Given a diameter of 6, the radius must equal 3. Next, the circumference = 2πr = 2π(3) = 6π. The area formula gives you πr^2 =
π(3)^2 = 9π. Adding these two values together results in 6 π + 9π = 15π.
J
In order to avoid a fraction in the denominator, begin by cross multiplying: 8 = xy. Next, substitute for x. Then
Multiply both sides of this equation by 8, and we have 64 = y, choice (K).
78.
C
With a perimeter of 16, each side of square ABCD has a length of and thus each small square has a side length of 2.
Each small square has a perimeter of 8, and the 4 small squares have a total perimeter of (4)(8) = 32, choice (C).
79.
J
Here, pay special attention to the definition of the foreign symbol. The sum of 4, 5, and 6 is 4 + 5 + 6 = 15 and the product of
4, 5, and 6 is 4 × 5 × 6 = 120. So (4, 5, 6) = 120 − 15 = 105. Now (1, 2, 3) = 1 × 2 × 3 − (1 + 2 + 3) = 6 − 6 = 0. So (4, 5, 6)
− (1, 2, 3) = 105 − 0 = 105.
80.
C
If the distance from O to the edge of the square is 3, then the length of each side must be 2 × 3 = 6. The perimeter (the
distance around the square) = 4 × the length of one side = 4 × 6 = 24.
81.
J
Substitute the value of −3 for x in the given expression:
Recall that when working with absolute value, the result of the expression within the bars must become nonnegative when the
bars are removed.
82.
C
The perimeter of a square is the distance around it. Hence, because all four sides are equal, the length of one side is simply
the perimeter divided by
83.
G
The charge for delivering one pound is $25, so the charge for delivering of a pound is of $25, or $20. The tax is 8% of
$20, or 0.08 of $20, or $1.60. The total cost of delivering the item is $20 + $1.60, or $21.60, choice (G).
84.
E
There are several ways to solve this problem. One way is to figure out how many pages John plans to type in the morning and
afternoon. In the morning, he plans to type of the pages. Then of 70 is 14, so John plans to type 14 pages in the morning,
85.
