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torisob [31]
3 years ago
15

You are walking near a river. While standing at A, you measure an angle of 90° between B and C, as shown. You then walk to B and

measure
an angle of 61° between A and C. The distance between A and B is about 1 mile. How wide (in miles) is the river between A and C? Round your
answer to the nearest tenth.
61
B Im
A
The width is about
miles.

Mathematics
1 answer:
Dima020 [189]3 years ago
8 0

Answer:

3.6

Step-by-step explanation:

tan61=x/2

2(tan61)=x

3.6=x

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Suppose you invested $500 at an annual interest rate of 7.1 compounded continuously. How much will you have in the account after
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\bf \qquad \textit{Continuous Interest Earned Amount}\\\\
A=Pe^{rt}\qquad 
\begin{cases}
A=\textit{compounded amount}\\
P=\textit{original amount deposited}\to& \$500\\
r=rate\to 7.1\%\to \frac{7.1}{100}\to &0.071\\
t=years\to &10
\end{cases}
\\\\\\
A=500e^{0.071\cdot 10}
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A population of bees is decreasing. The population in a particular region this year is 1,250. After 1 year, it is estimated tha
8_murik_8 [283]
To model this situation, we are going to use the decay formula: A=Pe^{rt}
where 
A is the final pupolation
P is the initial population 
e is the Euler's constant
r is the decay rate 
t is the time in years

A. We know for our problem that the initial population is 1,250, so P=1250; we also know that after a year the population is 1000, so A=1000 and t=1. Lets replace those values in our formula to find r:
A=Pe^{rt}
1000=1250e^{r}
e^{r}= \frac{1000}{1250}
e^{r}= \frac{4}{5}
ln(e^{r})=ln( \frac{4}{5} )
r=ln( \frac{4}{5} )
r=-02231

Now that we have r, we can write a function to model this scenario:
A(t)=1250e^{-0.2231t}.

B. Here we are going to use a graphing utility to graph the function we derived in the previous point. Please check the attached image.

C. 
- The function is decreasing
- The function doe snot have a x-intercept 
- The function has a y-intercept at (0,1250)
- Since the function is decaying, it will have a maximum at t=0: 
A(0)=1250e^{(-0.2231)(0)
A_{0}=1250e^{0}
A_{0}=1250
- Over the interval [0,10], the function will have a minimum at t=10:
A(10)=1250e^{(-0.2231)(10)
A_{10}=134.28

D. To find the rate of change of the function over the interval [0,10], we are going to use the formula: m= \frac{A(0)-A(10)}{10-0}
where 
m is the rate of change 
A(10) is the function evaluated at 10
A(0) is the function evaluated at 0
We know from previous calculations that A(10)=134.28 and A(0)=1250, so lets replace those values in our formula to find m:
m= \frac{134.28-1250}{10-0}
m= \frac{-1115.72}{10}
m=-111.572
We can conclude that the rate of change of the function over the interval [0,10] is -111.572.

7 0
3 years ago
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