0.73>0.37
0.37 is closer to zero and 0.73 is closer to one so this is the answer
<span>(5x2 + 3x + 4) − (2x2 + 5x − 1)
(10 + 3x + 4) - (4 + 5x -1)
(14+3x) - (3 + 5x)
(14+3x) + (-3 -5x)
11 - 2x
11-2x is as simplified as it goes there is no possible answer without an equal sign given.
</span>
Given:
Current revenue = $6000
<span>R(f)= -100f^2 + 400f + 6000
where f is a whole number of $5 fee increases
We are told to find f, when R(f) </span><span>< 6000
Since </span>R(f)= -100f^2+400f+6000
R(f) < 6000 ⇒ -100f^2+400f+6000 < 6000
Subtract 6000 from both sides
-100f^2 + 400f + 6000 - 6000 < 6000 - 6000
-100f^2 + 400f < 0
⇒ 400f - 100f^2 < 0
Divide the equation by 100
400f/100 - 100f^2/100 < 0/100
4f - f^2 < 0
Add f^2 to both sides of the equation
4f - f^2 + f^2 < 0 + f^2
4f < f^2
Divide both sides by f
4f/f < (f^2)/f
4 < f
⇒ f > 4
⇒ f ≥ 5
Therefore, <span> for 5 or more numbers of $5 fee increases, the revenue from fees will actually be less than its current value.</span>