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2 years ago

2/7+1/7+1/7= answer plss

Mathematics

2 answers:

Gnom [1K]2 years ago

7 0

Answer:

2÷7+1÷7+1÷7 = 0.57142857142

Leya [2.2K]2 years ago

6 0

Answer:

$\frac{4}{7}$

Step-by-step explanation:

$\frac{2}{7}$ + $\frac{1}{7}$ + $\frac{1}{7}$

= $\frac{2+1+1}{7}$

= $\frac{4}{7}$

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Step-by-step explanation:

6 percent of 90 is 5.4

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3 years ago

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AfilCa [17]

Answer:

Option b) is correct.

The difference of given expression is

$\left(\frac{2x+5}{x^2-3x}\right)-\left(\frac{3x+5}{x^3-9x}\right)-\left(\frac{1x+1}{x^2-9}\right)=\frac{(x+2)(x+5)}{(x^3-9x)}$

Step-by-step explanation:

Given expression is

$\left(\frac{2x+5}{x^2-3x}\right)-\left(\frac{3x+5}{x^3-9x}\right)-\left(\frac{1x+1}{x^2-9}\right)$

To find their difference

$\left(\frac{2x+5}{x^2-3x}\right)-\left(\frac{3x+5}{x^3-9x}\right)-\left(\frac{x+1}{x^2-9}\right)$

The expression can be written as below

$\left(\frac{2x+5}{x^2-3x}\right)-\left(\frac{3x+5}{x^3-9x}\right)-\left(\frac{x+1}{x^2-9}\right)=\left(\frac{2x+5}{x (x-3)}\right)-\left(\frac{3x+5}{x(x^2-9)}\right)-\left(\frac{x+1}{x^2-9}\right)$

$=\left(\frac{2x+5}{x(x-3)}\right)-\left(\frac{3x+5}{x(x^2-3^2)}\right)-\left(\frac{x+1}{x^2-3^2}\right)$

$=\left(\frac{2x+5}{x(x-3)}\right)-\left(\frac{3x+5}{x(x+3)(x-3)}\right)-\left(\frac{x+1}{(x+3)(x-3)}\right)$ (using $a^2-b^2=(a+b)(a-b)$ )

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

$=\frac{2x^2+6x+5x+15-3x-5-x^2-x}{x(x+3)(x-3)}$

$=\frac{x^2+7x+10}{x(x+3)(x-3)}$

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

$=\frac{(x+2)(x+5)}{x(x^2-3^2)}$ (using $a^2-b^2=(a+b)(a-b)$ )

$=\frac{(x+2)(x+5)}{x(x^2-9)}$

$=\frac{(x+2)(x+5)}{(x^3-9x)}$

Therefore $\left(\frac{2x+5}{x^2-3x}\right)-\left(\frac{3x+5}{x^3-9x}\right)-\left(\frac{1x+1}{x^2-9}\right)=\frac{(x+2)(x+5)}{(x^3-9x)}$

Option b) is correct.

The difference of given expression is

$\left(\frac{2x+5}{x^2-3x}\right)-\left(\frac{3x+5}{x^3-9x}\right)-\left(\frac{1x+1}{x^2-9}\right)=\frac{(x+2)(x+5)}{(x^3-9x)}$

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Answer/Step-by-step explanation:

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Justification: There are exactly one y-value/output related to each input/x-value in the relation.

5. Function: YES

Justification: each x-value/input is mapped to exactly one y-value/output. No input value is mapped to more than one output value.

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3 years ago

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