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djverab [1.8K]
3 years ago
7

What is the LCM of two numbers that have no common factors greater than 1?

Mathematics
2 answers:
anyanavicka [17]3 years ago
5 0
This question was already answered before. I'm going to post someone else's answer with credit given of course:

"<span>The answer is their product. The least common multiple (LCM) is the smallest number which is a multiple of two numbers. If two numbers have no common factors greater than 1, then their LCM will be their product. For example, take numbers 7 and 9. They have none common factor greater than 1. So, LCM will be 7 * 9 = 63." - @W0lf93

Not sure if this answers your question, but hope this helped...</span>
Marizza181 [45]3 years ago
5 0
The numbers are m and n where m and n have no common factors

to find the LCM, factor them and group the common factors

since m and n have no common factors
m=1*m
n=1*n
LCM=1*m*n=mn

the LCM of 2 numbers with no common factors greater than 1 is the product of the 2 numbers
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Write the equation of the line parallel to y=4/5x+6/5 that passes through the point (-6,4).
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Answer:

y-4=4/5*(x+6)

Step-by-step explanation:

First you need to find the vaules for y-y₁=m(x-x₁) point slope form

y1 =4

x1=-6

m=4/5( because thats the slope)

Second plug in:

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An Olympic swimming pool is 50 m long, 25 m wide, and 1.5 m deep. How many Olympic pools would be filled by the dollar bills spe
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Answer:

Number of dollars bills that can be fitted in Olympic pool = 1.67 billion.

Step-by-step explanation:

The dimensions of a dollar bill are 15.5956 cm by 6.6294 cm by 0.010922 cm. The U.S. federal government spent $3.8 trillion in fiscal year 2011.

An Olympic swimming pool is 50 m long, 25 m wide, and 1.5 m deep. How many Olympic pools would be filled by the dollar bills spent by the U.S. federal government in 2011

From above,

The volume for the dimension of the dollar bill = 15.5956 cm × 6.6294 cm × 0.010922 cm

The volume for the dimension of the dollar bill = 1.12922 cm³

The volume for the dimension of the dollar bill = 1.12922 × 10⁻⁶ m³

The volume of the pool = 50 m  ×  25 m  ×  1.5 m

The volume of the pool = 1875 m³

Number of dollars bills that can be fitted in pool = volume of the pool/volume of the dimension of dollar bills

Number of dollars bills that can be fitted in Olympic pool =1875 m³ / 1.12922 × 10⁻⁶ m³

Number of dollars bills that can be fitted in Olympic pool = 1,660,438,180

Number of dollars bills that can be fitted in Olympic pool = 1.67 billion.

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