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2 years ago

15

Find the absolute value of a-b-c^2 when a=4, b=-3, and c=10

2 answers:

VashaNatasha [74]2 years ago

6 0

Answer:

93

Explanation:

4-(-3)-10^2

4+3-100

7-100

-93

Find the absolute value......

so therefore

|-93|=93

dlinn [17]2 years ago

3 0

Answer:-93

Step-by-step explanation:4-(-3)-10²

4+3-100

7-100

=-93

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Answer:

1. $P(x)$ ÷ $Q(x)$ ---> $\frac{-3x + 2}{3(3x - 1)}$

2. $P(x) + Q(x)$ ---> $\frac{2(6x - 1)}{(3x - 1)(-3x + 2)}$

3. $P(x) - Q(x)$ ---> $\frac{-2(12x - 5)}{(3x - 1)(-3x + 2)}$

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Step-by-step explanation:

Given that:

1. $P(x) = \frac{2}{3x - 1}$

$Q(x) = \frac{6}{-3x + 2}$

Thus,

$P(x)$ ÷ $Q(x)$ = $\frac{2}{3x - 1}$ ÷ $\frac{6}{-3x + 2}$

Flip the 2nd function, Q(x), upside down to change the process to multiplication.

$\frac{2}{3x - 1}*\frac{-3x + 2}{6}$

$\frac{2(-3x + 2)}{6(3x - 1)}$

$= \frac{-3x + 2}{3(3x - 1)}$

2. $P(x) + Q(x)$ = $\frac{2}{3x - 1} + \frac{6}{-3x + 2}$

Make both expressions as a single fraction by finding, the common denominator, divide the common denominator by each denominator, and then multiply by the numerator. You'd have the following below:

$\frac{2(-3x + 2) + 6(3x - 1)}{(3x - 1)(-3x + 2)}$

$\frac{-6x + 4 + 18x - 6}{(3x - 1)(-3x + 2)}$

$\frac{-6x + 18x + 4 - 6}{(3x - 1)(-3x + 2)}$

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$= \frac{2(6x - 1}{(3x - 1)(-3x + 2)}$

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$\frac{2(-3x + 2) - 6(3x - 1)}{(3x - 1)(-3x + 2)}$

$\frac{-6x + 4 - 18x + 6}{(3x - 1)(-3x + 2)}$

$\frac{-6x - 18x + 4 + 6}{(3x - 1)(-3x + 2)}$

$\frac{-24x + 10}{(3x - 1)(-3x + 2)}$

$= \frac{-2(12x - 5}{(3x - 1)(-3x + 2)}$

4. $P(x)*Q(x) = \frac{2}{3x - 1}* \frac{6}{-3x + 2}$

$P(x)*Q(x) = \frac{2*6}{(3x - 1)(-3x + 2)}$

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3 years ago

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nika2105 [10]

Answer:

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