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
__
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:
Step-by-step explanation:
Add the trinomials it becomes a binomial 7x removes. Becomes 2x^4-7(2x^2-7)(x^2+1)
A visual example of this is one my mother showed me. She drew a circle and laid out a piece of yarn around it so that it imitated the perfect shape of the circle. She laid out another piece that was as long as the circle's radius. She wrapped the longer piece around the shorter one, and I saw that it could wrap around it three times and still have at least 0.14 units left. I'm not sure if this helps, but the circumference of this circle is 18.84 cm ^2. I hope I answered your question.
I don't get it, what are you suppose to do?
Their total expenses are ...
.. $1.50 +3.00 +1.25 = $5.75
Their profit is the difference between their charge and their cost:
.. $25.00 -5.75 = $19.25
The correct choice is ...
B) $19.25
