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Firlakuza [10]
2 years ago
5

The weight of an object on moon is 1/6 of its weight on Earth. If weight of an object on Earth is 18/5 kg, what would be its wei

ght on moon? ​Please help me!!!!​
Mathematics
2 answers:
DiKsa [7]2 years ago
8 0

Answer:

Solution given:

weight of moon=1/6 th of Earth

If weight of Earth is 18/5 kg

weight of moon=?

we have

<u>weight of </u>moon: 1/6 of weight of Earth

=1/6*18/5

=3/5kg

<u>I</u><u>t</u><u>s</u><u> </u><u>w</u><u>e</u><u>i</u><u>g</u><u>h</u><u>t</u><u> </u><u>o</u><u>n</u><u> </u><u>m</u><u>o</u><u>o</u><u>n</u><u> </u><u>i</u><u>s</u><u> </u><u>3</u><u>/</u><u>5</u><u> </u><u>k</u><u>g</u>

IrinaVladis [17]2 years ago
4 0

Answer:

  • 3/5 kg

Step-by-step explanation:

The weight of an object on Moon is 6 times smaller than same on Earth.

<u>If it weights 18/5 kg on Earth then its weight  on Moon is:</u>

  • 18/5 * 1/6 = 3/5 kg
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Find the amount of time.<br> I=$450, P=$2400, r=7.5%<br> Help!
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2 years ago
4y+13=37 a 6 b 24 c 20 d 4​
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A) 6
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7 0
2 years ago
Read 2 more answers
Let f(x)=2.912345x^2+3.131579x-0.099999
Rudiy27
A.) Integers are positive and negative counting numbers. So, in order to find the integer coefficients, round off the coefficients in the equation to the nearest whole number. The function for g(x) is:

g(x) = 3x²+3x

B.) Substitute x=4 to the two functions.

f(x) = 2.912345x²<span>+3.131579x-0.099999
</span>f(4) = 2.912345(4)²+3.131579(4)-0.099999
f(4) = 59.023837

g(x) = 3x²+3x
g(4) = 3(4)²+3(4)
g(4) = 60

C.) The percentage error is equal to:

Percentage error = |g(4) - f(4)|/f(4)  * 100
Percentage error = |60 - 59.023837|/59.023837  * 100
Percentage error = 1.65%

D.) If x is a large number, for example x=10 or x=20, then g(x) would be an overestimate. This is because the value of x is raised to the power of 2. So, as the x increases, the corresponding function would increase exponentially. Even at x=4, g(x) is already an overestimate. What more for larger values of x? That means that the gap from the true answer f(x) would increase.
5 0
3 years ago
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