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3 years ago

What kind of transformation is illustrated in this figure

Mathematics

1 answer:

kramer3 years ago

7 0

Answer:

The transformation in this figure is rotation.

Step-by-step explanation:

The transformation in this figure is rotation.

There are four types of transformation.

Translation:

Movement of the object without rotating it or changing its size.

Reflection:

Flipping the object about a line of reflection. The mirror image is formed after transformation.

Dilation:

Changing the size of a figure without changing its essential shape.

Rotation:

A rotation is a transformation that turns a figure about a fixed point called the center of rotation.An object and its rotation are the same shape and size, but the figures may be turned in different directions....

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A solid is formed by adjoining two hemispheres to the ends of a right circular cylinder. An industrial tank of this shape must h

mestny [16]

Answer:

Radius =6.518 feet

Height = 26.074 feet

Step-by-step explanation:

The Volume of the Solid formed = Volume of the two Hemisphere + Volume of the Cylinder

Volume of a Hemisphere $=\frac{2}{3}\pi r^3$

Volume of a Cylinder $=\pi r^2 h$

Therefore:

The Volume of the Solid formed

$=2(\frac{2}{3}\pi r^3)+\pi r^2 h\\\frac{4}{3}\pi r^3+\pi r^2 h=4640\\\pi r^2(\frac{4r}{3}+ h)=4640\\\frac{4r}{3}+ h =\frac{4640}{\pi r^2} \\h=\frac{4640}{\pi r^2}-\frac{4r}{3}$

Area of the Hemisphere = $2\pi r^2$

Curved Surface Area of the Cylinder = $2\pi rh$

Total Surface Area=

$2\pi r^2+2\pi r^2+2\pi rh\\=4\pi r^2+2\pi rh$

Cost of the Hemispherical Ends = 2 X Cost of the surface area of the sides.

Therefore total Cost, C

$=2(4\pi r^2)+2\pi rh\\C=8\pi r^2+2\pi rh$

Recall: $h=\frac{4640}{\pi r^2}-\frac{4r}{3}$

Therefore:

$C=8\pi r^2+2\pi r(\frac{4640}{\pi r^2}-\frac{4r}{3})\\C=8\pi r^2+\frac{9280}{r}-\frac{8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{24\pi r^2-8\pi r^2}{3}\\C=\frac{9280}{r}+\frac{16\pi r^2}{3}\\C=\frac{27840+16\pi r^3}{3r}$

The minimum cost occurs at the point where the derivative equals zero.

$C^{'}=\frac{-27840+32\pi r^3}{3r^2}$

$When \:C^{'}=0$

$-27840+32\pi r^3=0\\27840=32\pi r^3\\r^3=27840 \div 32\pi=276.9296\\r=\sqrt[3]{276.9296} =6.518$

Recall:

$h=\frac{4640}{\pi r^2}-\frac{4r}{3}\\h=\frac{4640}{\pi*6.518^2}-\frac{4*6.518}{3}\\h=26.074 feet$

Therefore, the dimensions that will minimize the cost are:

Radius =6.518 feet

Height = 26.074 feet

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