a quarter back completed 35% of his passes last year if he completed 56 passes how many attempts did he make ?
1 answer:
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We know the following:
Cylinder volume: V₁ = π r² h
Ball (sphere) volume:V₂ =
π r³
where:
V - volume
r - radius of base of cylinder and diameter of ball
h - height of cylinder.
R = 13 cm ⇒ r = 13 ÷ 2 = 6.5
π = 3.14
a) Since balls touch all sides of cylinder (as shown in image), it can be concluded that height of cylinder is equal to sum of diameters of 3 balls and that radius of base of cylinder is equal to radius of ball:
h = 3 × r = 3 × 13 cm = 39 cm
r = 6.5 cm
So,
V₁ = <span>π r² h
</span><span>V₁ = </span>3.14 × (6.5 cm)² × 39 cm
V₁ = 5,173.9 cm³
b. The total volume of three balls is the sum of volumes of each ball:
Vₐ = 3 × <span>V₂
</span>Vₐ = 3 × <span> π r³
</span>Vₐ = 3 × 3.14<span> (6.5 cm)³</span>
Vₐ = 3,449.3 cm³
c. Percentage of the volume of the container occupied by three balls ould be expressed as ratio of volume of three balls and volume of cylinder:
V =
×100
V =
×100
V = 0.6666 ×100
V = 66.66%
Cylinder volume: V₁ = π r² h
Ball (sphere) volume:V₂ =
where:
V - volume
r - radius of base of cylinder and diameter of ball
h - height of cylinder.
R = 13 cm ⇒ r = 13 ÷ 2 = 6.5
π = 3.14
a) Since balls touch all sides of cylinder (as shown in image), it can be concluded that height of cylinder is equal to sum of diameters of 3 balls and that radius of base of cylinder is equal to radius of ball:
h = 3 × r = 3 × 13 cm = 39 cm
r = 6.5 cm
So,
V₁ = <span>π r² h
</span><span>V₁ = </span>3.14 × (6.5 cm)² × 39 cm
V₁ = 5,173.9 cm³
b. The total volume of three balls is the sum of volumes of each ball:
Vₐ = 3 × <span>V₂
</span>Vₐ = 3 × <span>
</span>Vₐ = 3 ×
Vₐ = 3,449.3 cm³
c. Percentage of the volume of the container occupied by three balls ould be expressed as ratio of volume of three balls and volume of cylinder:
V =
V =
V = 0.6666 ×100
V = 66.66%
Answer:
D.
Step-by-step explanation:
Hello!
We can group the first two terms and the last two terms.
<h3>Factor by Grouping</h3>Factoring by grouping is the process of breaking down larger polynomials to smaller ones to factor. We can then combine like factors.
In the second step, we can see that we can rewrite as
, as both the two terms share a common factor of
. We can pull out
from that expression. Similarly,
and
share a common factor of
, so we can pull that out.