The length of a rectangle is 2 more than three times the width. the area of the rectangle is 161 square inches. what are the dimensions of the rectangle?

if x = the width of the rectangle, which of the following equations is used in the process of solving this problem?

2x^2 + 3x - 161 = 0
3x^2 + 2x - 161 = 0
6x^2 - 161 = 0

The answer is (B)

Step-by-step explanation: Given that 'x' represents the width of the rectangle.

According to the given information, the length of a rectangle is 2 more than three times the width. Therefore, the length of the rectangle is (3x + 2).

Hence, the area of the rectangle will be

Since area of the rectangle is 161 square inches, so the equation is

Solving the above equation, we find

Since the width of the rectangle cannot be negative, so x = 21 inches.

And, length will be (3 × 21 +2) = 65 inches.

Thus, the length and width of the rectangle are 65 inches and 21 inches respectively.

And the correct equation is (B).

c.

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.

x = 2

step-by-step explanation:

if sr parallel to ba then the sides bc and ac are in proportion, that is

= ← substitute values

= ( cross- multiply )

27x = 6(4x + 1)

27x = 24x + 6 ( subtract 24x from both sides )

3x = 6 ( divide both sides by 3 )

x = 2