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dybincka [34]
3 years ago
11

The length of a rectangle is increasing at a rate of 8cm/s and its width is increasing at a rate of 3 cm/s. When the length is 2

0 cm and the width is 10 cm, how fast is the area of the rectangle increasing?
Mathematics
1 answer:
Aloiza [94]3 years ago
8 0

Answer:

Step-by-step explanation:

let length=l

width=w

Area A=lw

\frac{dA}{dt}=l \frac{dw}{dt}+w \frac{di}{dt}\\\frac{dl}{dt}=8 cm/s\\\frac{dw}{dt}= 3 cm/s\\\frac{dA}{dt}=20 \times 3+10 \times 8=60+80=140 cm ^{2} /s

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There are a couple of ways to do this.

<h3>1) </h3>

Look for the GCF of the numerators when a common denominator is used.

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<h3>2) </h3>

Use Euclid's algorithm. If the remainder from division of the larger by the smaller is zero, then the smaller is the GCF; otherwise, the remainder replaces the larger, and the algorithm repeats.

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The GCF is 1/63.

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* The quotient is 28/27 = 1 +1/27 = 1 +(1/27)(3/7)/(3/7) = 1 +(1/63)/(3/7) or 1 with a remainder of 1/63.

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<em>Additional comment</em>

3/7 = (1/63) × 27

4/9 = (1/63) × 28

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