Point is (4,7)
So y = 7 = 4^2 + c
c = 7 - 16
c = -9
3.2d - 4d = 2.3d + 3...simplify by combining like terms
-0.8d = 2.3d + 3....subtract 2.3d from both sides
-0.8d - 2.3d = 3 ...simplify again
-3.1d = 3...divide both sides by -3.1
d = 3/ -3.1
d = - 0.97
or
3.2d - 4d = 2.3d + 3....multiply the equation by 10, gets rid of the decimals
32d - 40d = 23d + 30....subtract 23d from both sides
32d - 40d - 23d = 30....simplify
-31d = 30...divide by -31
d = -30/31
d = - 0.97
Answer:
1. 1343 years
2. 9 hours
3. 39 years
Step-by-step explanation:
1. Given, half-life of carbon = 5730 years.
∴ λ = 0.693/half-life of carbon = 0.693/5730 = 0.000121
If N₀ = 100 then N = 85
Formula:- N = N₀*e^(-λt)
∴ 85 = 100 * e^(-0.000121t)
∴㏑(-0.85)=-0.000121t
∴ t = 1343 years
2. Given half-life of aspirin = 12 hours
λ = 0.693/12 = 0.5775
Also N₀ = 100 then 70 will disintegrate and N = 30 will remain disintegrated.
∴ 70 = 100 *e^(-0.05775t)
0.70 = e^(-0.05775t)
㏑(0.70) = -0.05775t
∴ t = 9 hours
3. The population of the birds as as A=A₀*e^(kt)
Given that the population of birds fell from 1400 from 1000, We are asked how much time it will take for the population to drop below 100, let that be x years.
The population is 1400 when f = 0, And it is 1000 when f = 5
We can write the following equation :
1400 = 1000e^(5t).
∴1400/1000 = e^(5k)
∴ k = ㏑(1.4)/5
We need to find x such that 1400/100 = e^(xk)
14 = e^(xk)
∴ x = 39 years
