Question4: C, question 5: A
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:
2x^2(x^3+3x+8)
Step-by-step explanation:
Factor out the gcf which in this case is -2x^2 because that's the greatest factor all of the terms have in common this results in -2x^2(x^3+3x+8)
Btw the ^ represents the exponent
Answer:
x= 22.5°
Step-by-step explanation:
∠DEA= ∠BAE (alt. ∠s, DE// AB)
Substitute ∠DEA= 2x:
∠BAE= 2x
∠AEB +∠BEF= 180° (adj. ∠s on a str. line)
Substitute ∠BEF= 4x:
∠AEB +4x= 180°
∠AEB= 180° -4x
∠ABE +∠CBE= 180° (adj. ∠s on a str. line)
Substitute ∠CBE= 6x:
∠ABE +6x= 180°
∠ABE= 180° -6x
∠BAE +∠AEB +∠ABE= 180° (∠ sum of triangle)
2x +180° -4x +180° -6x= 180°
-8x +360°= 180°
8x= 360° -180°
8x= 180°
x= 180° ÷8
x= 22.5°
Answer:
C
Step-by-step explanation:
Since GJ bisects ∠ FGH , then ∠ FGJ = ∠ JGH = x + 14
∠ FGH = ∠ FGJ + ∠ JGH , substitute values
4x + 16 = x + 14 + x + 14 = 2x + 28 ( subtract 2x from both sides )
2x + 16 = 28 ( subtract 16 from both sides )
2x = 12 ( divide both sides by 2 )
x = 6
Thus
∠ FGJ = x + 14 = 6 + 14 = 20° → C