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3 years ago

What is the least common denominator of the rational expressions below? 1/x^2 - 4/3x^2+15x

Mathematics

2 answers:

mylen [45]3 years ago

7 0

Answer:3x^2(x+5)

Step-by-step explanation: A P E X

vladimir1956 [14]3 years ago

5 0

Answer:

The LCD is 3x²

The other two denominators can be factors of 3x²

Step-by-step explanation:

To get the equivalent fractions, divide the original denominator into the LCD, then multiply the original numerator by the quotient.

For the first term: 1/x²

3x² ÷ x² = 3 3×1=3 The equivalent is 3/3x²

Second term: Leave it as it is.

For the third term: 15x

The implied denominator is 1

3x² ÷ 1 = 3x² 3x² × 15 = 135x²

The equivalent is 135x²/3x²

To check the validity, substitute a value for x and calculate the numbers.

For example let x = 3

The original expression becomes

1/3² + 4/3(3²) - 15(3)

1/9 + 4/27 - 45

With the LCD it is

3/27 + 4/27 - 1215/27

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3 years ago

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4 years ago

Jasmine had a carrot cake with cream cheese frosting for her birthday .she and her friend each ate 1/4 of the cake and jasmine's

zloy xaker [14]

Hey there!

To solve this problem, you first would need to solve for how much of the cake is left after Jasmine, her friend, and Jasmine's brother got their slices. Then, you would need to split the remaining amount in half to get the answer of what each parent ate. We would need to make all of the denominators the same, so let's make all of the fractions out of 12, a common denominator of 3 and 4.

$\frac{12}{12} - \frac{1}{4} - \frac{1}{4} - \frac{1}{3}$
$\frac{12}{12} - \frac{3}{12} - \frac{3}{12} - \frac{4}{12}$
$\frac{12}{12} - \frac{10}{12} = \frac{2}{12}$

It can then be concluded that each parent ate $\frac{1}{12}$ of the cake, since there was $\frac{2}{12}$ remaining.

You can check this answer by adding up all of the parts we used in this problem:

$\frac{3}{12} + \frac{3}{12} + \frac{4}{12} + \frac{1}{12} + \frac{1}{12} = \frac{12}{12}$

Hope this helped you out! :-)

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3 years ago

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Ksju [112]

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Step-by-step explanation:

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tekilochka [14]

What is your question?

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3 years ago