3 years ago

(3 1/8) (2/5) divided by 2 1/4. Help please due tomorrow

Mathematics

1 answer:

LUCKY_DIMON [66]3 years ago

6 0

Answer:

5/9

Step-by-step explanation:

5/9

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

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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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