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Julli [10]
3 years ago
14

PLEASE HELP!!!!! I NEED THIS FAST!!!!!!

Mathematics
1 answer:
ioda3 years ago
5 0
Who says V1=V2?
if we simplify we get
(2/3)pir₁³=12pir₂²
for V1 to equal V2


a.
solve for r₁ to find r₁ as a function of r₂
(2/3)pir₁³=12pir₂²
times 3/2 both sides and divide by pi
r₁³=18r₂²
cube root both sides
r₁=∛(18r₂²)

if solve for r₂
(2/3)pir₁³=12pir₂²
divide by 12pi both sides
(1/18)r₁³=r₂²
squer root both sides
√((1/18)r₁³)=r₂



double radius of pond which is r1
√((1/18)r₁³)=r₂
r₁ turns to 2r₁ to double radius
√((1/18)(2r₁)³)=r₂double
√(8(1/18)(r₁)³)=r₂double
(√8)(√((1/18)(r₁)³))=r₂double
√((1/18)r₁³)=r₂ so
(√8)(r₂)=r₂double
(2√2)(r₂)=r₂double
the radius of the tank is multipled by 2√2
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Answer:

The cost of the material for the full-sized pad is;

d) 222,750.00

Step-by-step explanation:

The parameters of the steel pad and the scale model are;

The volume of the deck = 18 yd³

The dimensions of the model of the same material = 6 feet by 4.5 feet and 4.5 inches thick

The weight of the sample, m₂ = 75 lbs

The cost of the steel, P = $55/lb

The dimensions of the sample in yards are;

6 feet = 2 yards, 4.5 feet = 1.5 yards, and 4 inches = 0.\overline 1 yards

The volume of the sample, V₂ = 2 yd. × 1.5 yd. × 0.\overline 1 yd. =  (1/3) yd.³

The density of the sample material, ρ = m₂/V₂

∴ The density of the sample material, ρ = 75 lbs/((1/3)yd.³) = 225 lbs/yd³

Therefore, the density of the material of the pad, ρ = 225 lbs/yd³

The mass of the steel  pad, m₁ = ρ × V₁

∴ The mass of the steel  pad, m₁ = 225 lbs/yd³ × 18 yd³ = 4,050 lbs

The cost of the material for the full-sized steel pad, C = m₁ × P

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3 years ago
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Please see the attached file for explanation

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If f(x)=3x+1 and g(x)=x^2-6 find (f-g)(x)
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Hi there!

• f(x) = 3x + 1
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Then,
According to th' question :-

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~ Hope it helps!
6 0
3 years ago
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