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3 years ago

12

Find ( f - g)(x). f (x) = -2x2 - 2x - 3 and g(x) = 3x2 - 4x - 4

1 answer:

Kruka [31]3 years ago

3 0

Answer:

$\large\boxed{(f-g)(x)=-5x^2+2x+1}$

Step-by-step explanation:

$(f-g)(x)=f(x)-g(x)\\\\\text{We have}\ f(x)=-2x^2-2x-3\ \text{and}\ g(x)=3x^2-4x-4.\\\\\text{Substitute:}\\\\(f-g)(x)=(-2x^2-2x-3)-(3x^2-4x-4)\\\\(f-g)(x)=-2x^2-2x-3-3x^2+4x+4\qquad\text{combine like terms}\\\\(f-g)(x)=(-2x^2-3x^2)+(-2x+4x)+(-3+4)\\\\(f-g)(x)=-5x^2+2x+1$

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3 years ago

Read 2 more answers

A competitive knitter is knitting a circular place mat. The radius of the mat is given by the formula

Ilia_Sergeevich [38]

Answer: A. $A'(t)=\frac{4356\pi}{(t+11)^{3}}-\frac{527076\pi}{(t+11)^{5}}$

B. A'(5) = 1.76 cm/s

Step-by-step explanation: <u>Rate</u> <u>of</u> <u>change</u> measures the slope of a curve at a certain instant, therefore, rate is the derivative.

A. Area of a circle is given by

$A=\pi.r^{2}$

So to find the rate of the area:

$\frac{dA}{dt}=\frac{dA}{dr}.\frac{dr}{dt}$

$\frac{dA}{dr} =2.\pi.r$

Using $r(t)=3-\frac{363}{(t+11)^{2}}$

$\frac{dr}{dt}=\frac{726}{(t+11)^{3}}$

Then

$\frac{dA}{dt}=2.\pi.r.[\frac{726}{(t+11)^{3}}]$

$\frac{dA}{dt}=2.\pi.[3-\frac{363}{(t+11)^{2}}].\frac{726}{(t+11)^{3}}$

Multipying and simplifying:

$\frac{dA}{dt}=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}$

The rate at which the area is increasing is given by expression $A'(t)=\frac{4356\pi}{(t+11)^{3}} -\frac{527076\pi}{(t+11)^{5}}$ .

B. At t = 5, rate is:

$A'(5)=\frac{4356\pi}{(5+11)^{3}} -\frac{527076\pi}{(5+11)^{5}}$

$A'(5)=\frac{4356\pi}{4096} -\frac{527076\pi}{1048576}$

$A'(5)=\frac{2408693760\pi}{4294967296}$

$A'(5)=1.76268$

At 5 seconds, the area is expanded at a rate of 1.76 cm/s.

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Dmitriy789 [7]

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