<span>Here let the quadratic equation be ax^2 + bx + c
We know that a=5 from the question.
Since the roots are 6 and 2, the quadratic equation would take the form of a product like (a1x-b1)(a2x-b2).
However, let's assume that a2=1 and b2=6,
Since a=5, a1=5, then 5x-b1=5(x-2). Solving this shows that b1=10
So, the equation is (5x-10)(x-6)</span>
Answer:
See below.
Step-by-step explanation:
First, notice that this is a composition of functions. For instance, let's let 
and 
. Then, the given equation is essentially 
. Thus, we can use the chain rule.
Recall the chain rule: ![\frac{d}{dx}[f(g(x))]=f'(g(x))\cdot g'(x)](https://tex.z-dn.net/?f=%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28g%28x%29%29%5D%3Df%27%28g%28x%29%29%5Ccdot%20g%27%28x%29)
. So, let's find the derivative of each function:
We can use the Power Rule here:
Now:
Again, use the Power Rule and Sum Rule
Now, we can put them together:
Let "a" and "b" represent the values of the first and second purchases, respectively.
0.40*(original price of "a") = $10
(original price of "a") = $10/0.40 = $25.00 . . . . divide by 0.40 and evaluate
a = (original price of "a") - $10 . . . . . . Julia paid the price after the discount
a = $25.00 -10.00 = $15.00
At the other store,
$29 = 0.58b
$29/0.58 = b = $50 . . . . . . . divide by the coefficient of b and evaluate
Then Julia's total spending is
a + b = $15.00 +50.00 = $65.00
Julia spent $65 in all at the two stores.
The negative in front of the second bracket means that it's times by -1 so each of their signs switches:
3m^2+2n-m^2-2n
Add like terms
2m^2
the 2n cancels out
