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Effectus [21]
3 years ago
7

Suppose that the length of a certain rectangle is two centimeters more than three times its width. If the area of the rectangle

is 85 square centimeters, find its length and width
Mathematics
1 answer:
Angelina_Jolie [31]3 years ago
8 0

Answer:

width = 5 in

length = 17 in

Step-by-step explanation:

Area of a rectangle = length x width

width = w

length = 2 + 3w

85 = (w) x (2 + 3w)

85 = 2w + 3w³

this can be solved using quadratic equation

3w³ + 2w - 85 = 0

factors of 3w³ x -85 that add up to 2w are : +17 w and -15w

(3w³ - 15w) (17w - 85) = 0

3w(w - 5) 17(w - 5) 0

w = 5

or w = - 17/3

the width cannot be a negative number. the correct width is 5 in

Length = 2 x 3(5) = 17 inches

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The pressure is changing at \frac{dP}{dt}=3.68

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Suppose we have two quantities, which are connected to each other and both changing with time. A related rate problem is a problem in which we know the rate of change of one of the quantities and want to find the rate of change of the other quantity.

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The equation that model this situation is

PV^{1.4}=k

Differentiate both sides with respect to time t.

\frac{d}{dt}(PV^{1.4})= \frac{d}{dt}k\\

The Product rule tells us how to differentiate expressions that are the product of two other, more basic, expressions:

\frac{d}{{dx}}\left( {f\left( x \right)g\left( x \right)} \right) = f\left( x \right)\frac{d}{{dx}}g\left( x \right) + \frac{d}{{dx}}f\left( x \right)g\left( x \right)

Apply this rule to our expression we get

V^{1.4}\cdot \frac{dP}{dt}+1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}=0

Solve for \frac{dP}{dt}

V^{1.4}\cdot \frac{dP}{dt}=-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}\\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot V^{0.4} \cdot \frac{dV}{dt}}{V^{1.4}} \\\\\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}

when P = 23 kg/cm2, V = 35 cm3, and \frac{dV}{dt}=-4 \:{\frac{cm^3}{min}} this becomes

\frac{dP}{dt}=\frac{-1.4\cdot P \cdot \frac{dV}{dt}}{V}}\\\\\frac{dP}{dt}=\frac{-1.4\cdot 23 \cdot -4}{35}}\\\\\frac{dP}{dt}=3.68

The pressure is changing at \frac{dP}{dt}=3.68.

7 0
4 years ago
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