Answer:
<em>The maximum area of the second board must be
square inches.</em>
Step-by-step explanation:
The total area of two different size boards cannot exceed
square inches.
The area of one board is
square inches.
Suppose, the area of the second board is
square inches.
That means......
![y+(2x^2-9x+8)\leq 7x^2-6x+2\\ \\ y\leq (7x^2-6x+2)-(2x^2-9x+8)\\ \\y \leq 7x^2-6x+2-2x^2+9x-8\\ \\ y\leq 5x^2+3x-6](https://tex.z-dn.net/?f=y%2B%282x%5E2-9x%2B8%29%5Cleq%207x%5E2-6x%2B2%5C%5C%20%5C%5C%20y%5Cleq%20%287x%5E2-6x%2B2%29-%282x%5E2-9x%2B8%29%5C%5C%20%5C%5Cy%20%5Cleq%207x%5E2-6x%2B2-2x%5E2%2B9x-8%5C%5C%20%5C%5C%20y%5Cleq%205x%5E2%2B3x-6)
So, the maximum area of the second board must be
square inches.
Answer:12, 7
Step-by-step explanation:Using the process of elimination, I went through each number and subtracted five and added it until I got 12 and 7.
Answer:
16.67%
Step-by-step explanation:
initial rate=30
final rate =25
decreased rate = 30 - 25= 5
Now,
Decreased % = (5 / 30) * 100%
= 16.67%
The answer is 1/6. To solve, simply set up an equation. It will be x/6. As you know, x can stand for anything. 1 will be the reasonable number for variable, x. Why? Well since x=1x. You don't actually write "1" Hope this makes your understanding clear.