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kirza4 [7]
2 years ago
10

Determine whether the fractions 3 4 and 4 5 are equivalent. Question content area bottom Part 1 Select the correct choice below

and fill in the answer​ box(es) to complete your choice. ​(Simplify your​ answers.) A. The fractions are not equivalent because the product of the numerator of the first fraction and the denominator of the second fraction is enter your response here ​, which is not equal to the product of the numerator of the second fraction and the denominator of the first​ fraction, enter your response here . B. The fractions are equivalent because the product of the numerator of the first fraction and the denominator of the second fraction is equal to the product of the numerator of the second fraction and the denominator of the first fraction. That​ is, the cross product for each multiplication is enter your response here .
Mathematics
1 answer:
VMariaS [17]2 years ago
5 0

Answer: Choice A

The fractions are not equivalent because the product of the numerator of the first fraction and the denominator of the second fraction is <u>   15   </u> ​, which is not equal to the product of the numerator of the second fraction and the denominator of the first​ fraction <u>    16    </u>

===================================================

Explanation:

We have the template \frac{P}{Q} = \frac{R}{S} which cross multiplies to PS = QR

So if \frac{3}{4} = \frac{4}{5} was true, then 3*5 = 4*4 must be true as well, and vice versa.

The left side 3*5 turns into 15, while the right hand side 4*4 becomes 16. The 15 and 16 don't match up.

In short, 3*5 = 4*4 turns into 15 = 16 which is a false statement. So the original claim that \frac{3}{4} = \frac{4}{5} is also false.

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marishachu [46]
1. You have that:

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 2. Therefore, the tota area is:

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 L2 is the lenght of the walkway (L2=L1+3+3⇒L2=(W1+4+6)⇒L2=W1+10).
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2 years ago
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Answer:

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Step-by-step explanation:

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sesenic [268]

keeping in mind that perpendicular lines have negative reciprocal slopes, hmmm what's the slope of the equation above anyway?

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\bf \stackrel{\textit{perpendicular lines have \underline{negative reciprocal} slopes}} {\stackrel{slope}{\cfrac{2}{3}}\qquad \qquad \qquad \stackrel{reciprocal}{\cfrac{3}{2}}\qquad \stackrel{negative~reciprocal}{-\cfrac{3}{2}}}

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\bf (\stackrel{x_1}{0}~,~\stackrel{y_1}{0})~\hspace{10em} \stackrel{slope}{m}\implies -\cfrac{3}{2} \\\\\\ \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-\stackrel{y_1}{0}=\stackrel{m}{-\cfrac{3}{2}}(x-\stackrel{x_1}{0})\implies y=-\cfrac{3}{2}x

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Answer:

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Answer:

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