Easy peasy
just subsitute I(x) for the x in the h(x) so
h(I(s))=-(2s+3)^2-4
distribute and simplify
h(I(s))=-(4s^2+12s+9)-4
h(I(s))=-4s^2-12s-9-4
h(I(s))=-4s^2-12s-13
1.6 divided by x =4
we can exchange the 4 and the x
1.6 divided by 4 =x
.4 =x
We conduct long division, cancelling each term in x³ + 6x² + 3x + 1. As so:
x²(x -2) = x³ - 2x²
Subtracting from the original equation, we get:
8x² + 3x + 1
8x(x - 2) = 8x² - 16x
Subtracting from the new equation, we get:
21x + 1
21(x - 2) = 21x - 42
Subtracting from the new equation, we get:
43
The answer is x² + 8x + 21 with a remainder of 43.
Answer:
E
Step-by-step explanation:
Answer:
I guess that you want to calculate the difference between the annual healt care costs in 1950 and today.
The equation is:
C(x) = $620*(1.082)^x
in 1950, we have x = 0 (because is the initial year)
then the cost is:
C(0) = $620*1 = $620.
Now, today, in 2020, we have:
x = 2020 - 1950 = 70
then we have:
C(70) = $620*(1.082)^70 = $154,276.7
So, according to this equation, te difference is:
$154,276.7 - $620 = $153,656.6
