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S_A_V [24]
2 years ago
13

Kyle wants to put tile in his living room, dining room, and kitchen. How many square feet of tile does he need to buy to cover t

he entire area if there is a 7 foot by 10 foot island in the middle of the kitchen? Tile can only be sold in multiples of 10 square feet. Do not include units in your answer.

Mathematics
1 answer:
Dmitrij [34]2 years ago
8 0

Answer: 800  (Including the units: 800\ ft^2)

Step-by-step explanation:

The missing figure is attached.

1. Find the area of the semi-circle with this formula:

A=\frac{\pi r^2}{2}

Where "r" is the radius.

You can observe in the figure that the diameter of the semi-circle is:

d=15\ ft

Since the radius is half the diameter:

r=\frac{15\ ft}{2}=7.5\ ft

Substituting the radius into the formula, you get:

A_1=\frac{\pi (7.5\ ft)^2}{2}=88.35\ ft^2

2. The area of a rectangle can be calculated with this formula:

A=lw

Where "l" is the length and "w" is the width.

Since the tile won't be put in the island located at kitchen, you need to subtract the area of the smaller rectangle from the area of the larger rectangle. Then:

A_2=(36\ ft)(20\ ft)-(10\ ft)(7\ ft)=720\ ft^2-70\ ft^2=650\ ft^2

Therefore, the total area is:

A_t=88.35\ ft^2+650\ ft^2=738.35

Since tile can only be sold in multiples of 10 square feet, then he needs to buy  800\ ft^2

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To Find: Which payment option to recommend to a new employee.

Solution: I would recommend being a salaried employee.

Explanation:

We begin by calculating the typical number of hours worked per week.

Adding up the hours from the table, we have 0+8+8.5+9.5+10+8+3=47.

The payment for an hourly employee must be calculated as $26 per hour for working till 40 hours, and $39 per hour when they work more than 40 hours.

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As there are 52 weeks in a year, the yearly payment for an hourly emplyee would be (1313)(52) = 68276. That is, an hourly employee would earn $68276.

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Therefore, I would recommend a new emplyee to be paid a salary rather than work on an hourly basis.


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Answer:

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\frac{2}{3}

b)

\frac{2}{3}

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a) The first part requires that we use line integral to evaluate directly.

The line integral is

\int_C xydx +  {x}^{2}  {y}^{3} dy

where C is counterclockwise around the triangle with vertices (0, 0), (1, 0), and (1, 2)

The boundary of integration is shown in the attachment.

Our first line integral is

L_1 = \int_ {(0,0)}^{(1,0)} xydx +  {x}^{2}  {y}^{3} dy

The equation of this line is y=0, x varies from 0 to 1.

When we substitute y=0 every becomes zero.

\therefore \: L_1 =0

Our second line integral is

L_2 = \int_ {(1,0)}^{(1,2)} xydx +  {x}^{2}  {y}^{3} dy

The equation of this line is:

x = 0 \implies \: dx = 0

y varies from 1 to 2.

We substitute the boundary and the values to get:

L_2 = \int_ {1}^{2}1 \cdot y(0) +  {1}^{2}   \cdot \: {y}^{3} dy

L_2 = \int_ {1}^2 {y}^{3} dy =  \frac{8}{3}

The 3rd line integral is:

L_3 = \int_ {(1,2)}^{(0,0)} xydx +  {x}^{2}  {y}^{3} dy

The equation of this line is

y = 2x \implies \: dy = 2dx

x varies from 0 to 1.

We substitute to get:

L_3 = \int_ {1}^{0} x \cdot \: 2xdx +  {x}^{2}  {(2x)}^{3}(2 dx)

L_3 = \int_ {1}^{0} 8 {x}^{5}  + 2 {x}^{2} dx  =  - 2

The value of the line integral is

L = L_1 + L_2 + L_3

L = 0 +  \frac{8}{3}  +  - 2 =  \frac{2}{3}

b) The second part requires the use of Green's Theorem to evaluate:

\int_C xydx +  {x}^{2}  {y}^{3} dy

Since C is a closed curve with counterclockwise orientation, we can apply the Green's Theorem.

This is given by:

\int_C \: Pdx +Q  \: dy =  \int \int_ R \: Q_y -  P_x \: dA

\int_C \: xydx + {x}^{2} {y}^{3}   \: dy =  \int \int_ R \: 3 {x}^{2}  {y}^{2}  -  y \: dA

We choose our region of integration parallel to the y-axis.

\int_C \: xydx + {x}^{2} {y}^{3}   \: dy =  \int_ 0^{1} \int_ 0^{2x}  \: 3 {x}^{2}  {y}^{2}  -  y \: dydx

\int_C \: xydx + {x}^{2} {y}^{3}   \: dy =  \int_ 0^{1} \:  {x}^{2}  {y}^{3}  -   \frac{1}{2}  {y}^{2} |_ 0^{2x}  dx

\int_C \: xydx + {x}^{2} {y}^{3}   \: dy =  \int_ 0^{1} \:  8{x}^{5} -  2 {x}^{2}   dx =  \frac{2}{3}

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