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ycow [4]
3 years ago
9

What is 12 1/2% of 256

Mathematics
1 answer:
s344n2d4d5 [400]3 years ago
6 0
To determine this, we need to set up fractional proportions.
First, put 12.5% (what we're looking for) over 100%, which is the total (256).
12.5/100 should be your first fraction.
Now, put x over 256, x being 12.5% of 256.
x/256 should be your second fraction.
Now put these two fractions as an equation.
x/256 = 12.5/100
This may be where things get tricky if you don't pay attention.
Cross multiply the top number (numerator) of x/256 with the bottom number (denominator) of 12.5/100.
You should end up with 100x.
Now, cross multiply the top number of 12.5/100 with the bottom number of x/256.
You should end up with 3200.
Now our equation is:
100x = 3200
Divide both sides by 100 to get x.
100x/100 = x
3200/100 = 32
x = 32 is now your simplified equation.
Your final answer is:
32 is 12.5% of 256.
Also, since 12.5% is 1/8 of 100%, we can test our answer.
Multiply 32 by 8 to get 8/8 (100%).
32 x 8 = 256.
Your answer is 32.
I hope this helps!

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The total distance phoebe has to drive each day (round trip) while her usual route is closed is 42 miles

<h3>What is a Pythagoras theorem?</h3>

The square of the longest side is equal to the square of the sum of the othersides.

Before we can calculate the total distance, we will need to get the hypotenuse on  both sides using the Pythagoras theorem as shown:

H^2= 12^2 + 9^2

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The total distance = 5 + 15 + 9 + 13

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The total distance phoebe has to drive each day (round trip) while her usual route is closed is 42 miles

Learn more on Pythagoras theorem here: brainly.com/question/12306722

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Use the change-of-basis identity,

\log_x(y) = \dfrac{\ln(y)}{\ln(x)}

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xyz = \log_a(bc) \log_b(ac) \log_c(ab) = \dfrac{\ln(bc) \ln(ac) \ln(ab)}{\ln(a) \ln(b) \ln(c)}

Use the product-to-sum identity,

\log_x(yz) = \log_x(y) + \log_x(z)

to write

xyz = \dfrac{(\ln(b) + \ln(c)) (\ln(a) + \ln(c)) (\ln(a) + \ln(b))}{\ln(a) \ln(b) \ln(c)}

Redistribute the factors on the left side as

xyz = \dfrac{\ln(b) + \ln(c)}{\ln(b)} \times \dfrac{\ln(a) + \ln(c)}{\ln(c)} \times \dfrac{\ln(a) + \ln(b)}{\ln(a)}

and simplify to

xyz = \left(1 + \dfrac{\ln(c)}{\ln(b)}\right) \left(1 + \dfrac{\ln(a)}{\ln(c)}\right) \left(1 + \dfrac{\ln(b)}{\ln(a)}\right)

Now expand the right side:

xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} \\\\ ~~~~~~~~~~~~+ \dfrac{\ln(c)\ln(a)}{\ln(b)\ln(c)} + \dfrac{\ln(c)\ln(b)}{\ln(b)\ln(a)} + \dfrac{\ln(a)\ln(b)}{\ln(c)\ln(a)} \\\\ ~~~~~~~~~~~~ + \dfrac{\ln(c)\ln(a)\ln(b)}{\ln(b)\ln(c)\ln(a)}

Simplify and rewrite using the logarithm properties mentioned earlier.

xyz = 1 + \dfrac{\ln(c)}{\ln(b)} + \dfrac{\ln(a)}{\ln(c)} + \dfrac{\ln(b)}{\ln(a)} + \dfrac{\ln(a)}{\ln(b)} + \dfrac{\ln(c)}{\ln(a)} + \dfrac{\ln(b)}{\ln(c)} + 1

xyz = 2 + \dfrac{\ln(c)+\ln(a)}{\ln(b)} + \dfrac{\ln(a)+\ln(b)}{\ln(c)} + \dfrac{\ln(b)+\ln(c)}{\ln(a)}

xyz = 2 + \dfrac{\ln(ac)}{\ln(b)} + \dfrac{\ln(ab)}{\ln(c)} + \dfrac{\ln(bc)}{\ln(a)}

xyz = 2 + \log_b(ac) + \log_c(ab) + \log_a(bc)

\implies \boxed{xyz = x + y + z + 2}

(C)

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