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amm1812
4 years ago
8

Mr. Keller bought a car for $20,000. A study shows that a car will depreciate (go DOWN in value) by 15% each year. How much is M

r. Keller’s car worth after 5 years?
A) Formula used:
B)Substitute values:
C) Final answer
D) Does this final answer make sense compared to the original cost of the car? Why?
E) DESCRIBE (using a complete sentence or two) how you can solve this problem graphically.
Please answer all

Mathematics
1 answer:
Aneli [31]4 years ago
7 0
Average annual value lost: $7,390.65First-year depreciation: $3,000.00Total depreciation: $11,125.89Total depreciation percentage: 55.63%Value of vehicle at end of ownership period: $8,874.11
see attachment for graph


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Tom can buy 12 marbles.
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3 years ago
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Raise 5 to the 3rd power, then find the quotient of the result and s
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Answer:

125/s

Step-by-step explanation:

5^3 = 5 x 5 x 5 = 25 x 5 = 125

Find the quotient of the result and s

125 ÷ s

= 125/s

Hope this helps :)

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3 years ago
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5 0
3 years ago
The amount of money a worker makes varies directly with the hourly rate of pay. Worker A earns $168 for an 8 hour period. Worker
Nataly_w [17]
\bf \qquad \qquad \textit{direct proportional variation}\\\\
\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad  y=kx\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad  variation
\end{array}\\\\
-------------------------------

\bf \textit{money varies directly with hourly rate of pay}\qquad \stackrel{money}{m}=k\stackrel{hour~pay}{h}
\\\\\\
\textit{we also know that }
\begin{cases}
m=168\\
h=8
\end{cases}\implies 168=k8\implies \cfrac{168}{8}=k
\\\\\\
21=k\qquad therefore\qquad \boxed{m=21h}
\\\\\\
\textit{now, when h = 6, what is \underline{m}?}\qquad m=21(6)
3 0
3 years ago
How many 5 digit numbers are there whose digits sum to 39?
xenn [34]
The distribution of the possible digits of the numbers are
1.) 9, 9, 9, 9, 3  [Number of arrangements = 5! / 4! = 120 / 24 = 5]
2.) 9, 9, 9, 8, 4  [Number of arrangements = 5! / 3! = 120 / 6 = 20]
3.) 9, 9, 9, 7, 5  [Number of arrangements = 5! / 3! = 120 / 6 = 20]
4.) 9, 9, 9, 6, 6  [Number of arrangements = 5! / (3! x 2!) = 120 / 12 = 10]
5.) 9, 9, 8, 8, 5  [Number of arrangements = 5! / (2! x 2!) = 120 / 4 = 30]
6.) 9, 9, 8, 7, 6  [Number of arrangements = 5! / 2! = 120 / 2 = 60]
7.) 9, 9, 7, 7, 7  [Number of arrangements = 5! / (3! x 2!) = 120 / 12 = 10]
8.) 9, 8, 8. 8, 6  [Number of arrangements = 5! / 3! = 120 / 6 = 20]
9.) 9, 8, 8, 7, 7  [Number of arrangements = 5! / (2! x 2!) = 120 / 4 = 30]
10.) 8, 8, 8, 8, 7  [Number of arrangements = 5! / 4! = 120 / 24 = 5]

Number of 5 digit numbers whose digit sum up to 39 = 5 + 20 + 20 + 10 + 30 + 60 + 10 + 20 + 30 + 5 = 210
6 0
3 years ago
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