Answer: blue, to love me, to nice, to i love you, to be a good boy and suck.
Step-by-step explanation:
hi I am girl
Just use y2-y2/x2-x1
so
1+5 (because subtracting a negative is the same as adding a positive)
4-2
6/2
or 3/1 which is the same as 3
Answer:
see explanation
Step-by-step explanation:
Using the rules of exponents
•
⇔ ![\frac{1}{a^{m} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Ba%5E%7Bm%7D%20%7D)
•
= 1
•
×
⇔ ![a^{(m+n)}](https://tex.z-dn.net/?f=a%5E%7B%28m%2Bn%29%7D)
Given
(a)
= ![\frac{1}{8^{2} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B8%5E%7B2%7D%20%7D)
(b)
= 3³
(c)
×
× ![6^{0}](https://tex.z-dn.net/?f=6%5E%7B0%7D)
=
× 1
= ![6^{-4}](https://tex.z-dn.net/?f=6%5E%7B-4%7D)
= ![\frac{1}{6^{4} }](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B6%5E%7B4%7D%20%7D)
Answer:
Q(t) = Q_o*e^(-0.000120968*t)
Step-by-step explanation:
Given:
- The ODE of the life of Carbon-14:
Q' = -r*Q
- The initial conditions Q(0) = Q_o
- Carbon isotope reaches its half life in t = 5730 yrs
Find:
The expression for Q(t).
Solution:
- Assuming Q(t) satisfies:
Q' = -r*Q
- Separate variables:
dQ / Q = -r .dt
- Integrate both sides:
Ln(Q) = -r*t + C
- Make the relation for Q:
Q = C*e^(-r*t)
- Using initial conditions given:
Q(0) = Q_o
Q_o = C*e^(-r*0)
C = Q_o
- The relation is:
Q(t) = Q_o*e^(-r*t)
- We are also given that the half life of carbon is t = 5730 years:
Q_o / 2 = Q_o*e^(-5730*r)
-Ln(0.5) = 5730*r
r = -Ln(0.5)/5730
r = 0.000120968
- Hence, our expression for Q(t) would be:
Q(t) = Q_o*e^(-0.000120968*t)