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svlad2 [7]
3 years ago
6

Blake simplified the expression (startfraction x superscript 12 baseline over x superscript negative 3 baseline endfraction) sup

erscript 5 to startfraction 1 over x superscript 20 baseline endfraction. What was blake’s mistake? he added 5 to the exponent in the numerator instead of multiplying. He subtracted the exponents in the parentheses instead of dividing. He multiplied only the exponent in the numerator of the fraction by 5. He divided the exponents in the parentheses instead of subtracting.
SAT
1 answer:
ra1l [238]3 years ago
5 0

Blake's mistake is (d) He divided the exponents in the parentheses instead of subtracting.

<h3>How to determine the mistake</h3>

The expression is given as:

(\frac{x^{12}}{x^{-3}})^5 = \frac{1}{x^{-20}}

The actual solution is as follows:

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

Add the exponents

(\frac{x^{12}}{x^{-3}})^5 = (x^{15})^5

Multiply 15 by 5

(\frac{x^{12}}{x^{-3}})^5 = x^{75}

Blake's simplified expression showed that:

He divided 12 by -3 and then multiplied the result by 5

Hence, Blake's mistake is (d) He divided the exponents in the parentheses instead of subtracting.

Read more about expressions at:

brainly.com/question/4344214

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Answer:

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Explanation:

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4 years ago
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Based on the standard formula and data in the question, the frequency of a sound source that produces the same wavelength in 20∘c air is "<u>115.87Hz</u>."

<h3>What is the frequency of a sound source?</h3>

The <u>frequency of a sound source</u> is the quantity of waves that move through a point per unit of time.

Typically, a sound wave arises from a source vibrating at a frequency f, propagates at v, and has a wavelength λ.

Therefore, in this case, we have

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Frequency (f) =?

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=> 115.87Hz

Hence, in this case, it is concluded that the correct answer is 115.87Hz.

Learn more about Frequency of Sound Source here: brainly.com/question/1418576

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