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kumpel [21]
3 years ago
7

Rodney drops a ball from the bridge at an initial height of 220 feet. How many seconds after the ball is released will it hit th

e ground? Write an equation to model this situation. Solve for time in seconds. Show work
Mathematics
1 answer:
bixtya [17]3 years ago
4 0

Answer:

t = √(2d/g); 3.70 s  

Step-by-step explanation:

The formula for the distance (d) travelled by a falling object in time (t) is  

d = ½gt²

Multiply each side by 2                    2d = gt²

Divide each side by g                 (2d/g) = t²

Take the square root of each side     t = √(2d/g)

Data:

d = 220 ft

g = 32.17 ft·s⁻²

Calculation:

t = √ [(2 × 220)/32.17] = √13.68 = 3.70 s

The ball will hit the ground 3.70 s after it is released.

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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
Sketch a cylinder with radius 8 feet and height 3 feet, then find the volume.
4vir4ik [10]

Answer: 339.29ft^3

Step-by-step explanation:

The formula for a cylinder is πr^2h

In your scenario, r=8 and h=3 so put those into the equation -

π8^2(3) and just solve!

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Answer: He still has to mow 45 lawns, or 75% of the lawns before he is finished.

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X= 2 I believe my friend
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A coupon subtracts $5.16 from the price (p) of a shirt. You pay $15.48 for the shirt after using the coupon. Write and solve an
lbvjy [14]

The equation would look like: p - 5.16 = 15.48

because the coupon subtracts 5.16 from p and the price you pay is 15.48

Solve the equation by adding 5.16 to both sides

p = 20.64

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