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2 years ago

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A pile of earth removed from an excavation is a cone measuring 9 ft high and 24 ft across its base. How many trips will it take

to haul away the earth using a dump truck with a capacity of 9 cubic yards?

1 answer:

nydimaria [60]2 years ago

6 0

Answer: It would take 6 trips

Step-by-step explanation: The pile of earth has formed into a cone with the following dimensions, height is 9 feet and base diameter is 24 feet across.

However the haulage truck has its capacity measured in yards, and so we use a standard rate of conversion of yards to feet and vice versa. The standard rate is;

1 Yard = 3 Feet

Hence, the height of the pile of earth an now be given as;

h yards = 9/3

h yards = 3

height = 3 yards

Similarly, the the base of the pile of earth can be calculated as;

(Note that the radius is derived as diameter divided by 2, that is 24/2)

r yards = 12/3

r yards = 4

radius = 4 yards

The volume of a cone is given as;

Volume = πr²h/3

Volume = 3.14 x 4² x (3/3)

Volume = 3.14 x 16

Volume = 50.24 cubic yards

For a haulage truck that can haul 9 cubic yards per trip, the total number of trips required to haul the entire volume of dirt is given as follows;

No of trips = 50.24 ÷ 9

No of trips = 5.5822

No of trips ≈ 6

Therefore it would take the truck 6 trips to haul the entire pile of earth.

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Answer:

There was a 25% increase.

Step-by-step explanation:

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The yearly cost in dollars, y, for a member at a video game arcade based on total game tokens purchased, x, is y =y equals Start

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Answer:

$y = \frac{1}{5}x$

Step-by-step explanation:

Given

Yearly Cost for members; $y = \frac{1}{10}x + 6$

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From the question, we understand that:

<em>A non member pays $0.20 for each game;</em>

Yearly cost is then calculated as thus;

y = Amount * Number of Games

Where:

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2 years ago

Read 2 more answers

Pierre lifts a 60-kg crate onto a truck bed 1 m high in 3 s. Marie lifts twenty-four 2.5-kg boxes onto the same truck bed in a t

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Answer:

Pierre operates at a greater power level than Marie.

Marie and Pierre do the same amount of work

Step-by-step explanation:

Mass of crate=60kg

Height of truck bed=1 m

Time=3s

Pierre's work done= $60\times 1\times 9.8=588J$

Using formula Work done=mgh

Where m=mass of object

g=Acceleration due to gravity=9.8 $m/s^2$

Power= $\frac{w}{t}$

Using the formula

Pierre's power= $\frac{588}{3}=196watt$

Mass of box=2.5Kg

Total number of boxes=24

Total mass of boxes= $2.5\times 24=60kg$

Time=2 min= $2\times 60=120s$

1 min=60 s

Marie;s work done= $60\times 1\times 9.8=588J$

Marie's power= $\frac{588}{120}=4.9watt$

Pierre operates at a greater power level than Marie.

Marie and Pierre do the same amount of work

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2 years ago

Qué aspectos de la formación intelectual del escritor destaca la autora ?

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1 year ago

answer A radio station located 120 miles due east of Collinsville has a listening radius of 100 miles. A straight road joins Col

melamori03 [73]

Answer:

168.7602 miles

Step-by-step explanation:

One way to solve this problem is by using an equation that describes the listening radius of the station, and another for the road, then the points where this two-equation intersect each other will represent when the driver starts and stops listening to the station, and the distance between the points is the miles that the driver will receive the signal.

The equation for the listening radius (the radio station is at (0,0)):

$x^2+y^2=100^2$

The equation for the road that past through the points (-120,0) and (80,100) (Collinsville and Harmony respectively):

$m=\frac{y_2-y_1}{x_2-x_1} =\frac{100-0}{80-(-120)}=\frac{100}{200}=\frac{1}{2}$

$y-y_1=m(x-x_1)\\y-0=\frac{1}{2}(x-(-120))\\ y=\frac{1}{2}x+60$

Substitutes the value of y in the equation of the circle:

$x^2+(\frac{1}{2}x+60)^2=100^2\\x^2+\frac{1}{4} x^2+60x+3600=10000\\\frac{5}{4} x^2+60x+3600=10000\\\frac{5}{4} x^2+60x+3600-10000=0\\\frac{5}{4} x^2+60x-6400=0\\5 x^2+240x-25600=0\\x^2+48x-5120=0\\$

The formula to solve second-degree equations:

$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac} }{2a} \\x_{1,2}=\frac{-48\pm\sqrt{48^2-4(1)(-5120)} }{2(1)}\\x_{1,2}=\frac{-48\pm\sqrt{2304+20480} }{2}\\x_{1,2}=\frac{-48\pm\sqrt{22784} }{2}\\x_{1,2}=\frac{-48\pm16\sqrt{89} }{2}\\x_{1,2}=-24\pm8\sqrt{89} \\x_1=-24+8\sqrt{89}\approx51.4718\\x_2=-24-8\sqrt{89}\approx-99.4718\\$

Using the values in x to find the values in y:

$y_1=\frac{1}{2}x_1+60\\y_1=\frac{1}{2}(-24+8\sqrt{89} )+60\\y_1=-12+4\sqrt{89}+60\\ y_1=48+4\sqrt{89}\approx85.7359$

$y_2=\frac{1}{2}x_2+60\\y_2=\frac{1}{2}(-24-8\sqrt{89} )+60\\y_1=-12-4\sqrt{89}+60\\ y_1=48-4\sqrt{89}\approx10.2641$

The distance between the points (51.4718,85.7359) and (-99.4718,10.2641) :

$d=\sqrt{(x_1 -x_2 )^2+(y_1 -y_2)^2} \\d=\sqrt{(-24+8\sqrt{89} -(-24-8\sqrt{89}) )^2+(48+4\sqrt{89} -(48-4\sqrt{89}) )^2}\\d=\sqrt{(-24+8\sqrt{89} +24+8\sqrt{89} )^2+(48+4\sqrt{89} -48+4\sqrt{89} )^2}\\d=\sqrt{(16\sqrt{89} )^2+(8\sqrt{89} )^2}\\d=\sqrt{22784+5696}\\d=\sqrt{28480}\\d=8\sqrt{445}\approx168.7602miles$

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2 years ago