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DIA [1.3K]
3 years ago
10

Solve for a missing factor,

Mathematics
2 answers:
Eva8 [605]3 years ago
6 0

Answer:

1. 32

2. 170

3. 29

Step-by-step explanation:

1. is just 10 divided by 320, which gives you 32

2. is 10 times 17 which is 170

3. you need 29 tens to get 290.

BARSIC [14]3 years ago
5 0
1. 32
2. 170_________
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Just take the original fraction, and multiply it by some other
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Each fraction you multiply it by must have the value of ' 1 ' so
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The second fraction is equal to ' 1 ', because the top and the bottom
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Multiply the fractions:  (4 miles x 1 hour) / (2 hour x 60 minutes)

Now you can cancel 'hour' from the top and the bottom, and you have

             (4 miles x 1) / (2 x 60 minutes)

               = (4 miles) / (120 minutes) 

               =          (4 / 120) mile/minute = 0.0333... mile / minute .

Let's do it again, go a little farther, and get an answer that
might mean more and feel more like an answer. 

   (4 miles) / (2 hours) x (5280 feet / mile) x (1 hour / 60 minutes)

The 2nd and 3rd fractions both have the value of ' 1 ', because
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Multiply all three fractions: 

     (4 miles x 5280 feet x 1 hour) / (2 hours x 1 mile x 60 minutes)

You can cancel both 'mile' and 'hour' out of the top and bottom,
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Answer:

13) (5x)^{-\frac{5}{4} ⇒ \frac{1}{\sqrt[4]{(5x)^5}}

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Given expression:

13) (5x)^{-\frac{5}{4}

15) (10n)^{\frac{3}{2}

Write the expressions in radical form.

Solution:

For an expression with exponents as fraction like

(x)^{\frac{m}{n}

the numerator m represents the power it is raised to and the denominator n represents the nth root of the expression.

For an expression with exponents as negative  fraction like

(x)^{-\frac{m}{n}

We take the reciprocal of the term by rule for negative exponents.

So it is written as:

\frac{1}{(x)^{\frac{m}{n}}}

using the above properties we can write the given expressions in radical form.

13) (5x)^{-\frac{5}{4}

⇒ \frac{1}{(5x)^{\frac{5}{4}}}   [Using rule of negative exponents]

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⇒ \sqrt{(10n)^3}     [Since 2nd root is given as \sqrt{} in radical form]

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