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stepan [7]
2 years ago
15

What is the quotient of startfraction 7 superscript negative 1 baseline over 7 superscript negative 2 baseline endfraction?

Mathematics
1 answer:
Vika [28.1K]2 years ago
5 0

The expression 7⁻¹ ÷ 7⁻²  is a quotient expression, and the value of the quotient 7⁻¹ ÷ 7⁻²  is 7

<h3>How to evaluate the exponents?</h3>

The expression is given as:

7⁻¹ ÷ 7⁻²

Rewrite the expression using the product sign

7⁻¹ ÷ 7⁻²  = 7⁻¹ * 7²

Apply the product rule of indices

7⁻¹ ÷ 7⁻²  = 7⁻¹⁺²

Evaluate the sum of the exponents

7⁻¹ ÷ 7⁻²  = 7¹

Evaluate the exponent

7⁻¹ ÷ 7⁻²  = 7

Hence, the value of the quotient 7⁻¹ ÷ 7⁻²  is 7

Read more about quotient at:

brainly.com/question/629998

#SPJ4

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\bf (\stackrel{x_1}{1}~,~\stackrel{y_1}{4})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{-1}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-1-4}{6-1}\implies \cfrac{-5}{5}\implies -1 \\\\\\ \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-4=-1(x-1) \\\\\\ y-4=-x+1\implies y=-x+5


9)


\bf (\stackrel{x_1}{-12}~,~\stackrel{y_1}{14})\qquad (\stackrel{x_2}{6}~,~\stackrel{y_2}{-1}) \\\\\\ slope = m\implies \cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-1-14}{6-(-12)}\implies \cfrac{-15}{6+12}\implies \cfrac{-15}{18}\implies -\cfrac{5}{6}


\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-14=-\cfrac{5}{6}[x-(-12)] \\\\\\ y-14=-\cfrac{5}{6}(x+12)\implies y-14=-\cfrac{5}{6}x-10\implies y=-\cfrac{5}{6}x+4

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3 years ago
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