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Soloha48 [4]
2 years ago
13

In the year 1985, a house was valued at $106,000. By the year 2005, the value had appreciated to $146,000. What was the annual g

rowth rate between 1985 and 2005? Assume that the value continued to grow by the same percentage. What was the value of the house in the year 2010?
Mathematics
1 answer:
N76 [4]2 years ago
8 0

The value of the house in the year 2010 is 150056.47.

<h3>What is growth rate?</h3>

The growth rate of a value (GDP, turnover, wages, etc.) measures its change from one period to another (month, quarter, year)

Value

v(t) = V0 ( 1 + r)^t

let V0 =  106, v(20) = 146

So,

146 = 110(1 + r )^20

146/110 = (1 + r )^20

1.327 = (1 + r )^20

1+ r = 1.014

r= 0.014

So, annual growth rate is 1.4%.

In the year 2010

v(t) = 106(1.014)^25

v(t) = 106(1.014)^25

v(t) - 150056.47

Learn more about this concept here:

brainly.com/question/15383938

#SPJ1

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Answer:

6 billion years.

Step-by-step explanation:

According to the decay law, the amount of the radioactive substance that decays is proportional to each instant to the amount of substance present. Let P(t) be the amount of ^{235}U and Q(t) be the amount of ^{238}U after t years.

Then, we obtain two differential equations

                               \frac{dP}{dt} = -k_1P \quad \frac{dQ}{dt} = -k_2Q

where k_1 and k_2 are proportionality constants and the minus signs denotes decay.

Rearranging terms in the equations gives

                             \frac{dP}{P} = -k_1dt \quad \frac{dQ}{Q} = -k_2dt

Now, the variables are separated, P and Q appear only on the left, and t appears only on the right, so that we can integrate both sides.

                         \int \frac{dP}{P} = -k_1 \int dt \quad \int \frac{dQ}{Q} = -k_2\int dt

which yields

                      \ln |P| = -k_1t + c_1 \quad \ln |Q| = -k_2t + c_2,

where c_1 and c_2 are constants of integration.

By taking exponents, we obtain

                     e^{\ln |P|} = e^{-k_1t + c_1}  \quad e^{\ln |Q|} = e^{-k_12t + c_2}

Hence,

                            P  = C_1e^{-k_1t} \quad Q  = C_2e^{-k_2t},

where C_1 := \pm e^{c_1} and C_2 := \pm e^{c_2}.

Since the amounts of the uranium isotopes were the same initially, we obtain the initial condition

                                 P(0) = Q(0) = C

Substituting 0 for P in the general solution gives

                         C = P(0) = C_1 e^0 \implies C= C_1

Similarly, we obtain C = C_2 and

                                P  = Ce^{-k_1t} \quad Q  = Ce^{-k_2t}

The relation between the decay constant k and the half-life is given by

                                            \tau = \frac{\ln 2}{k}

We can use this fact to determine the numeric values of the decay constants k_1 and k_2. Thus,

                     4.51 \times 10^9 = \frac{\ln 2}{k_1} \implies k_1 = \frac{\ln 2}{4.51 \times 10^9}

and

                     7.10 \times 10^8 = \frac{\ln 2}{k_2} \implies k_2 = \frac{\ln 2}{7.10 \times 10^8}

Therefore,

                              P  = Ce^{-\frac{\ln 2}{4.51 \times 10^9}t} \quad Q  = Ce^{-k_2 = \frac{\ln 2}{7.10 \times 10^8}t}

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                                          \frac{P(t)}{Q(t)} = 137.7

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Solving for t yields t \approx 6 \times 10^9, which means that the age of the  universe is about 6 billion years.

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