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:
$100.00
Step-by-step explanation:
%20 off of $125 is $100 dollars
Answer:
The correct answer is first option
24
Step-by-step explanation:
From the figure we get, mAXM = 72° and m<AMR = 38°
Also it is given that, all triangles are isosceles triangles and
m<FXA = 96°
<u>To find the measure of <FXM</u>
From the figure we get,
m<FXA = m<AXM + m<FXM
m<FXM = m<FXA - m<AXM
= 96 - 72
= 24
Therefore the correct answer is first option
24
