Apply concepts of density in modeling situations

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The graphic below shows a blueprint for a swimming pool.

Swimming pool dimensions

If the pool is going to be 66 inches deep, how many cubic feet of water will it be able to hold? (1 ft = 12 in)

Explanation

Notice that the outer dimensions of the blueprint are the dimensions for the entire pool, including the concrete, while the inner dimensions are for the part of the pool that will be filled with water. Therefore, we want to focus on just the inner dimensions.

Notice that the depth is given in inches, while the dimensions are in feet. Convert 66 inches to feet by dividing 66 by 12, since 12 inches makes a foot:

$66;in\cdot\frac{1 ft}{12 in}=5.5 ft$

The inch units cancel out and leave us with just the feet units. 66 in is 5.5 ft.

Now we have all of the information we need to solve for the volume of the pool. The pool is a rectangular prism, and the formula for volume of a rectangular prism is

$length\cdot width\cdot height$

(In this case, the "height" of the swimming pool is its depth.)

The blueprint shows that the pool is 40 ft long and 30 ft wide. Plugging in the measurements from the problem, we get

$40,ft\cdot30,ft\cdot5.5,ft$

Multiplying this out, we get .

The graphic below shows a blueprint for a swimming pool.

Swimming pool dimensions

If the pool is going to be 66 inches deep, how many cubic feet of water will it be able to hold? (1 ft = 12 in)

Explanation

Notice that the depth is given in inches, while the dimensions are in feet. Convert 66 inches to feet by dividing 66 by 12, since 12 inches makes a foot:

$66;in\cdot\frac{1 ft}{12 in}=5.5 ft$

The inch units cancel out and leave us with just the feet units. 66 in is 5.5 ft.

Now we have all of the information we need to solve for the volume of the pool. The pool is a rectangular prism, and the formula for volume of a rectangular prism is

$length\cdot width\cdot height$

(In this case, the "height" of the swimming pool is its depth.)

The blueprint shows that the pool is 40 ft long and 30 ft wide. Plugging in the measurements from the problem, we get

$40,ft\cdot30,ft\cdot5.5,ft$

Multiplying this out, we get .

Find the area of a circle with the following radius:

$36\pi$

$6\pi$

$12\pi$

$66\pi$

$18\pi$

Explanation

The area of a circle is found using the following formula:

$A=\pi r^2$

In this formula the variable, , is the radius. Let's substitute in our known values and solve for the area.

$A=\pi 6^2$

$A=36\pi$

Find the area of a square with the following side length:

Explanation

We can find the area of a circle using the following formula:

In this equation the variable, , represents the length of a single side.

Substitute and solve.

A farmer has to make a square pen to hold chickens. If each chicken has to have $1 ;\textup{m}^2$ of area to roam and there are chickens total, what is the length of the amount of fencing required to pen in the chickens?

$20;\textup{m}$

$25 ;\textup{m}$

$50 ;\textup{m}$

$75 ;\textup{m}$

$83;\textup{m}$

Explanation

A square has area formula

The total area required for the chickens will be

$25 ;\textup{m}^2$

since each chicken requires $1;\textup{m}^2$ of space and there are chickens.

Thus, we have

for the length of our chicken fence.

$\sqrt{25}=\sqrt{l^2}^$

Since there are four sides of a square and each side has a length of 5, the total length of fence required is

$5\times4 = 20;\textup{m}$ .

$20;\textup{m}$

$25 ;\textup{m}$

$50 ;\textup{m}$

$75 ;\textup{m}$

$83;\textup{m}$

Explanation

A square has area formula

The total area required for the chickens will be

$25 ;\textup{m}^2$

since each chicken requires $1;\textup{m}^2$ of space and there are chickens.

Thus, we have

for the length of our chicken fence.

$\sqrt{25}=\sqrt{l^2}^$

Since there are four sides of a square and each side has a length of 5, the total length of fence required is

$5\times4 = 20;\textup{m}$ .

Give the area of the circle on the coordinate plane with equation

$x^{2}+y^{2}+ 6x-20y+ 9=0$ .

$100 \pi$

$50 \pi$

$10 \pi$

$20 \pi$

$40 \pi$

Explanation

We must first rewrite the equation of the circle in standard form

$(x-h)^{2} + (y-k)^{2}= r^{2}$ ;

will be the radius.

$x^{2}+y^{2}+ 6x-20y+ 9=0$

Subtract 9 from both sides, and rearrange the terms remaining on the left as follows:

$x^{2}+y^{2}+ 6x-20y+ 9- 9 =0- 9$

$x^{2}+ 6x + \underline{; ; ; ; ; ; }+y^{2}-20y+ \underline{; ; ; ; ; ; } = - 9$

Note that blanks have been inserted after the linear terms. In these blanks, complete two perfect square trinomials by dividing linear coefficients by 2 and squaring:

$(6 \div 2)^{2} = 3^{2} = 9$

$(20 \div 2)^{2} = 10^{2} = 100$

Add to both sides, as follows:

$x^{2}+ 6x + 9+y^{2}-20y+ 100 = - 9+9+ 100$

$x^{2}+ 6x + 9+y^{2}-20y+ 100 = 100$

Rewrite the expression on the left as the sum of the squares of two binomials.

$(x+3)^{2} + (y-10)^{2} = 100$

The equation is now in standard form.

$r^{2} = 100$

The area of this circle can be found using this formula:

$A = \pi r^{2} = \pi (100) = 100 \pi$ ,

the correct response.

A cube has surface area 6. Give the surface area of the sphere that is inscribed inside it.

$\pi$

$\frac{1}{3} \pi$

$\frac{2}{3} \pi$

$\frac{1}{6} \pi$

$\frac{1}{2} \pi$

Explanation

A cube with surface area 6 has six faces,each with area 1. As a result, each edge of the cube has length the square root of this, which is 1.

This is the diameter of the sphere inscribed in the cube, so the radius of the sphere is half this, or $r= \frac{1}{2}$ . Substitute this for in the formula for the surface area of a sphere:

$A = 4\pi r^{2} = 4\pi \left ( \frac{1}{2} \right )^{2} = 4\pi \left ( \frac{1}{4} \right ) = \pi$ ,

the correct choice.

A circle on the coordinate plane has center and passes through . Give its area.

$100 \pi$

$10 \pi$

$20 \pi$

$25 \pi$

The information given is insufficient to answer the question

Explanation

The radius of the circle is the distance between its center and a point that it passes through. This radius can be calculated using the distance formula:

$r = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }$ .