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Art [367]
3 years ago
9

Each side of a square is increasing at a rate of 5 cm/s. At what rate is the area of the square increasing when the area of the

square is 36 cm2
Mathematics
2 answers:
Alexus [3.1K]3 years ago
6 0

Answer: 60

Step-by-step explanation:

A = s^2\\\\frac{dA}{dt}=2s \frac{ds}{dt}\\36=s^2 \\ s=6\\\frac{dA}{dt} = 2(6)(5)=60 \frac{cm^2}{s}

masya89 [10]3 years ago
3 0

Answer:

The area of the square is increasing at a rate of 60 cm²/s

Step-by-step explanation:

Let l be the length of the side of the square

Hence, The area of the square is given by

A = l^{2}

Where A is the area of the square

From the question, each side of a square is increasing at a rate of 5 cm/s, that is,

\frac{dl}{dt} = 5 cm/s

Now, to determine the rate at which the area of the square is increasing, That is \frac{dA}{dt}

To find \frac{dA}{dt}, differentiate A = l^{2} with respect to t

From, A = l^{2}

\frac{dA}{dt} = \frac{d(l^{2}) }{dl} ×\frac{dl}{dt}

\frac{dA}{dt} = 2l ×\frac{dl}{dt}

Now, to determine the rate at which the area of the square is increasing when the area of the square is 36 cm²

First, we will determine the length at this instance,

From, A = l^{2}

36 = l^{2}\\ \sqrt{36}  = l\\6 = l\\l = 6cm

The length at this instance is 6cm

To determine the rate at which the area of the square is increasing at this instance

l = 6cm\\

From the question,  \frac{dl}{dt} = 5 cm/s

Hence,

\frac{dA}{dt} = 2l ×\frac{dl}{dt}

\frac{dA}{dt} = 2(6) × 5

\frac{dA}{dt} = 60 cm^{2}/s

Hence. the area of the square is increasing at a rate of 60 cm²/s

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Answer:

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Step-by-step explanation:

Since you are already given the slope for this equation all you have to do is plug in the ordered pair in this equation: y = mx+b

Here is what this equation stands for:

y = the y in the order pair you are given, in this case it would be -6

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I Hope This Helped You Understand this Problem Better/More!!! Stay Safe and Healthy!

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