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3 years ago

6

Each side of a square is increasing at a rate of 6cm/s. At what rate is the area of the square increasing when the area of the s

quare is 16cm^(2). Show step-by-step solution

1 answer:

diamong [38]3 years ago

7 0

Area of a square = length²
A = x²
16 = x²; x = 4
dA/dx = 2x cm²/s
dx/dt = 6 cm/s
Using chain rule:
dA/dt = dA/dx * dx/dt
dA/dt = 2x * 6
dA/dt = 12x
At x = 4,
dA/dt = 12(4) = 48 cm²/s

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Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

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<em>From the above expression, the power of (5) is 2</em>

<em>Express 2 as 6 - 4</em>

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

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By direct comparison of $b^r= (-\frac{1}{2})^4$

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Substitute values for a, b, n and r in

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<em>Check the list of options for the expression on the left hand side</em>

<em>The correct answer is </em> $(5-\frac{1}{2})^6$ <em />

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