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SVETLANKA909090 [29]
3 years ago
10

If n is the square root of a positive integer, and is divisible by both 2 and 6, what is the least possible value of n?

Mathematics
1 answer:
Paraphin [41]3 years ago
6 0
6 is the square root of 36 and is divisible by 2 and 6. 
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ANSWER ASAP!!! Find the length of the segment indicated
trapecia [35]

Answer:

14.45

Step-by-step explanation:

x^2+(7.1)^2=(16.1)^2

×^2+50.41=259.21

x^2=208.8

sqrt(x^2)=sqrt(208.8)

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check.

14.45^2 + 7.1^2 = 16.1^2

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What is the distance between point A(−1, 3) and point B(−8, 3) ?
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HELP PLEASE ITS DUE TODAY!!! Express 14 hours in 2 weeks as a unit rate
user100 [1]

Answer:

14/336 ----------> 0.41

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7 0
3 years ago
Evaluate the given integral by changing to polar coordinates. 8xy dA D , where D is the disk with center the origin and radius 9
BabaBlast [244]

Answer:

0

Step-by-step explanation:

∫∫8xydA

converting to polar coordinates, x = rcosθ and y = rsinθ and dA = rdrdθ.

So,

∫∫8xydA = ∫∫8(rcosθ)(rsinθ)rdrdθ = ∫∫8r²(cosθsinθ)rdrdθ = ∫∫8r³(cosθsinθ)drdθ

So we integrate r from 0 to 9 and θ from 0 to 2π.

∫∫8r³(cosθsinθ)drdθ = 8∫[∫r³dr](cosθsinθ)dθ

= 8∫[r⁴/4]₀⁹(cosθsinθ)dθ

= 8∫[9⁴/4 - 0⁴/4](cosθsinθ)dθ

= 8[6561/4]∫(cosθsinθ)dθ

= 13122∫(cosθsinθ)dθ

Since sin2θ = 2sinθcosθ, sinθcosθ = (sin2θ)/2

Substituting this we have

13122∫(cosθsinθ)dθ = 13122∫(1/2)(sin2θ)dθ

= 13122/2[-cos2θ]/2 from 0 to 2π

13122/2[-cos2θ]/2 = 13122/4[-cos2(2π) - cos2(0)]

= -13122/4[cos4π - cos(0)]

= -13122/4[1 - 1]

= -13122/4 × 0

= 0

5 0
3 years ago
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