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ZanzabumX [31]
2 years ago
11

Cindy works at Jurassic Park and has been tasked to design a container in the shape of a rectangular prism for the incoming baby

dinosaurs. The scaled model of the container has dimensions 2m by 4m by 6m. Cindy has decided to increase each dimension of the scaled model by the same amount in order to produce a container with a volume of 84 times the volume of the scale model. By what amount should Cindy increase each dimension of the scaled model?
Mathematics
1 answer:
svetoff [14.1K]2 years ago
6 0

The amount Cindy should increase in each dimension of the scaled model is 12 m.  

<h3>What is a rectangular prism?</h3>

It is defined as the six-faced shape, a type of hexahedron in geometry.

It is a three-dimensional shape. It is also called a cuboid.

We have:

The scaled model of the container has dimensions 2m by 4m by 6m.

Volume of the scaled model = 2×4×6 = 48 cubic m

Let x be the amount by which each dimension is increased.

(2 + x)(4 + x)(6 + x) = 84(48)

\rm 48+44x+12x^2+x^3=4032

\rm x^3+12x^2+44x-3984=0

\rm x-12=0\quad \mathrm{or}\quad \:x^2+24x+332=0

The quadratic equation has no real solution so,

x = 12 m

Thus, the amount Cindy should increase in each dimension of the scaled model is 12 m.  

Learn more about the rectangular prism here:

brainly.com/question/21308574

#SPJ1

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3|2z-4|-6&gt;12<br> Solve for z
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Answer:

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2 years ago
Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2
Verdich [7]

Answer:

(a) f'(x)=-\frac{2}{x^3}

(b) y=-0.25x+0.75

Step-by-step explanation:

The given function is

f(x)=\frac{1}{x^2}                  .... (1)

According to the first principle of the derivative,

f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}

f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}

f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}

f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}

f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}

f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}

Cancel out common factors.

f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}

By applying limit, we get

f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}

f'(x)=\frac{-2x)}{x^4}

f'(x)=\frac{-2)}{x^3}                         .... (2)

Therefore f'(x)=-\frac{2}{x^3}.

(b)

Put x=2, to find the y-coordinate of point of tangency.

f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}

Substitute x=2 in equation 2.

f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

y-y_1=m(x-x_1)

y-0.25=-0.25(x-2)

y-0.25=-0.25x+0.5

y=-0.25x+0.5+0.25

y=-0.25x+0.75

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

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zhuklara [117]

we have

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If c is the independent variable

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Write the equation in function notation

f(c)=3c+5

therefore

the answer is

the equation in function notation is equal to f(c)=3c+5

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