Ahot air balloon is 10 meters above the ground and rising at a rate of 15 meters per minute. another balloon is 150 meters above the ground and descending at a rate of 20 meters per minute. what is the solution of the system of equations? make sure you write the answer as an ordered pair.

(4, 70)

Step-by-step explanation:

Let's call

x: time, in minutes

y: height of the balloon, in meters

The height of the rising balloon is modeled as follows:

y = 15x + 10 (eq. 1)

The height of the descending balloon is modeled as follows:

y = -20x + 150 (eq. 2)

Combining equations 1 and 2:

15x + 10 = -20x + 150

15x + 20x  = 150 - 10

35x = 140

x = 140/35 = 4

Replacing it in equation 1:

y = 15(4) + 10 = 70

So, the solution is (4, 70), that is, after 4 minutes both balloons will have the same height, which is 70 meters.

answer: the answer is (b) amie should have multiplied

step-by-step explanation: given a sphere and a cylinder with same radius and height.

let, 'r' be the radius of the cylinder and sphere and 'h' be the height of the cylinder. according to the question,

r = h.

now, the volume of the sphere will be

and the volume of the cylinder will be

therefore, we have

thus, ammie should have multiplied the volume of the sphere

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step-by-step explanation: vsfdvsfvd