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Nady [450]
1 year ago
15

Tom takes exactly 30 minutes to rake a lawn and his son Mike takes exactly 60 minutes to rake the same lawn. If Tom and Mike dec

ide to rake the lawn together, and both work at the same rate that they did previously, how many minutes will it take them rake the lawA. 16
b. 20


c. 36


d. 45


e. 90
Mathematics
1 answer:
____ [38]1 year ago
6 0

Answer:

b) 20 minutes

Step-by-step explanation:

Tom=0.5Mike

in 15 minutes, Tom will have done 1/2 of the lawn

in 15 minutes, Mike will have done 1/4 of the lawn

in 15 minutes, they will have done 3/4 of the lawn

15/3=5 minutes for 1/4 of the lawn to be done

5x4=20 minutes for the entire lawn

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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 = 4 − x2 − y2 and the
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The paraboloid meets the x-y plane when x²+y²=9. A circle of radius 3, centre origin. 

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

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

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

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


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

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

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

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3 0
1 year ago
Three out of seven students in the cafeteria line are chosen to answer survey questions. How many different combinations of thre
Andreyy89

Answer:

The number of different combinations of three students that are possible is 35.

Step-by-step explanation:

Given that three out of seven students in the cafeteria line are chosen to answer a survey question.

The number of different combinations of three students that are possible is given as:

7C3 (read as 7 Combination 3)

xCy (x Combination y) is defines as

x!/(x-y)!y!

Where x! is read as x - factorial or factorial-x, and is defined as

x(x-1)(x-2)(x-3)...2×1.

Now,

7C3 = 7!/(7 - 3)!3!

= 7!/4!3!

= (7×6×5×4×3×2×1)/(4×3×2×1)(3×2×1)

= (7×6×5)/(3×2×1)

= 7×5

= 35

Therefore, the number of different combinations of three students that are possible is 35.

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Answer:1: Row 3

2: Row 2

3:6 and 10

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