I took the liberty of finding for the complete question.
And here I believe that the problem asks for the half life of Curium. Assuming
that the radioactive decay of Curium is of 1st order, therefore the
rate equation is in the form of:
A = Ao e^(-kt)
where,
A = amount after t years = 2755
Ao = initial amount = 3312
k = rate constant
t = number of years passed = 6
Therefore the rate constant is:
2755/3312 = e^(-6k)
-6k = ln (2755/3312)
k = 0.0307/yr
The half life, t’, can be calculated using the formula:
t’ = ln 2 / k
Substituting the value of k:
t’ = ln 2 / 0.0307
t’ = 22.586 years
or
t’ = 22.6 years
1.9 is greater bc there is a whole number
Hope this helped!!!!
Answer:
Step-by-step explanation:
The diagram of the triangles are shown in the attached photo.
1) Looking at ∆AOL, to determine AL, we would apply the sine rule
a/SinA = b/SinB = c/SinC
21/Sin25 = AL/Sin 105
21Sin105 = ALSin25
21 × 0.9659 = 0.4226AL
AL = 20.2839/0.4226
AL = 50
Looking at ∆KAL,
AL/Sin55 = KL/Sin100
50/0.8192 = KL/0.9848
50 × 0.9848 = KL × 0.8192
KL = 49.24/0.8192
KL = 60
AK/Sin25 = AL/Sin 55
AKSin55 = ALSin25
AK × 0.8192 = 0.4226 × 50
AK = 21.13/0.8192
AK = 25.8
2) looking at ∆AOC,
Sin 18 = AD/AC = 18/AC
AC = 18/Sin18 = 18/0.3090
AC = 58.25
Sin 85 = AD/AB = 18/AB
AB = 18/Sin85 = 18/0.9962
AB = 18.1
To determine BC, we would apply Sine rule.
BC/Sin77 = 58.25/Sin85
BCSin85 = 58.25Sin77
BC = 58.25Sin77/Sin85
BC = 58.25 × 0.9744/0.9962
BC = 56.98
The 3rd one. (where the round part of the graph is at 0, 0) because that is where the y axis meets
93
Because basically u are adding 3 and then u multiply by 3
