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3 years ago

The mass of the moon is 7.36*10^25 grams to the nearest tenth how many moons would it take to equal the mass of earth

Mathematics

1 answer:

trapecia [35]3 years ago

6 0

Nearly 81 moons will be required to equate the mass of moon to the mass of earth.

Step-by-step explanation:

Mass of earth is 5.972*10^24 kg.

Mass of the moon is 7.36*10^25 g = 7.36*10^22 kg

As mass of the Earth is given as 5.972 * 10^24 kg and mass of the moon is given as 7.36 * 10^22 kg, then the number of moons required to make it equal to the mass of earth can be calculated by taking the ratio of mass of earth to moon.

Mass of Earth = Number of moons * Mass of Moon

Number of Moons = Mass of Earth/Mass of moon

Number of moons = 5.972 * 10^24/7.36*10^22= 81 moons.

So nearly 81 moons will be required to equate the mass of moon to the mass of earth.

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Answer:

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F(x)=x² G(x)=(x-6)² Complete the description of the transformation

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Read 2 more answers

To find 10, Beau found 32 = 9 and 42 = 16. He said that since 10 is between 9 and 16, 10 is between 3

AlexFokin [52]

Given that:

Consider it is $\sqrt{10}$ instead of 10 on two places.

$\sqrt{10}$ is between 3 and 4. So, Beau thinks a good estimate for $\sqrt{10}$ is = 3.5.

Solution:

To find $\sqrt{10}$ , Beau found 3² = 9 and 4² = 16.

He said that since 10 is between 9 and 16.

Since 10 is close to 9, therefore $\sqrt{10}$ must be close to 3. So, Beau's estimate is high.

Now,

$(3.1)^2=9.61$

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Since, 10 lies between 9.61 and 10.24, therefore $\sqrt{10}$ must be lies between 3.1 and 3.2.

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4 0

3 years ago

5. The recursive algorithm given below can be used to compute gcd(a, b) where a and b are non-negative integer, not both zero.

s2008m [1.1K]

Implementating the given algorithm in python 3, the greatest common divisors of (124 and 244) and (4424 and 2111) are 4 and 1 respectively.

The program implementation is given below and the output of the sample run is attached.

def gcd(a, b):

#initialize a function named gcd which takes in two parameters

if a>b:

#checks if a is greater than b

return gcd (b, a)

#if true interchange the Parameters and Recall the function

elif a == 0:

return b

elif a == 1:

return 1

elif((a%2 == 0)and(b%2==0)):

#even numbers leave no remainder when divided by 2, checks if a and b are even

return 2 * gcd(a/2, b/2)

elif((a%2 !=0) and (b%2==0)):

#checks if a is odd and B is even

return gcd(a, b/2)

else :

return gcd(a, b-a)

#since it's a recursive function, it recalls the function with new parameters until a certain condition is satisfied

print(gcd(124, 244))

print()

#leaves a space after the first output

print(gcd(4424, 2111))

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3 years ago