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faust18 [17]
3 years ago
6

The number forty one to the power of negative start fraction two over five end fraction end power can be written in the form sta

rt fraction one over root of order cap a of forty one to the power of cap b end power end fraction. What is the value of B?
Mathematics
1 answer:
Fantom [35]3 years ago
3 0
From the description we can infer that we have the expression: 41^{- \frac{2}{5} }.
Now, to write our expression as a as a root, we are going to apply the law of exponents: x^{-n}= \frac{1}{x^n} first
41^{- \frac{2}{5} }= \frac{1}{41^{ \frac{2}{5} }}

Next, we are going to apply the law about fractional exponents: x^{ \frac{m}{n}}= \sqrt[n]{x^m}
\frac{1}{41^{ \frac{2}{5} }}= \frac{1}{ \sqrt[5]{41^2} }

We can conclude that the value of B is 2.

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3.  P(x) - Q(x)---> \frac{-2(12x - 5)}{(3x - 1)(-3x + 2)}

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Q(x) = \frac{6}{-3x + 2}

Thus,

P(x) ÷ Q(x) = \frac{2}{3x - 1} ÷ \frac{6}{-3x + 2}

Flip the 2nd function, Q(x), upside down to change the process to multiplication.

\frac{2}{3x - 1}*\frac{-3x + 2}{6}

\frac{2(-3x + 2)}{6(3x - 1)}

= \frac{-3x + 2}{3(3x - 1)}

2. P(x) + Q(x) = \frac{2}{3x - 1} + \frac{6}{-3x + 2}

Make both expressions as a single fraction by finding, the common denominator, divide the common denominator by each denominator, and then multiply by the numerator. You'd have the following below:

\frac{2(-3x + 2) + 6(3x - 1)}{(3x - 1)(-3x + 2)}

\frac{-6x + 4 + 18x - 6}{(3x - 1)(-3x + 2)}

\frac{-6x + 18x + 4 - 6}{(3x - 1)(-3x + 2)}

\frac{12x - 2}{(3x - 1)(-3x + 2)}

= \frac{2(6x - 1}{(3x - 1)(-3x + 2)}

3. P(x) - Q(x) = \frac{2}{3x - 1} - \frac{6}{-3x + 2}

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\frac{-6x - 18x + 4 + 6}{(3x - 1)(-3x + 2)}

\frac{-24x + 10}{(3x - 1)(-3x + 2)}

= \frac{-2(12x - 5}{(3x - 1)(-3x + 2)}

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P(x)*Q(x) = \frac{2*6}{(3x - 1)(-3x + 2)}

P(x)*Q(x) = \frac{12}{(3x - 1)(-3x + 2)}

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