Consider the following graph of a piecewise-defined function.

A rectangular swimming pool is twice as long as it is wide. A small concrete walkway surrounds the pool. The walkway is a 2 feet

Ksivusya [100]

Answer:

The width and the length of the pool are 12 ft and 24 ft respectively.

Step-by-step explanation:

The length (L) of the rectangular swimming pool is twice its wide (W):

$L_{1} = 2W_{1}$

Also, the area of the walkway of 2 feet wide is 448:

$W_{2} = 2 ft$

$A_{T} = W_{2}*L_{2} = 448 ft^{2}$

Where 1 is for the swimming pool (lower rectangle) and 2 is for the walkway more the pool (bigger rectangle).

The total area is related to the pool area and the walkway area as follows:

$A_{T} = A_{1} + A_{w}$ (1)

The area of the pool is given by:

$A_{1} = L_{1}*W_{1}$

$A_{1} = (2W_{1})*W_{1} = 2W_{1}^{2}$ (2)

And the area of the walkway is:

$A_{w} = 2(L_{2}*2 + W_{1}*2) = 4L_{2} + 4W_{1}$ (3)

Where the length of the bigger rectangle is related to the lower rectangle as follows:

$L_{2} = 4 + L_{1} = 4 + 2W_{1}$ (4)

By entering equations (4), (3), and (2) into equation (1) we have:

$A_{T} = A_{1} + A_{w}$

$A_{T} = 2W_{1}^{2} + 4L_{2} + 4W_{1}$

$448 = 2W_{1}^{2} + 4(4 + 2W_{1}) + 4W_{1}$

$224 = W_{1}^{2} + 8 + 4W_{1} + 2W_{1}$

$224 = W_{1}^{2} + 8 + 6W_{1}$

By solving the above quadratic equation we have:

W₁ = 12 ft

Hence, the width of the pool is 12 feet, and the length is:

$L_{1} = 2W_{1} = 2*12 ft = 24 ft$

Therefore, the width and the length of the pool are 12 ft and 24 ft respectively.

I hope it helps you!

8 0

3 years ago

I need to know the answer to number one

Law Incorporation [45]

Supplementary angles are two angles whose measures sum to a 180 degrees and complementary are the sum have to add up to 90 degrees.So 180÷2=90;90÷2=45;180÷6=30 and 90÷6=15.Altough the answer is A.

5 0

3 years ago

The volume of this rectangular prism is 36 cm^2. How many ½ cm cubes will fit in this rectangular prism? Show your reasoning.

nata0808 [166]

Answer:

Liquid R has a mass of of 1 kg at a temperature of 30°c kept in a refrigerator to freeze . Given the specific heat capacity is 300 J kg-¹ °c-1 and the freezing point is 4°c . Calculate the heat release by liquid R.

Step-by-step explanation:

Liquid R has a mass of of 1 kg at a temperature of 30°c kept in a refrigerator to freeze . Given the specific heat capacity is 300 J kg-¹ °c-1 and the freezing point is 4°c . Calculate the heat release by liquid R.

7 0

3 years ago

31/9 as a mix number

Nataly [62]

3 and 4/9 would be the answer to this.

8 0

3 years ago

Read 2 more answers

She is serving hot coffee and cold coffee in two rectangular-shaped tanks, each with dimensions width 8.75 inches, length 8.75 i

Blababa [14]

The volume of the coffee in each tank will be 1760. in³. The mass of cold coffee and hot coffee will be 6163.29 oz and 10213.45 oz. Then the number of the tank placed on a table is one only.

<h3>How to find the density and volume?</h3>

She is serving hot coffee and cold coffee in two rectangular-shaped tanks, each with dimensions width of 8.75 inches, a length of 8.75 inches, and a height of 23 inches.

The hot coffee has a density of 5.8 oz/in³ and the cold coffee has a density of 3.5 oz/ in³.

Then the volume of the coffee in each tank will be

We know that the dimension of both tanks is equal. Then the volume will remains equal.

Volume = L × B × H

Volume = 8.75 × 8.75 × 23

Volume = 1760.9375 ≅ 1760.94 in³

Then the mass of the coffee in each the tank will be

The mass of hot coffee will be

Mass = Density × Volume

Mass = 5.8 × 1760.94

Mass = 10213.45 oz

The mass of cold coffee will be

Mass = Density × Volume

Mass = 3.5 × 1760.94

Mass = 6163.29 oz

Then the number of the tank placed on a table is one only.

More about the density and volume link is given below.

brainly.com/question/952755

#SPJ1

3 0

2 years ago