I WILL MARK BRAINLIEST Jared also gathered data from a competing real estate broker’s office. The competitor’s sales were equal to the average number of sales 8 months before Jared began gathering data and again 2 and 6 months after he began gathering data. The relative minimum number of sales was lower than the relative minimum number of sales of Jared’s office. Use the sliders for a, b, c, and d to create a graph that could model the competitor’s number of sales.

answer: bea's uncle is 25 years old.

step-by-step explanation:

given the equation , we can say that the variable "x" represents the age of bea's uncle. then to find his age, we need to solve for "x":

we must add 15 to both sides of the equation:

now we need to divide both sides of the equation by 3:

therefore, we can conclude that bea's uncle is 25 years old.

i dont know the answer but here's a picture of a dog

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.