Algebra 2 Absolute Value, Piece-wise and Linear Programming: Spencer is an artist getting ready for a gallery opening. He paints both watercolors and pastels. Each type of picture takes him about the same amount of time to paint. He figures he has time to paint at most 16 pictures. The materials for each watercolor will cost him $15 and the materials for each pastel will cost him $5. He has $180 to spend on materials. Spencer makes $100 profit on each watercolor and $40 profit on each pastel. How many of each type should Spencer paint to achieve maximum profit?

a. 12.5 kg

step-by-step explanation:

f = ma

so

m = f/a = (2500 n)/(200 m/s²) = 12.5 kg

1 n = 1 kg·m/s²

step-by-step explanation:

graph using a few selected

x         y

-1       15

0         6

1         2.4

2       0.96

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.