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:
81.68 is the correct answer. Hoped this helped.
3* (a + 1,5) = - 1,5 [:3
a + 1.5 = - 0,5
a = - 0,5 - 1.5 = -2
Answer:
Base = 24 cm or 10cm
Step-by-step explanation:
REMEMBER:
An isosceles triangle ABC with base BC = ‘b' & height AD = ‘h' & its equal sides =13 cm & area = 60 cm²
Using the formulas


There are 2 solutions for 
≈ 
Less complex:
Area of a triangle = 1/2 * b * h = 60
=> h = 120/b
In right triangle ABD
13² = h² + b² /4 ( by Pythagoras law)
=>169 = 120²/b² + b²/4
=>676 b² = 57600 + b^4
=> b^4 - 676 b² + 57600 = 0
=> b² = 676 +- √(676² - 4*57600) / 2
=> b²= 676 +- √(226576) /2
=> b² = (676 +- 476 )/2
=> b² = 1152/2 , 200 /2
=> b² = 576 , 100
=> b = 24, 10
So, Base = 24 cm or 10cm