Question

Please answer the attached question by selecting of the the given answers.

Answer 1

Answer:

$Perimeter\,\,of \,\,triangle =\frac{3x^2+7x+21}{3x^2}$ so, Option D is correct.

Step-by-step explanation:

Perimeter of triangle = Sum of all sides

Perimeter of triangle = a+ b+ c

a= 1/3x, b= x+2/x and c = 7/x^2

$Perimeter\,\,of\,\,triangle =\frac{1}{3x}+\frac{x+2}{x}+\frac{7}{x^2}$

LCM of 3x, x and x^2 : 3x^2

$Perimeter\,\,of\,\,triangle =\frac{x+3x(x+2)+7(3)}{3x^2}\\Perimeter\,\,of \,\,triangle =\frac{x+3x^2+6x+21}{3x^2}\\Perimeter\,\,of\,\,triangle =\frac{3x^2+7x+21}{3x^2}$

So, $Perimeter\,\,of\,\,triangle =\frac{3x^2+7x+21}{3x^2}$ so, Option D is correct.

Answer 2

Answer:

The correct answer is option D.  [3x² + 7x + 21]/3x²

Step-by-step explanation:

Perimeter of a triangle = sum of all sides

To find the perimeter of triangle

Here sides are,

7/x² , 1/3x and (x + 2)/x

Perimeter = 7/x² + 1/3x + (x + 2)/x

Here LCM = 3x²

Perimeter = (7 * 3)/3x² + (1 * x)/3x² + (x + 2)3x/3x²

 = [21 + x + 3x² +6x]/3x²

 = [3x² + 7x + 21]/3x²

The correct answer is option D.  [3x² + 7x + 21]/3x²