In this scenario we know that carbon14 at a given time A(t) = A0e^(-kt), where A0 is the original carbon present, t is time in years and k is the constant.
As we are working with the half life of carbon being 5730 years, we assume original carbon-14 content, A0 = 1, and carbon-14 at half life 5730 years, A(t) = 0.5.
i.e.
0.5 = 1e^(-5730k)
apply Ln to both sides of equation to cancel e
ln(0.5) = -5730k
k = ln(0.5) / -5730
k = -0.69315 / - 5730 = 1.20968 x 10^-4
Answer:
y = 4.375x + 2.5
Step-by-step explanation:
The equation of the line is expressed as y = mx+c
m is the slope
c is the intercept
Using the coordinate points (0, 2.5), (4, 20)
Slope m = 20-2.5/4-0
m = 17.5/4
m = 4.375
Get the intercept
Substitute (0, 2.5) and m = 4.375 into y - mx+c
2.5 = 4.375(0) + c
2.5 = c
Get the equation required
y = 4.375x + 2.5
The answer is x^2 - 3x - 8 - (33/(x - 3))
Answer:
x^2 -x -2
Step-by-step explanation:
Use polynomial long division.
0 + x^2 -x -2
___________________________________
(2x+3) | 2x^3 +x^2 -7x -6
-2x^3 -3x^2
_________
-2x^2 -7x
+2x^2 +3x
__________
-4x -6
+2x +6
_______
0
It’s between B or C
Sorry if I’m wrong