You're going to need to be a bit more specific. What is it you're asking us?
Answer:
a. 215.6 in^3
b. 1.51 lb
Step-by-step explanation:
The area of each hand grip hole is that of a circle of radius 0.6 in together with a rectangle 2 in long and 1.2 in wide. So, that area is ...
π·(0.6 in)^2 + (2 in)(1.2 in) = (0.36π +2.4) in^2
The area of the kickboard before the hand grip holes are put in is that of a semicircle of radius 5.5 in together with a rectangle 12 in long and 11 in wide. So, that area is ...
(1/2)·π·(5.5 in)^2 + (12 in)(11 in) = (15.125π +132) in^2
Taking the hand grip holes out, the top area of the board is ...
((15.125π +132) -2(0.36π +2.4)) in^2
= (14.405π + 127.2) in^2
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a. The volume is the product of the area and the thickness, so is ...
((14.405π +127.2) in^2)·(1.25 in) ≈ 215.568 in^3
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b. The weight of the kickboard is the product of its volume and its density:
(215.568 in^3)(0.007 lb/in^3) ≈ 1.509 lb
Answer:
D. 60
Step-by-step explanation:
4(3x + 3) = 5(x + 8) 12(4) + 12 5(4) + 40
12x + 12 = 5x + 40 48 + 12 20 + 40
-5x -5x 60 60
7x + 12 = 40
-12 -12
7x = 28
x = 4
To make calculation easier, we first multiply
1.35 × 100 = 135
then we need to find how many groups of 5 are there in 135.
to do so, we simply take
135 ÷ 5 = 27
therefore, the answer is <u>27.</u>
