\left \{ {{y=2} \atop {x=2}} \right. \leq \geq x^{2} \sqrt{x} \neq \pi \sqrt[n]{x} \frac{x}{y} \alpha \lim_{n \to \infty} a_n \int\limits^a_b {x} \, dx \geq x^{2} \\ \left[\begin{array}{ccc}1&2& ;3\\4&5&6\\7&8&9\en d{array}\right] \neq \left[\begin{array}{ccc}1&2& ;3\\4&5&6\\7&8&9\en d{array}\right]

the answer is y < -3 or   y > 3

-3y > 7 + 2

-3 y > 9

y < -3

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204

answer: can you show the options ?

step-by-step explanation: