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gladu [14]
3 years ago
15

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.

Mathematics
1 answer:
lord [1]3 years ago
7 0

Answer:

A)  F(x) = 50 + ( x - 10 )*9  

B) Graph

C) lim ( x ⇒ 10) F(x) = 50

Step-by-step explanation:

A) F(x) = 50 + ( x - 10 )*9       since the first 10 hours are included in initial fee

B) In Annex

C) lim (x ⇒ 10) F(x) = 50 + ( 10 -10)*9

lim ( x ⇒ 10) F(x) = 50 + 0

lim ( x ⇒ 10) F(x) = 50

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VLD [36.1K]

Answer:

  $169.33

Step-by-step explanation:

Each of the friends spent ...

  $14.95 +6.99 +2.25 = $24.19

The seven friends together spent ...

  7($24.19) = $169.33

5 0
2 years ago
Need answers ASAP
andrew-mc [135]
Data:
r (radius) = 14 m
h (height) = 6 m
v (volume) = ?

Formula:
V =  \pi *r^2*h

Solving:
V = \pi *r^2*h
V =  \pi *14^2*6
V =  \pi *196*6
\boxed{\boxed{V = 1176 \pi\:m^3}}\end{array}}\qquad\quad\checkmark
4 0
3 years ago
I have no idea how to do this
Dimas [21]

Answer:

can u send me example of relatable solved question so I can try to help u

Step-by-step explanation:

sorry

7 0
3 years ago
4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.
nikitadnepr [17]

Answer:

\sum_{n=0}^9cos(\frac{\pi n}{2})=1

\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0

\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}

Step-by-step explanation:

\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))

=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})

=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1

2nd

\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}

=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0

3th

\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))

=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}

What we use?

We use that

e^{i\pi n}=cos(\pi n)+i sin(\pi n)

and

\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}

6 0
3 years ago
A football team gained 10 yards on one play and then lost 22 yards on the next. What was the overall change in field position.
grandymaker [24]

Answer:12 yards

Step-by-step explanation: 22-10=12

4 0
3 years ago
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