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Afina-wow [57]
3 years ago
7

alyssa bought a microwave $30,000 in the year 2007 it loses it value by 5% oer year what is the value of the microwave in the ye

ar 2011 please show working thank you
Mathematics
1 answer:
DedPeter [7]3 years ago
4 0
5% of 30,000 is 1,500
2007 til 2011 is 4 years so
1,500 times the 4 years would be 6,000
so 30,000 minus the 6,000 lost value would be 24,000.

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After losing the __________, the Chinese government was forced to make reparation payments to Britain.
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Option A. Anglo-Chinese War.

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An office manager reported that he spent $350 on gifts for employees. He said that $240 was spent on clothing with each man rece
Greeley [361]

Answer:

The correct answer is the solution of the system of equations are w = 10 and m = -8. The answer is not at all reasonable.

Step-by-step explanation:

The two system of equations are m + 2w = 12 and m + 3w = 22 where m and w represent the number of men and women working in an office.

We subtract both the sides with each other in order to solve the system of equation.

⇒ m + 3w - m - 2w = 22 - 12

⇒ w = 10.

⇒ m = -8.

The values of w and m are not reasonable at all. Since m and w represent the number of men and women in an office, they cannot be less than zero.

3 0
3 years ago
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riadik2000 [5.3K]

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Step-by-step explanation:

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The length of each side of a square is 3 inches more than the length of each side of a smaller square the sum of the areas of th
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The length of each side of a square is 3 in. more than the length of each side of a smaller square. The sum of the areas of the squares is 149 in. $^{2} .$ Find the lengths of the sides of the two squares.
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3 years ago
PLEASE HURRY IM TIMED!!!
andriy [413]

<u>Answer-</u>

<em>A. Brandon’s sound intensity level is 1/4th as compared to Ahmad’s.</em>

<u>Solution-</u>

Given that, loudness measured in dB is

L=10\log \frac{I}{I_0}

Where,

I   = Sound intensity,

I₀ = 10⁻¹² and is the least intense sound a human ear can hear

Given in the question,

I₁ = Intensity at Brandon's = 10⁻¹⁰

I₂ = Intensity at Ahmad's  = 10⁻⁴

Then,

L_1=10\log \frac{I_1}{I_0}=10\log \frac{10^{-10}}{10^{-12}}=10\log \frac{1}{10^{-2}}=10\log 10^{2}=2\times10\log 10=20

L_2=10\log \frac{I_2}{I_0}=10\log \frac{10^{-4}}{10^{-12}}=10\log \frac{1}{10^{-8}}=10\log 10^{8}=8\times10\log 10=80

\therefore \frac{L_1}{L_2} =\frac{20}{80} =\frac{1}{4}

5 0
3 years ago
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