Step-by-step explanation:
Given:
and 
We can solve for f(x) by writing

Let 

Then


We know that f(0) = 0 so we can find the value for k:

Therefore,

Answer:
Step-by-step explanation:
X= (-7+radical 13)/6
X=(-7-radical 13)/6
If it's addition you do the opposite which is subtraction and that's the answer that you get so 6m_2=m+13 so add 6-2=4 and then 13-4=9
a = total amount invested at 5%
b = total amount invested at 8%
we know 5% was earned in interest from "a", so that'd be (5/100) * a or 0.05a.
we know 8% was earned in interest from "b", so that'd be (8/100) * a or 0.08b.
we also know that the total for both is 1000, a + b = 100.

1. Given that the width of the rectangle is x, and the area of the rectangle may be represented by the equation x^2 + 5x = 300, we can solve this equation for the width (x) as such:
x^2 + 5x = 300
x^2 + 5x - 300 = 0 (Subtract 300 from both sides)
(x - 15)(x + 20) = 0 (Factorise x^2 + 5x - 300)
From this, we get: x = 15 or x = -20
Since the width must be a positive length (ie. more than 0), -20 would be an invalid answer in the given context and thus the width is given by x = 15.
2. If we know that the length is 5 inches more than the width, we simply need to add 5 to the width we found above to obtain the length:
Length = x + 5
Length = 15 + 5 = 20
Thus, the width of the rectangle is 15 inches and the length of the rectangle is 20 inches.