If you have 1912 copies x 305 runs per copy. Would I estimate by multiplying 1912 copies per run by 305 runs.
1 answer:
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Answer:
The half-life of the radioactive substance is 135.9 hours.
Step-by-step explanation:
The rate of decay is proportional to the amount of the substance present at time t
This means that the amount of the substance can be modeled by the following differential equation:
Which has the following solution:
In which Q(t) is the amount after t hours, Q(0) is the initial amount and r is the decay rate.
After 6 hours the mass had decreased by 3%.
This means that . We use this to find r.
So
Determine the half-life of the radioactive substance.
This is t for which Q(t) = 0.5Q(0). So
The half-life of the radioactive substance is 135.9 hours.
Answer:
1.1) B=12.42°, C=146.58°, c=15.37
1.2) B=8°, C=124°, c=35.7
2) In the case of SSA, when the given angle A is acute (<90°), the side "a" (opposite to the given angle "A") is less than the other given side "b" (a<b), and a > b sinA.
3.1) Perimeter = 31.37 units, Area = 16.5 square units
3.2) Perimeter = 73.70 units, Area = 75.6 square units
4.1) Magnitude of vector AB=5.1
4.2) Direction of vector AB=348.69°
5.1) The dot product of two vectors is equal to the sum of the product of their components
5.2) a . b = 10
6.1) Trigonometric form: z = 3√2 (cos 315°+ i sin 315°)
6.2) The 4th roots are:
w1 = 18^(1/8) [cos(78.75°) + i sin(78.75°)]
w2 = 18^(1/8) [cos(168.75°) + i sin(168.75°)]
w3 = 18^(1/8) [cos(258.75°) + i sin(258.75°)]
w4 = 18^(1/8) [cos(348.75°) + i sin(348.75°)]
Step-by-step explanation:
3.1) Perimeter: P = a+b+c → P = 10+6+15.37 → P = 31.37
Area: A = a b sinC / 2
Replacing the known values:
A = (10)(6) sin 146.58325418° / 2 → A = 60(0.55072472) / 2 →
A = 16.52174152 → A = 16.52 square units
3.2) Perimeter: P = a+b+c → P = 32+6+35.70 → P=73.70
Area: A = a b sin C /2
Replacing the known values:
A = (32)(6) sin 123.99036327° / 2 → A = 192(0.82913161) / 2 →
A = 75.59663485 → A = 75.6 square units
