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The quadratic function f(x) = 2x² – 44x + 185 is written in vertex form is f(x) = 2(x – 11)² – 57.
<h3>What is a quadratic equation?</h3>
It is a polynomial that is equal to zero. Polynomial of variable power 2, 1, and 0 terms are there. Any equation having one term in which the power of the variable is a maximum of 2 then it is called a quadratic equation.
The quadrattic equation is f(x) = 2x² – 44x + 185.
Take common 2 from the equation, we have

Add and subtract 121, we have
![f(x) = 2(x^2 - 22x +121 - 121 )+ 185\\\\f(x) = 2[(x^2 - 11)^2-121]+ 185\\\\f(x) = 2(x-11)^2 - 242 + 185\\\\f (x ) = 2(x-11)^2 -57](https://tex.z-dn.net/?f=f%28x%29%20%3D%202%28x%5E2%20-%2022x%20%2B121%20-%20121%20%29%2B%20185%5C%5C%5C%5Cf%28x%29%20%3D%202%5B%28x%5E2%20-%2011%29%5E2-121%5D%2B%20185%5C%5C%5C%5Cf%28x%29%20%3D%202%28x-11%29%5E2%20-%20242%20%2B%20185%5C%5C%5C%5Cf%20%28x%20%29%20%3D%202%28x-11%29%5E2%20-57)
And we know that the standard equation
f(x) = a(x - h)² + k
On comparing, we have
The vertex (h, k) is (11, -57).
More about the quadratic equation link is given below.
brainly.com/question/2263981
Sure! So this is ready as "the cube root of 125". This basically means, "what number cubed can get me 125?"
Let's go through our options.
We can rule out D, as D cubed would be unreasonably big.
We can also rule out C, because 375 cubed is easily over 10000, you know this even if you haven't computed it all, just compute the 300 cubed.
We can rule out B, too. 41 squared is already over 125, therefore it can't be the answer.
Therefore our answer is A, 5. We can check that by cubing 5, and that indeed gets us 125.
Hope this helps!