You're a manager in a company that produces rocket ships. Machine A and Machine B both produce cockpits and propulsion systems. Machine A ran for 22 hours and produced 3 cockpits and 5 propulsion systems. Machine B ran for 44 hours and produced 6 cockpits and 10 propulsion systems. Assume both machines produce cockpits at the same rate and both produce propulsion systems at the same rate.
Can we use a system of linear equations in two variables to solve for a unique amount of time that it takes each machine to produce a cockpit and to produce a propulsion system?

you bought 31 v bucks

step-by-step explanation:

25+6=31

step-by-step explanation:

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.