Hunter and Brian are creating a poster for a project and they need to combine two different size boards to create their project. If the total area cannot exceed 7x to the second power -6x+2 square inches and one board Is 2x to the second power -9x+8 square inches, then what must be the maximum area of the second board?
2 answers:
Answer:
5x² + 3x - 6
Step-by-step explanation:
Data:
Area of one board: 2x² - 9x + 8 Area of the other board: f(x) The addition of the area of the two boards cannot exceed the total area. Then:
f(x) + 2x² - 9x + 8 ≤ 7x² - 6x + 2
f(x) ≤ 7x² - 6x + 2 - 2x² + 9x - 8
f(x) ≤ (7x² - 2x²) + (-6x + 9x) + (2 - 8)
f(x) ≤ 5x² + 3x - 6 in square inches
Answer:
<em>The maximum area of the second board must be </em>
Step-by-step explanation:
The total area of two different size boards cannot exceed
The area of one board is
Suppose, the area of the second board is
That means......
So, the maximum area of the second board must be
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The answer is X=4.
It is both parallel to x=-3 and passes through (4,2)
Here y=1/2 x=pi/3=60°
You could multiply 19.45 by 0.2 to solve for it, but if you're not using a calculator, then 19.45/5 would be much more convenient and easier.
0.2 is 1/5 in fraction form, which is why we can do that.
19.45/5 = 3.89
Dino should leave $3.89 as a tip.
