Question

mika is designing a rectangular blanket where the width is \\(\frac{4}{9}\\) of the length. the area where she will place her design must be smaller than 4850 square inches. write the quadratic inequality representing the relationship between the area of the blanket design and the blankets maximum area. use the keypad to enter the answers in the boxes. additional symbols can be found using the drop - down arrow at the top of the keypad. the inequality representing the relationship between the area of the blanket design and the blankets maximum area is \\(\square\\).

Explanation:

Step1: Define variables for length

Let $l$ = length of blanket (in inches)

Step2: Express width in terms of length

Width $w = \frac{4}{9}l$

Step3: Write area formula for blanket

Area of blanket $A = l \times w = l \times \frac{4}{9}l = \frac{4}{9}l^2$

Step4: Set up inequality for design area

Let $A_d$ = area of the design. The design area must be smaller than 4850 square inches, so:
$A_d < \frac{4}{9}l^2$
Or, using the blanket's area variable:
$A_d < A$ where $A \leq 4850$, so combining:
$A_d < 4850$ and $A = \frac{4}{9}l^2$, so the inequality relating design area to maximum blanket area is:
$A_d < \frac{4}{9}l^2 \leq 4850$
To write the inequality for the relationship between design area and maximum blanket area directly:
Let $x$ = area of the design.

Answer:

$x < 4850$ where the blanket's area is $\frac{4}{9}l^2 \leq 4850$, or the core quadratic inequality is $\frac{4}{9}l^2 > x$ and $\frac{4}{9}l^2 \leq 4850$, simplified to $x < \frac{4}{9}l^2 \leq 4850$

If only the inequality between design area and maximum area is needed:
$x < 4850$ where $x$ is the design area, and the blanket's area follows $\frac{4}{9}l^2 \leq 4850$