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zepelin [54]
3 years ago
10

Rewrite the following expression.

Mathematics
2 answers:
faust18 [17]3 years ago
6 0

We have been given the expression

x^{\frac{9}{7}}

We have the exponent rule

x^{\frac{a}{b}}= x^{a^{(\frac{1}{b})}}

Using this rule, we have

x^{\frac{9}{7}}= x^{9^{(\frac{1}{7})}}

Now, using the fact that x^{\frac{1}{n}}=\sqrt[n]{x}, we get

x^{\frac{9}{7}}= \sqrt[7]{x^9}\\
\\
x^{\frac{9}{7}}=\sqrt[7]{x^7\times x^2}\\
\\
x^{\frac{9}{7}}=x\sqrt[7]{x^2}

D is the correct option.

Ostrovityanka [42]3 years ago
4 0

<u>Answer</u>

D.  x (⁷√x²)


<u>Explanation</u>

x⁹/⁷  = ⁷√(x⁹)

      = ⁷√(x⁷ × x²)

       = (⁷√x⁷) × (⁷√x²)

        = x (⁷√x²)

The answer is D.

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Answer:

Given: \frac{x}{6} + 2 = 15                             ......[1]

To prove : x =78

Subtraction property states that you subtract the same number to both sides of an equation.

Subtract 2 from both sides of an equation [1];

\frac{x}{6} + 2 - 2= 15 -2

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Multiplication property states that you multiply the same number to both sides of an equation.

Multiply 6 to both sides of an equation [2];

\frac{x}{6} \times 6 = 13 \times 6

Simplify:

x = 78                      proved!

Statement                                           Reason

1.  \frac{x}{6} + 2 = 15                 Given

2.  \frac{x}{6} = 13                 Subtraction property of equality

3. x = 78                              Multiplication property of equality

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The addison see to the horizon at 2 root 2mi.

We have given that,Kaylib’s eye-level height is 48 ft above sea level, and addison’s eye-level height is 85 and one-third ft above sea level.

We have to find the how much farther can addison see to the horizon

<h3>Which equation we get from the given condition?</h3>

d=\sqrt{\frac{3h}{2} }

Where, we have

d- the distance they can see in thousands

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For Kaylib

d=\sqrt{\frac{3\times 48}{2} }\\\\d=\sqrt{{3(24)} }\\\\\\d=\sqrt{72}\\\\d=\sqrt{36\times 2}\\\\\\d=6\sqrt{2}....(1)

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To learn more about the eye level visit:

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