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3 years ago

A swimmer swam the 50-meter freestyle in a time of 21 seconds. What was his speed to the nearest tenth of a meter per second?

2 answers:

6 0

I believe that the correct answer would be B:2.4

You would want to do 50/21 which would equal 2.38 which you would round to the nearest tenth which is 2.4.

dimaraw [331]3 years ago

3 0

50 / 21 = 2.38 rounds to 2.4 meters per second

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Find the average value of the function f(x, y, z) = 5x2z 5y2z over the region enclosed by the paraboloid z = 9 − x2 − y2 and the

Katyanochek1 [597]

The paraboloid meets the x-y plane when x²+y²=9. A circle of radius 3, centre origin.

Use cylindrical coordinates (r,θ,z) so paraboloid becomes z = 9−r² and f = 5r²z. 

If F is the mean of f over the region R then F ∫ (R)dV = ∫ (R)fdV 

∫ (R)dV = ∫∫∫ [θ=0,2π, r=0,3, z=0,9−r²] rdrdθdz 

= ∫∫ [θ=0,2π, r=0,3] r(9−r²)drdθ = ∫ [θ=0,2π] { (9/2)3² − (1/4)3⁴} dθ = 81π/2 

∫ (R)fdV = ∫∫∫ [θ=0,2π, r=0,3, z=0,9−r²] 5r²z.rdrdθdz 

= 5∫∫ [θ=0,2π, r=0,3] ½r³{ (9−r²)² − 0 } drdθ 

= (5/2)∫∫ [θ=0,2π, r=0,3] { 81r³ − 18r⁵ + r⁷} drdθ 

= (5/2)∫ [θ=0,2π] { (81/4)3⁴− (3)3⁶+ (1/8)3⁸} dθ = 10935π/8 

∴ F = 10935π/8 ÷ 81π/2 = 135/4

4 0

3 years ago

8h = 48 can yiu please solve this

lord [1]

What it is asking is 8 times what equals 48. The best way to do this is by doing the opposite operation. 48 divided by 8 equals 6. Then if you put it in the other way you get 8(6)=48 which works out. So the answer is 6.

8 0

3 years ago

Read 2 more answers

A heavy rope, 50 ft long, weighs 0.6 lb/ft and hangs over the edge of a building 120 ft high. Approximate the required work by a

Anastasy [175]

Answer:

Exercise (a)

The work done in pulling the rope to the top of the building is 750 lb·ft

Exercise (b)

The work done in pulling half the rope to the top of the building is 562.5 lb·ft

Step-by-step explanation:

Exercise (a)

The given parameters of the rope are;

The length of the rope = 50 ft.

The weight of the rope = 0.6 lb/ft.

The height of the building = 120 ft.

We have;

The work done in pulling a piece of the upper portion, ΔW₁ is given as follows;

ΔW₁ = 0.6Δx·x

The work done for the second half, ΔW₂, is given as follows;

ΔW₂ = 0.6Δx·x + 25×0.6 × 25 = 0.6Δx·x + 375

The total work done, W = W₁ + W₂ = 0.6Δx·x + 0.6Δx·x + 375

∴ We have;

W = $2 \times \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= 2 \times \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 750$

The work done in pulling the rope to the top of the building, W = 750 lb·ft

Exercise (b)

The work done in pulling half the rope is given by W₂ as follows;

$W_2 = \int\limits^{25}_0 {0.6 \cdot x} \, dx + 375= \left[0.6 \cdot \dfrac{x^2}{2} \right]^{25}_0 + 375 = 562.5$

The work done in pulling half the rope, W₂ = 562.5 lb·ft

6 0

2 years ago

A school sing games played 160 games and won 65 percent of them how many games did the team win?

jonny [76]

The team won 104 games
160 * 0.65 = 104

8 0

3 years ago

Do i solve this?? need help???

Arlecino [84]

A =
5 and -5
-1 and 3

¹/₃B =

-6 and 12
-7 and 7

¹/₃B + A =

-1 and 7
-8 and 10

7 0

3 years ago