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Artist 52 [7]
3 years ago
6

Luis purchases a collectible model car from a dealer. The dealer advises him that the car’s value (The car value is $75 now) wil

l grow by about 9% each year. What is the estimated value of the model car after 10 years?
Mathematics
2 answers:
bearhunter [10]3 years ago
5 0

Answer: $178

Step-by-step explanation:

We know that the exponential growth equation with rate of growth r in time period x is given by :-

f(x)=A(1+r)^x, A is the initial value .

Given: The initial value of car = $75

Rate of growth : 9%=0.09

Now, the function represents the car's value after x years is given by ;-

f(x)=75(1+0.09)^x=75(1.09)^x

Thus, the value of the model car after 10 years is given by :-

f(10)=75(1.09)^{10}=177.552275594\approx178   [To the nearest dollar]

Hence, the estimated value of the model car after 10 years = $178

serious [3.7K]3 years ago
4 0
Judging by the question I generated the equation y=75(1.09^x)
x is the amount of years. 
So the equation you should get for 10 years is y=75(1.09^10)
The answer you get should get is $177.55
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Select all the equations that are true when “x” is (-2).
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Answer:

The first one, the third one, and the fourth one.

Step-by-step explanation:

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1.

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Read 2 more answers
Estimate the difference 8 7/12 - 4 7/8 - 4/10
Gnesinka [82]
When you are told to estimate, you should be able to do the problem in your head.
8 7/12 is close to 8 1/2
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Now take away the 5


8 - 5 = 3

So the answer is very nearly
3 <<<<< answer.

Just for fun, I'll get the exact answer. Don't hand this one  in. It is 3 37/120
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4 0
3 years ago
Show me how you solve it
julia-pushkina [17]

Answer:

5^20

Step-by-step explanation:

<u>L</u><u>a</u><u>w</u><u> </u><u>o</u><u>f</u><u> </u><u>E</u><u>x</u><u>p</u><u>o</u><u>n</u><u>e</u><u>n</u><u>t</u><u> </u><u>I</u>

\displaystyle \large{ \frac{ {a}^{m} }{ {a}^{n} }  =  {a}^{m - n} }

Therefore:

\displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =  ({5}^{8 - 3})^{4}  } \\  \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =  ({5}^{5})^{4}  }

<u>L</u><u>a</u><u>w</u><u> </u><u>o</u><u>f</u><u> </u><u>E</u><u>x</u><u>p</u><u>o</u><u>n</u><u>e</u><u>n</u><u>t</u><u> </u><u>I</u><u>I</u>

\displaystyle \large{( {a}^{m} ) ^{n} =  {a}^{m \times n}  }

Thus:

\displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =  ({5}^{8 - 3})^{4}  } \\  \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =  ({5}^{5})^{4}  } \\  \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  = {5}^{5 \times 4}   } \\  \displaystyle \large{( \frac{ {5}^{8} }{ {5}^{3} } )^{4}  =   {5}^{20} }

7 0
3 years ago
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