What is the domain of f(x) = 2|x − 1| + 3?
A.
{x | x ≥ 3}
B.
{x | x < 1}
C.
{x | x ≤ -1}
D.
{x | x = all real numbers}

7/2

(decimal: 3.5)

welcome ; )

step-by-step explanation:

mnkjn

step-by-step explanation:

both the histogram and the stem and leaf plot will visually show how many shoppers were of various ages within each range of data.   the histogram does not give us the specific ages of these shoppers unlike the stem and leaf plot.a histogram does not give us a true range of the data because we do not know exactly how old the shoppers were. a stem and leaf plot will give exact ages, so we can determine the exact range along with all measures of center and the others measures of center.   with the histogram, we cannot find any of these.

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.