t^2 - 4t - 5 = 0[/tex]

Which is a quadratic equation with coefficients [tex]a = 1, b = -4, c = -5[/tex].

Then:

[tex]\Delta = (-4)^2 - 4(1)(-5) = 36[/tex]

[tex]x_{1} = \frac{-(-4) + \sqrt{36}}{2} = 5[/tex]

[tex]x_{2} = \frac{-(-4) - \sqrt{36}}{2} = -1[/tex]  

We want the positive value, thus, the ball hits the ground after 5 seconds.

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