let g(x)= 2x and h(x)=X^2+4 (x^2+40 x2x)(-5)
let g(x)= 2x and h(x)=X^2+4 (x^2+40 x2x)(-5)
stop using "x" for both multiplication and variables!
I suppose you mean
h(x)*g(x) = (x^2+4)*(2x)
If you want that at x=5, just plug in x=5 !
(5^2+4)(2*5) = 29*10 = 290
By this time in Algebra II evaluating a function should not still be a problem ...
oops. I used 5 instead of -5
so plug in x = -5 instead.
To evaluate the expression `(x^2 + 40 * x^2 * 2x)(-5)` using the given functions `g(x) = 2x` and `h(x) = x^2 + 4`, we need to replace each `x` in the expression with `h(x)` and then multiply the result by `-5`.
Let's break down the process step by step:
Step 1: Replace `x` with `h(x) = x^2 + 4`:
`(x^2 + 40 * x^2 * 2x)(-5)` becomes `((x^2 + 40 * x^2 * 2(x^2 + 4)))(-5)`
Step 2: Simplify the expression inside the parentheses:
Taking care of the multiplication and distribution, the expression inside the parentheses simplifies to:
`(x^2 + 40 * 2 * x^2 * 2 * 4)(-5)`
`= (x^2 + 160x^2)(-5)`
Step 3: Combine like terms:
Adding the terms with the same exponent `x^2`, we get:
`(161x^2)(-5)`
Step 4: Multiply the result by `-5`:
Finally, we multiply `(161x^2)` by `-5`:
`= -805x^2`
Therefore, the expression `(x^2 + 40 * x^2 * 2x)(-5)` simplifies to `-805x^2`.