That’s hard , I don’t even know
Answer:
-2a³+9a²+45a+6ab²+18b²
Step-by-step explanation:
(a+3)(-2a²+15a+6b²)
Distribute the a:
-2a³+15a²+6ab²
Then distribute the 3:
-6a²+45a+18b²
Now add both of the results together:
-2a³+15a²+6ab²-6a²+45a+18b²
Combine like terms:
-2a³+9a²+45a+6ab²+18b²
R(x) = 60x - 0.2x^2
The revenue is maximum when the derivative of R(x) = 0.
dR(x)/dx = 60 - 0.4x = 0
0.4x = 60
x = 60/0.4 = 150
Therefore, maximum revenue is 60(150) - 0.2(150)^2 = 9000 - 4500 = $4,500
Maximum revenue is $4,500 and the number of units is 150 units
Answer: 6(x²-4x+4-4)+1=0, 6(x-2)²-24+1=0, 6(x-2)²=23, x-2=±√(23/6), x=2±√(23/6)=2±1.95789, so x=3.95789 or 0.04211 approx. these are the zeros.
step by step explanation:
\boxed{\boxed{\dfrac{12+\sqrt{138}}{6},\ \dfrac{12-\sqrt{138}}{6}}}
Solution-
The quadratic function is,
6x^2-24x + 1
a = 6, b = -24, c = 1
x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}
=\dfrac{-(-24)\pm \sqrt{-24^2-4\cdot 6\cdot 1}}{2\cdot 6}
=\dfrac{24\pm \sqrt{576-24}}{12}
=\dfrac{24\pm \sqrt{552}}{12}
=\dfrac{24\pm 2\sqrt{138}}{12}
=\dfrac{12\pm \sqrt{138}}{6}
=\dfrac{12+\sqrt{138}}{6},\ \dfrac{12-\sqrt{138}}{6}
