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2 years ago

The diameter of a semicircle is 6 kilometres. What is the semicircle's radius?

Mathematics

1 answer:

bixtya [17]2 years ago

3 0

Answer:

3km

Step-by-step explanation:

d = 2r so r = 1/2 * d = 1/2 * 6 = 3km

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Learning Theory In a typing class,the averege number N of words per minutes typed after t weeks of lessons can be modeled by N =

Jet001 [13]

Answer:

a) $N(t=10) = \frac{95}{1+8.5 e^{-0.12(10)}}= \frac{95}{1+ 8.5 e^{-1.2}} = 26.684$

b) $N(t=20) = \frac{95}{1+8.5 e^{-0.12(20)}}= \frac{95}{1+ 8.5 e^{-2.4}} = 53.639$

c) $70 =\frac{95}{1+8.5 e^{-0.12t}}$

$1+ 8.5 e^{-0.12t} = \frac{95}{70}= \frac{19}{14}$

$8.5 e^{-0.12t} = \frac{19}{14}-1= \frac{5}{14}$

$e^{-0.12t} = \frac{\frac{5}{14}}{8.5}= \frac{5}{119}$

$ln e^{-0.12t} = ln (\frac{5}{119})$

$-0.12 t = ln(\frac{5}{119})$

$t = \frac{ln(\frac{5}{119})}{-0.12} = 26.414 weeks$

d) If we find the limit when t tend to infinity for the function we have this:

$lim_{t \to \infty} \frac{95}{1+8.5 e^{-0.12t}} = 95$

So then the number of words per minute have a limit and is 95 as t increases without bound.

Step-by-step explanation:

For this case we have the following expression for the average number of words per minutes typed adter t weeks:

$N(t) = \frac{95}{1+8.5 e^{-0.12t}}$

Part a

For this case we just need to replace the value of t=10 in order to see what we got:

$N(t=10) = \frac{95}{1+8.5 e^{-0.12(10)}}= \frac{95}{1+ 8.5 e^{-1.2}} = 26.684$

So the number of words per minute typed after 10 weeks are approximately 27.

Part b

For this case we just need to replace the value of t=20 in order to see what we got:

$N(t=20) = \frac{95}{1+8.5 e^{-0.12(20)}}= \frac{95}{1+ 8.5 e^{-2.4}} = 53.639$

So the number of words per minute typed after 20 weeks are approximately 54.

Part c

For this case we want to solve the following equation:

$70 =\frac{95}{1+8.5 e^{-0.12t}}$

And we can rewrite this expression like this:

$1+ 8.5 e^{-0.12t} = \frac{95}{70}= \frac{19}{14}$

$8.5 e^{-0.12t} = \frac{19}{14}-1= \frac{5}{14}$

Now we can divide both sides by 8.5 and we got:

$e^{-0.12t} = \frac{\frac{5}{14}}{8.5}= \frac{5}{119}$

Now we can apply natural log on both sides and we got:

$ln e^{-0.12t} = ln (\frac{5}{119})$

$-0.12 t = ln(\frac{5}{119})$

And then if we solve for t we got:

$t = \frac{ln(\frac{5}{119})}{-0.12} = 26.414 weeks$

And we can see this on the plot 1 attached.

Part d

If we find the limit when t tend to infinity for the function we have this:

$lim_{t \to \infty} \frac{95}{1+8.5 e^{-0.12t}} = 95$

So then the number of words per minute have a limit and is 95 as t increases without bound.

8 0

3 years ago

What does the equation x^2+y^2=14 represent?

sdas [7]

Your answer is b. circle

8 0

3 years ago

If a = 4 and b = 5, what is the value of the following expression? 6a + 3b - 8

castortr0y [4]

Simple...

you have: 6a+3b-8

a=4 and b=5

Plug in what you know-->>

6(4)+3(5)-8

24+15-8

39-8

31

Thus, your answer.

5 0

3 years ago

ANswer this question!!!!!

4vir4ik [10]

<u><em>Its Going to Be Table B! Good Luck :)⇔</em></u>

6 0

3 years ago

HELP PLEASE I really don’t understand it

valina [46]

Answer:

i think its b

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers