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Nady [450]
1 year ago
9

How do I find the absolute value

Mathematics
1 answer:
pychu [463]1 year ago
8 0

Absolute value of function = - 9 / 3 .

The absolute value or modulus of a real number x, denoted by |x|, is the non-negative value of x without regard to its sign. Namely, |x|=x if x is a positive number, and |x|=-x if x is negative (in which case negating x makes -x positive) , and |0|=0.

For example, the absolute value of 3 is 3, and the absolute value of −3 is also 3. The absolute value of a number may be thought of as its distance from zero.

x ≥ 1.52 / 3 x - 1 = 2 / 3 x + 4⇒ -1 = 4 which is not possible .Now for all x ≤ 1.5- 2 /3 x + 1 = 2 / 3 x + 4⇒ 4 / 3 x = -3  x = - 9 /4 Hence absolute value = - 9 / 3 .

To learn more on absolute value follow link :

brainly.com/question/13041863

#SPJ9

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(3xy/3x^2-12)*(x^2+3x+2/xy+y)
TiliK225 [7]

Value of expression (3xy/3x^2-12)*(x^2+3x+2/xy+y) is  \frac{x}{x-2} .

<u>Step-by-step explanation:</u>

Here we need to evaluate expression : (3xy/3x^2-12)*(x^2+3x+2/xy+y)  or ,

(3xy/3x^2-12)*(x^2+3x+2/xy+y)

Let's simplify this

⇒ (\frac{3xy}{3x^2-12})(\frac{x^2+3x+2}{xy+y})

Factorizing the terms we get:

⇒ (\frac{3xy}{3(x^2-4)})(\frac{x^2+2x+x+2}{y(x+1)})              { a^2-b^2=(a+b)(a-b)      }

⇒ (\frac{xy}{(x-2)(x+2)})(\frac{x(x+2)+1(x+2)}{y(x+1)})

⇒ (\frac{xy}{(x-2)(x+2)})(\frac{(x+1)(x+2)}{y(x+1)})

Cancelling similar terms we get:

⇒ \frac{x}{x-2}

Therefore , Value of expression (3xy/3x^2-12)*(x^2+3x+2/xy+y) is  \frac{x}{x-2} .

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Step-by-step explanation:

Rectangles

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