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A car bought for $13,000 depreciates at a rate of 15% per year. Use an exponential function to find the value of the car after 5

anygoal [31]

Answer:

$ 11093

Explanation:

Recall that the formula for compound interest is:

A = P ( 1 + r/ n ) n ( t )

−−−−−−−−−−−−−−−−−−−−−−

where:

A = future value

P = principal (starting) amount

r = annual interest rate, expressed as a decimal

n = number of times the interest is compounded in a year

t = number of years compound interest occurs for

1 . Start by substituting your known values into the formula. Note that the annual interest rate is − 0.15 since the car depreciates (becomes lower in value) each year.

A = 25000 ( 1 + − 0.15 1 ) 1 ( 5 )

2 . Solve for A .

A = 25000 ( 0.85 ) )5

A = 11092.63

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

A ≈ $ 11093 a

−−−−−−−−−−−−−−−

3 0

3 years ago

Read 2 more answers

Can someone plz help me on this I’m bad at doing these plz I bed u I beg u plz

Pepsi [2]

Answer:

I'm sorry me dont know kfnkdks

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3 years ago

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Find the area of this parallelogram. Be sure to include the correct unit in your answer.

Slav-nsk [51]

182 ft squared is the answer

5 0

3 years ago

Mathematics 13 don’t understand

UNO [17]

Answer:

1/6^3

Step-by-step explanation:

The applicable rules of exponents are ...

a^-b = 1/a^b

(a^b)(a^c) = a^(b+c)

Your expression can be simplified as follows:

$\dfrac{6^{-5}}{6^{-2}}=\dfrac{1}{(6^{-2})(6^5)}=\dfrac{1}{6^{-2+5}}=\boxed{\dfrac{1}{6^3}}$

_____

Additional comment

If you think of an exponent as signifying repeated multiplication, the rules of exponents may be easier to remember. The exponent tells you how many times the base is a factor in the product.

Consider multiplication:

$(x\cdot x\cdot x)\cdot(x\cdot x)=x^3\cdot x^2=x^{3+2}=x^5\\\\(x\cdot x\cdot x)\cdot(x\cdot x\cdot x)=(x^3)^2=x^{3\cdot2}=x^6$

Consider division:

$\dfrac{x\cdot x\cdot x}{x\cdot x}=x\quad\Longleftrightarrow\quad\dfrac{x^3}{x^2}=x^{3-2}=x^1\\\\\dfrac{x\cdot x}{x\cdot x\cdot x}=\dfrac{1}{x}\quad\Longleftrightarrow\quad\dfrac{x^2}{x^3}=x^{2-3}=x^{-1}$

This may help you see that a positive exponent in the denominator is equivalent to a negative exponent in the numerator (and vice versa).

6 0

3 years ago

7 ( 4 1/3 ) using distributive property can someone show me how to work this out?

scoray [572]

You’re gonna “distribute” the 7 into the parentheses by multiplying the 7 by the numbers inside the parentheses

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3 years ago