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gladu [14]
3 years ago
14

How can knowing the GCF and LCM help you when you add, subtract, and multiply fractions?

Mathematics
2 answers:
Ksenya-84 [330]3 years ago
8 0

<em><u>Explanation</u></em>:

The LCM, or least common multiple, is the smallest number that two other numbers can divide evenly into. The LCM is the basis of the LCD, or least common denominator, which is used while adding and subtracting fractions.

When adding or subtracting fractions, we must have a common denominator. In order to avoid simplifying more than necessary, it is helpful to use the least common denominator, or LCD. This is basically the LCM of the denominators.

The GCF, or greatest common factor, is what we use to simplify fractions. When we finish adding, subtracting or multiplying fractions, we often have a fraction that needs simplified. In order to do this, we want to divide the numerator and denominator by a common factor. However, using small factors results in more work; it is much easier to use the largest factor that will evenly divide into both numbers, or the GCF.

Stells [14]3 years ago
6 0
<span>When you add fractions, the denominators must be the same, so you may have to generate equivalent fractions by using the lowest common multiple of all the denominators. For subtraction of fractions, you use the same method. When multiplying fractions, you can only multiply the numerators together, and the denominators together. </span>
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Find the mass of the triangle with vertices (1; 0; 0), (0; 2; 0), (0; 1; 1) if the density function is given by (x; y; z)
Luda [366]

Answer:

=\frac{17}{3}

Step-by-step explanation:

Density of a function is \rho(x,y)=x^{2} +y^{2}

I have drawn the right angle triangle for visualization

equation for the line passing through (1,0) and (0,4) is

\frac{y}{4} =\frac{x-1}{-1}

y=4-4x

=$$\int_{0}^1\int_{0}^{4-4x} (x^{2}+y^{2})dydx$$

=$\int_{0}^{1}(x^{2}+\frac{y^{3} }{3})dx$  

=$\int_{0}^{1}(x^{2}(4-4x)+\frac{1}{3}(4-4x)^{3} )dx$

=$\int_{0}^{1}(\frac{-76}{3}x^{3} +68x^{2}-64x^{}+\frac{64}{3}  )dx$

==(\frac{-76}{3} \frac{x^{4} }{4})+\frac{68}{3}x^{3}-32x^{2}+\frac{64}{3}x^{})  

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3 0
3 years ago
Nevermind, the question has been solved.
san4es73 [151]

okay.... I guess??? lol

7 0
3 years ago
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Where is the hole for the following function located? mc005-1.jpg
alex41 [277]
The hole is located where the hole is part a

4 0
3 years ago
Consider the following system of equations.
Dmitrij [34]

Answer:

C. When the equations -10x + 5y = -60 and -3x − 5y = -5 are added together, a third linear equation, -13x = -65, is formed, and it shares a common solution with the original equations.

Step-by-step explanation:

2x-y=12

-3x-5y=-5

multiply the first equation by -5

-10x +5y = -60

add the 2 equations together

-3x-5y=-5

-10x +5y = -60

-------------------------

-13x = -65

We use the addition property of equality which says adding the same thing to both sides, doesn't change the equation, so we will get the same answer as the original equation

6 0
3 years ago
The temperature of lake water over a period of time is shown in the graph below.
castortr0y [4]

Answer:

Answers are 3 AM ; 10 , 54

Step-by-step explanation:

We are given a graph.

Recording of temperature starts from 9 PM.

So from Graph,

Temperature ate 9 PM = 59°F

Temperature ate 11 PM = 54°F

Temperature ate 1 AM = 51°F

Temperature ate 3 AM = 50°F

Temperature ate 5 AM = 51°F

Temperature ate 7 AM = 54°F

Temperature ate 9 AM = 59°F

So, At 3 AM the water is at its lowest temperature.

and

After 10 hours passed, the temperature of the lake is 54°F.

Therefore, Answers are 3 AM ; 10 , 54

7 0
3 years ago
Read 2 more answers
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