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torisob [31]
3 years ago
9

An inner city revitalization zone is a rectangle that is twice as long as it is wide. The width of the region is growing at a ra

te of 34 m per year at a time when the region is 260 m wide. How fast is the area changing at that point in time
Mathematics
1 answer:
lbvjy [14]3 years ago
3 0

Answer:

The rate of change of area is 35360 m /year

Step-by-step explanation:

Let the width of rectangle be w

We are given that  a rectangle that is twice as long as it is wide

So, Length of Rectangle  = 2w

Area of rectangle A = Length \times Breadth = w \times 2w= 2w^2

Differentiating area with respect to time

A=2w^2\\\frac{dA}{dt}=4w\frac{dw}{dt}

We are given that \frac{dw}{dt} = 34m/year and w = 260

\frac{dA}{dt}=4(260)(34)=35360

Hence the rate of change of area is 35360

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\begin{array}{ccccccc}&Natural&Whole &Integer&Rational&Irrational&Real\\-3\dfrac{3}{9}&&&& \checkmark&&\checkmark\\\sqrt{11} &&&&&\checkmark &\checkmark\\125&\checkmark&\checkmark&\checkmark&\checkmark&&\checkmark\\4\cdot \sqrt[3]{125} &\checkmark&\checkmark&\checkmark&\checkmark&&\checkmark\\7.\overline {17} &&&& \checkmark&&\checkmark\end{array}

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\begin{array}{ccccccc}&Natural&Whole &Integer&Rational&Irrational&Real\\-3\dfrac{3}{9}&&&& \checkmark&&\checkmark\\\sqrt{11} &&&&&\checkmark &\checkmark\\125&\checkmark&\checkmark&\checkmark&\checkmark&&\checkmark\\4\cdot \sqrt[3]{125} &\checkmark&\checkmark&\checkmark&\checkmark&&\checkmark\\7.\overline {17} &&&& \checkmark&&\checkmark\end{array}

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