Atruck hauled 150 cubic feet of sand to a contraction site. the sand is dumped into a cone shaped pile 10 feet in height. what is the approximate diameter of the pile of sand

Diameter of pile of sand is 7.57 feet.

Step-by-step explanation:

The volume of the cone =

Here, let π = 3.14,H = 10 ft, and V = 150 cubic feet.

So, putting the values in the formula, we get

or ,

or, =  14.33

or, r =

⇒ r = 3.78 ft

As we know, 2 x RADIUS = DIAMETER

So, diameter of pile of sand = 2x 3.78= 7.57 feet.

multiplication is multiplication.   it might feel harder depending on what level of math you are in, but generally it will be the same old stuff, while implementing some new concepts and methods of solving work.

hello!

to begin with, we're given that on earth, a rock weighs 20 pounds, and while on jupiter, that same rock weighs 42 pounds. now, we're trying to find the weight of a rock that weighs 10.5 pounds on earth, on jupiter.

to solve this, first, set up a proportion, cross-multiply, and solve it algebraically.

setting up the proportion is the hardest and the most confusing part of the problem, but it's simple.

if on earth, the rock weighs 20 lbs, and on jupiter it weighs 42 lbs, our first part of our proportion is 20/42 lbs. the second part has to be equal to each other. this means that 10.5 has to be on the numerator of the fraction, and x, the value we're trying to find, has to be on the denominator. remember that the rock weighs 10.5 lbs on earth, and we're trying to find the weight of it on jupiter (which is x).

20/42 = 10.5/x

next, you would cross-multiply.

20x = 42(10.5)

20x = 441

now, solve it algebraically by using division.

x = 22.05 lbs

this can be rounded to 22.1 lbs.

therefore, the weight of the rock on jupiter is 22.1 lbs, the fourth choice.