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joja [24]
3 years ago
6

What two numbers are located 5/8 of a unit from 1/6 on a number line?

Mathematics
1 answer:
Molodets [167]3 years ago
7 0
First find the common denominator. It would be 24. For 8 to get to 24 you have to multiply it by 3. So 8*3=24 and 5*3=15. For 6 to get to 24 you have to multiply it by 4. So 6*4=24 and 1*4=4. Our fractions are now at 15/24 and 4/24. 15+4 is 19 and 15-4 is 11. Your answers are 19/24 and 11/24.
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%70

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Add and Subtract Rational Expressions with a Common Denominator
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Answer:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{q^2+9}{q^2+6q+5}

Step-by-step explanation:

<u>Simplifying Rational Expressions</u>

If two or more rational expressions have the same denominator, the add and subtract operations are done only with the numerator. The final denominator will be the common of both.

The expression is:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}

Operating on the numerators:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{4q^2-q+3-(3q^2-q-6)}{q^2+6q+5}

Removing parentheses:

\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{4q^2-q+3-3q^2+q+6}{q^2+6q+5}

Simplifying:

\boxed{\displaystyle \frac{4q^2-q+3}{q^2+6q+5}-\frac{3q^2-q-6}{q^2+6q+5}=\frac{q^2+9}{q^2+6q+5}}

The expression cannot be further simplified.

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check the picture below.

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Answer:

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Step-by-step explanation:

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3 years ago
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9514 1404 393

Answer:

  the y-intercepts differ

Step-by-step explanation:

The x-coefficient is the same for each function, so parallel lines are described. The function g(x) has a y-intercept of -4; f(x) has a y-intercept of 0.

The graphs differ in their intercepts.

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<em>Additional comment</em>

g(x) can be considered to be a translation downward of f(x) by 4 units. The same graph of g(x) can be obtained by translating f(x) to the right by 2 units. That is, both the x-intercepts and y-intercepts differ between the two functions.

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