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ololo11 [35]
3 years ago
6

A sculpture at a park is in the shape of a rectangular pyramid. What is the volume of the sculpture if the length and width of t

he base of the pyramid are 11 feet and 5 feet respectively, while the height of the pyramid is 6 feet? ​
Mathematics
1 answer:
sweet [91]3 years ago
4 0

Answer:

40

Step-by-step explanation:

11x6=60

60-5x4= 20

=40

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Field book of an land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilatera
Kazeer [188]

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete question.

Field book of an agricultural land is given in the figure. It is divided into 4 plots. Plot I is a right triangle, plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

From the figure attached,

Area of the right triangle I = \frac{1}{2}(\text{Base})\times (\text{Height})

Area of ΔADC = \frac{1}{2}(\text{CD})(\text{AD})

                        = \frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})

                        = \frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)

                        = \frac{1}{2}(\sqrt{169-144})(12)

                        = \frac{1}{2}(5)(12)

                        = 30 cm²

Area of equilateral triangle II = \frac{\sqrt{3} }{4}(\text{Side})^2

Area of equilateral triangle II = \frac{\sqrt{3}}{4}(13)^2

                                                = \frac{(1.73)(169)}{4}

                                                = 73.0925

                                                ≈ 73.09 cm²

Area of rectangle III = Length × width

                                 = CF × CD

                                 = 7 × 5

                                 = 35 cm²

Area of trapezium EFGH = \frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})

Since, GH = GJ + JK + KH

17 = \sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}

12 = \sqrt{81-x^2}+\sqrt{225-x^2}

144 = (81 - x²) + (225 - x²) + 2\sqrt{(81-x^2)(225-x^2)}

144 - 306 = -2x² + 2\sqrt{(81-x^2)(225-x^2)}

-81 = -x² + \sqrt{(81-x^2)(225-x^2)}

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = \frac{1}{2}(5+17)(9)

                                = 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

                                    = 237.09 cm²

7 0
3 years ago
2/x multiplied by what equals 1
Varvara68 [4.7K]

Answer:

Multiplied by2x

2/x ×2x =1

5 0
3 years ago
Bradley cut a square hole out of a block of wood in wood shop. If the block was cube-shaped with side lengths of 9 inches, and t
MatroZZZ [7]
To solve this problem, you must follow the proccedure below:

 1. T<span>he block was cube-shaped with side lengths of 9 inches and to calculate its volume (V1), you must apply the following formula:

 V1=s</span>³
<span>
 s is the side of the cube (s=9)

 2. Therefore, you have:

 V1=s</span>³
 V1=(9 inches)³
 V1=729 inches³
<span>
 3. The lengths of the sides of the hole is 3 inches. Therefore, you must calculate its volume (V2) by applying the formula for calculate the volume of a rectangular prism:

 V2=LxWxH

 L is the length (L=3 inches).
 W is the width (W=3 inches).
 H is the heigth (H=9 inches).

 4. Therefore, you have:

 V2=(3 inches)(3 inches)(9 inches)
 V2=81 inches
</span><span>
 5. The amount of wood that was left after the hole was cut out, is:
</span>
 Vt=V1-V2
 Vt=648 inches³
7 0
3 years ago
a how much will he have in 4. At the age of 30, Jasmine started a retirement account with $50,000 which compounded interest semi
Natalka [10]

Answer:

6795.70

Step-by-step explanation:

smart people

6 0
2 years ago
Mr. Douglas drove 2,482 miles. He spent 42.5 hours in the car driving. What was his avrage speed for the trip?
Readme [11.4K]

Use the formula speed=distance/time, or s=d/t.

Substitute 2,482 for d and 42.5 for t.

s=2,482/42.5

Simply divide for the answer:

Mr. Douglas' average speed for the trip was 58.4 miles per hour.

I hope this helps! ^_^

4 0
3 years ago
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