Exercise 48 Page 33

Practice makes perfect

a The recommended weight of a soccer ball is

430

grams. The difference between the recommended weight and the actual weight

x

can be expressed as

x - 430 .

Since we are not interested in the sign, we will only consider an absolute value of the expression.

∣ x - 430 ∣

In the exercise we are told that the weight can vary up to

20

grams. Since we are looking for the minimum and maximum weights of the ball, we need to find for what

x

the expression above is equal to

20 .

∣ x - 430 ∣ = 20

To solve for

x,

we will split our absolute value equation into two separate cases.

∣ x - 430 ∣ = 20

$x - 430 \geq 0 : x - 430 = 20 x - 430 < 0 : x - 430 = - 20 (I) (II)$

x - 430 = 20 x - 430 = - 20 (I) (II)

$(I), (II):$ $LHS + 430 = RHS + 430$

x = 450 x = 410

The weight cannot be less than

410

grams, and it cannot be more than

450

grams.

b We know that the weight of a ball that was originally

423

grams decreased by

16

grams. To find out whether the weight is acceptable, we need to determine if it falls between the minimum and maximum acceptable weight. First, let's find its current weight.

423 - 16 = 407

The ball now weighs

407

grams. In Part A we found that the minimum acceptable weight is

410

grams, which is more than our ball weighs.

407 < 410

Therefore, the ball weighs less than the minimum acceptable weight and is not acceptable.

Subchapter links

Communicate Your Answer p.27

Monitoring Progress p.28-31

Exercises p.32-34

Exercise 48 Page 33

Hints

Check the answer

Practice exercises

Asses your skill