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life: people tell others all the time how they feelwhethe it is appropriate or not.
1 Simplify 7x+16+47x+16+4 to 7x+207x+20<span> 7x+20=x<span>7x+20=x</span></span>
2 Subtract 7x from both sides<span> 20=x-7x<span>20=x−7x</span></span>
3 Simplify <span>x-7x<span>x−7x</span></span> to <span>-6x<span>−6x</span></span>
<span>20=-6x<span>20=−6x</span></span>
4 Divide both sides by −6 -20/6=x
5 Simplify 20/6 to 10/3 -10/3=x
6 Switch sides
x=-10/3<span><span><span> </span></span></span>
<u>Answer:</u>
32 units
<u>Step-by-step explanation:</u>
We have a quadrilateral ABCD and we are given the following coordinates for these vertices:
A (-11,-6)
B (-3,0)
C (1,0)
D (1,-6)
AB = 
= 10 units
BC = 
= 4 units
CD = 
= 6 units
AD = 
= 12 units
Perimeter of ABCD = 
= 32 units
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)
