Solve for x in each of the following equations At least 2 of the 4 must be solved algebraically. When solving graphically, sketch or paste a digital image of the graph you used to solve for the question.
4)
140 = 200 * (1/2)^(3/h), where h = half-life.
Divide both sides by 200
0.7 = 1/2^(3/h)
Take the natural log of both sides
3/h =Ln(0.7) / Ln(1/2)
3/h =0.514573173
Cross multiply
0.514573173h = 3
Divide both sides by 0.514573173
h =5.83 hours - the half-life of this radioactive substance.
1.5 days x 24 hours = 36 hours.
Use the same above formula:
R =200 * 1/2^(36/5.83)
R =200 * 1/2^(6.175)
R =200 * 0.0138401175...
R =2.768 grams will remain after 1.5 days or 36 hours.
3) I modified your equation: e^-(xk)
L(x)=100e^-((k)x)
L(9) =100*e^-(0.2*9)
L(9) =100 * e^-1.8
L(9) =100 * 0.1653
L(9) =16.53% - percentage of sunlight that penetrates water at a depth of 9m.
5% = 100 * e^-(0.2x), solve for x
0.05 =100 * e^-(0.2x) divide both sides by 100
0.0005 = e^-(0.2x) take the natural log of both sides
-0.2x = -7.6 divide both sides by -0.2
x = 38 meters - at which only 5% of sunlight will penetrate the water.