Question

Write an absolute value inequality that
represents the situation. Then solve the
inequality. The difference between the
areas of the figures is at most 6.
Please I need help I don’t understand and I need help on problem 16 pleaseee

Answer 1

Answer:

All real numbers greater than or equal to -1 and less than or equal to 2

Step-by-step explanation:

Let

A_1 ------> the area of rectangle

A_2 ----> the area of triangle

we know that

$\left|A_1-A_2\right|\le 6$

Find out the area of rectangle

$A_1=4(x+2)=(4x+8)\ units^2$

Find out the area of triangle

$A_2=(1/2)(4)(5)=10\ units^2$

substitute

$\left|(4x+8)-10\right|\le 6$

Simplify

$\left|4x-2\right|\le 6$

First case (positive value)

$+(4x-2) \leq 6$

$4x \leq 6+2$

$4x \leq 8$

$x \leq 2$

Second case (negative value)

$-(4x-2) \leq 6$

Multiply by -1 both sides

$(4x-2) \geq -6$

$(4x \geq -6+2$

$4x \geq -4$

$x \geq -1$

therefore

The solution for x is the interval -----> [-1,2]

All real numbers greater than or equal to -1 and less than or equal to 2