A set of weights includes a 5 lb barbell and 4 pairs of weight plates. Each pair of plates weighs 20 Ib. If x pairs of plates are added to the barbell, the total weight
of the barbell and plates in pounds can be represented by f(x) = 20x+ 5.
What is the range of the function for this situation?
A {0, 1, 2, 3, 4}
B {5, 25, 45, 65, 85}
C {0, 20, 40, 60, 80}
D {5, 45, 85, 125}

The answer to this question would be B:

Based on the question, since the weight of the weight plates are 20 lbs, this would be represented by the 20x in the function. As well, the 5 lb barbell would be represented by the 5 in the function. The range of the function is determined by the amount of weight plates are added. So if I added one weight plate the equation would equal, f(x) = 20(1) + 5 = 25. This continues on the more and more weight plates you add.

Hope this reached you well :)

Step-by-step explanation:

20 = weight of the weight plates

x = amount of weight plates.

5 = weight of the barbell

f(x) = 20(0) + 5 = 5

f(x) = 20(1) + 5 = 25

f(x) = 20(2) + 5 = 45

f(x) = 20(3) + 5 = 65

f(x) = 20(4) + 5 = 85

24 units

step-by-step explanation:

b) 1/3 * 1/5

step-by-step explanation:

each smallest part is 1/15 since the written one is 1/5 and it's divided by 3 equal parts. (1/5 )/3 = 1/15 = 1/3 * 1/5