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3 years ago

15

During the first two hours of a 4-h trip, the force of the wind causes a sailboat to travel 68 km. During the next two hours, th

e wind speed changes. The boat sails 3 km in the next hour and 25 km in the last hour. What was the boat's AVERAGE speed?

1 answer:

mezya [45]3 years ago

5 0

Answer:

24 km/hr

Step-by-step explanation:

Speed is the rate of distance traveled in a defined time. Average speed is to take calculate the total distance travel at different speeds dividing by the total time taken.

Total Time = 4 hours

Total Distance = 68 km + 3 km + 25 km = 96 km

Average Speed = Total Distance / Total Time taken= 96 km / 4 hours

Average Speed = 24 km per hour

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Determine if the two expressions are equivalent and

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Answer:

yes. Commutative Property of Addition

Step-by-step explanation:

The commutative property of addition tells you that the order of the addends is immaterial. These expressions are equivalent.

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3 years ago

What describes this angle? Because I don’t know what the hell it means.

maw [93]

Answer:

Less than a right angle.

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If you really think about it, a right angle is 90 degrees, and from the looks of it, that angle shown is an acute angle. Acute angles are always smaller than a right angle, so your answer is less than a right angle. :)

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3 years ago

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 $m$ :
$m= \frac{134.28-1250}{10-0}$
$m= \frac{-1115.72}{10}$
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We can conclude that the rate of change of the function over the interval [0,10] is -111.572.

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2 years ago

the volume of a cylindrical watering can is 100cm^3. If the radius is doubled, then how much water can the new can hold?

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V=(pi)(2r^2)(h)
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3 years ago

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aliina [53]

Answer: True

Step-by-step explanation:

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2 years ago

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