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
___
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
__
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:
(1) x^2 is greater than 1.
(2) x is greater than -1.
(1) x^2 is greater than 1 --> x2>1x2>1 --> x<−1x<−1 or x>1x>1. Sufficient.
(2) x is greater than -1 --> x>−1x>−1. Not sufficient.
C = Wtc/1000
1000C = Wtc
Wtc = 1000C
W = (1000C) / (tc)
This is assuming that C and c are not the same.
If they are the same, we have:
W = (1000) / t, provided ,that C and c are not equal to 0.
