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MAVERICK [17]
4 years ago
10

Car rental company chargers it’s customers an initial fee of 30$ plus daily charge of 25$ to rent a car . Jacob has a budget of

200
Mathematics
2 answers:
nignag [31]4 years ago
6 0

Answer:

Jacob can rent a car for 6 days with $20 left to spare.

Step-by-step explanation:

200 - 30 = 170

170 / 25 = 6.8

25 * 6 = 150

170 - 150 = 20

sergij07 [2.7K]4 years ago
5 0

Answer:

55$ for spent and 145$ left

Step-by-step explanation:

He will spent 55$ and have 145$ left.

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Graph "-3n+4<25" on a line plot
Studentka2010 [4]

Answer:The solution is in the attached file

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3 years ago
This scatter plot shows the relationship between students’ scores on the first exam in a class and their corresponding scores on
son4ous [18]

Answer:

D) 2 points

Step-by-step explanation:

Compare a couple of points in the clumps of points and that is your answer

7 0
3 years ago
Find the value of 3a^2+2b, If a=3 and b = 4.<br><br>An explanation would be great.
vichka [17]

Answer:

=35

Step-by-step explanation:

3(3) ^2 + 2(4)

= 3(9) + 8

= 35

Where you see a, substitute with 3 and where you see b, substitute with 4

Just follow the steps above.

8 0
2 years ago
Read 2 more answers
Determine which relation is a function.
lys-0071 [83]
You know if something is a function or not because the X will never be used twice in a set. If it is, then it's not a function.
A. Uses -1 twice
B. Uses Nothing twice (good)
C. Uses 0 twice
D. Uses 0 twice
B is your answer


5 0
4 years ago
Why is the answer to this integral's denominator have 1+pi^2
ss7ja [257]

It comes from integrating by parts twice. Let

I = \displaystyle \int e^n \sin(\pi n) \, dn

Recall the IBP formula,

\displaystyle \int u \, dv = uv - \int v \, du

Let

u = \sin(\pi n) \implies du = \pi \cos(\pi n) \, dn

dv = e^n \, dn \implies v = e^n

Then

\displaystyle I = e^n \sin(\pi n) - \pi \int e^n \cos(\pi n) \, dn

Apply IBP once more, with

u = \cos(\pi n) \implies du = -\pi \sin(\pi n) \, dn

dv = e^n \, dn \implies v = e^n

Notice that the ∫ v du term contains the original integral, so that

\displaystyle I = e^n \sin(\pi n) - \pi \left(e^n \cos(\pi n) + \pi \int e^n \sin(\pi n) \, dn\right)

\displaystyle I = \left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n - \pi^2 I

\displaystyle (1 + \pi^2) I = \left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n

\implies \displaystyle I = \frac{\left(\sin(\pi n) - \pi \cos(\pi n)\right) e^n}{1+\pi^2} + C

6 0
2 years ago
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