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3 years ago

Unit 6 lesson 9: linear functions sample work

Mathematics

1 answer:

loris [4]3 years ago

4 0

Answer:

Initially, you have 2573 pieces.

You use 5 pieces each minute.

Right when you start, you have 2573 pieces.

One minute after that, you have 2573 - 5 pieces.

Another minute after, you have 2573 - 2*5 pieces.

Another minute after, you have 2573 - 3*5 pieces.

And so on.

You can see the pattern here, and with this, we can find the linear equation that represents the number of pieces that you have as a function of time.

Then, if the variable t represents the number of minutes that passed since you started, we can write the equation:

f(t) = 2573 - 5*t

That represents the number of pieces that you have after t minutes.

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Eighteen subtracted from half a number is 38.<br> Find the number.

Andrews [41]

Answer:

x=112

Step-by-step explanation:

x/2-18=38

first add +18

x/2=56

now multiply by 2

x=112

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3 years ago

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:

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