Answer:
Step-by-step explanation:
Formula to be used is N=N0e^(kt)
N0 - amount of bacteria at time 0, which is 10g
t=2, 30g
30 = 10*e^(k*2)
3 = e^(2k)
2k = ln3
k = 0.5ln3
At t=6:
N = 10*e^(0.5ln3*6)
N = 10e^(3ln3)
N = 10e^ln27
N = 10*27 =270
So the amount of bacteria at t=6 is 270
Answer:
5. 34°
6. 29.7
Step-by-step explanation:
5. Using sine law
25/sin(93) = 14/sin(C)
sin(C) = 0.5592325395
C = 34.00273934°
6. Using cosine law
AC² = 23.2² + 18.1² - 2(23.2)(18.1)cos(91)
AC² = 880.507229
AC = 29.67334206
The coefficient C-term is 1 in problem. The coefficient is the number before the variable of said equation. Example, the coefficient of X for thus problem 4x-4y is 4 the number BEFORE the variable, never variable itself it won’t ever be 4x
If the half-life is t, then every t days, the amount of the radioactive isotope will be cut in half.
(1/2)^(number of half-lives) = 3%
number of half-lives = ln(0.03) / ln(0.5)
This gives the number of half-lives as 5.06.
Then 300 days = (5.06)(length of 1 half-life)
length of 1 half-life = 300 / 5.06 = 59.29 days
