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Angelina_Jolie [31]
3 years ago
11

Daniel Potter bought a new car for $20,000.00. Two years later, he wanted to sell it. He was offered $14,650.00 for it. If he so

ld it for that amount, what was his depreciation rate?
Mathematics
2 answers:
olga_2 [115]3 years ago
5 0
Answer: 13.375% per year

Explanation:

1) Depreciation is the loss of value: $ 20,000.00 - $ 14,650.00 = $ 5,350

2) The percent of depreciation is amount of the depreciation divided by the value of the car when purchased, times 100.

That is (5,350 / $ 20,000) * 100 = 26.75 %

2) The rate is percent of depreciation per year:

depreciation rate = % of depreciation / number of years = 26.75%  / 2 = 13.375% per year.
Oksana_A [137]3 years ago
5 0
First compute the different of the price like this:
 $20,000.00-$14,650.00=5350$<span>
Then divide like this:
</span>\frac{5350.00}{20,000}=75
Now compute the percentage like this:
0,26\times100=26.75\%
In order to compute the rate, divide by the number of years we get:
\dfrac{26.75}{2}=13.37\%
The rate of depreciation is 13.37\%. 
<span>

</span>
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3 years ago
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7 0
2 years ago
Which expression is equivalent??? help!
dexar [7]

Answer:

The equivalent expression for the given expression \sqrt[3]{256x^{10}y^{7} } is

4x^{3} y^{2}(\sqrt[3]{4xy} )

Step-by-step explanation:

Given:

\sqrt[3]{256x^{10}y^{7} }

Solution:

We will see first what is Cube rooting.

\sqrt[3]{x^{3}} = x

Law of Indices

(x^{a})^{b}=x^{a\times b}\\and\\x^{a}x^{b} = x^{a+b}

Now, applying above property we get

\sqrt[3]{256x^{10}y^{7} }=\sqrt[3]{(4^{3}\times 4\times (x^{3})^{3}\times x\times (y^{2})^{3}\times y   )} \\\\\textrm{Cube Rooting we get}\\\sqrt[3]{256x^{10}y^{7} }= 4\times x^{3}\times y^{2}(\sqrt[3]{4xy}) \\\\\sqrt[3]{256x^{10}y^{7} }= 4x^{3}y^{2}(\sqrt[3]{4xy})

∴ The equivalent expression for the given expression \sqrt[3]{256x^{10}y^{7} } is

4x^{3} y^{2}(\sqrt[3]{4xy} )

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3 years ago
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3 years ago
X to the 2 power -6 when X = -14
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3 years ago
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