The slope of the line has to be the same the the equation given to be parallel.
The slope of the line given is... 2/3.
If the line passes through point (1,3) then...
3=2/3(1)+b
b=2 1/3
answer: y=2/3x+2 1/3
Answer:
360
Step-by-step explanation: A full angle is 360. 90*4=360.
Answer:
the answer is the 2nd one :)
Step-by-step explanation:
Answer:
D. (25π - 50) sq m
Step-by-step explanation:
Since ABCD is a square with the area of 100, we know that the dimensions are 10 by 10. With that, we can find that triangle ABC = 50 sq m because of (h*l)/2. Since A is the center of the circle that contains arc BD, we know that 10 is the radius of the quarter-circle ABC. Using (πr^2)/4 to get the area of the quarter-circle, we get that the quarter-circle ABC = 25π sq m. But the original triangle ABC in the quarter-circle is not shaded. Therefore, we subtract 50 from 25π, which leads to (25π - 50) sq m.
I hope this helped! :D
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)
