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Alecsey [184]
3 years ago
11

The formula for the sum of an infinite geometric series, S=a1/1-r, may be used to convert

Mathematics
2 answers:
schepotkina [342]3 years ago
7 0

Answer:

  D.  a1=23/100, r=1/100

Step-by-step explanation:

The repeating fraction can be written as the sum ...

0.\overline{23}=0.23+0.0023+0.000023+\dots

The first term is a1 = 0.23 = 23/100, and each successive term is shifted 2 decimal places to the right, so is multiplied by the common ratio r=1/100.

qwelly [4]3 years ago
6 0

Answer:

Step-by-step explanation:

Here, a1 = 0.23 and r = 0.01.  Thus, the sum of this infinite series will be

 a1          0.23          0.23

------- = ------------- = ----------- = 23/99.

1 - r      1 - 1/100      99/100

Check this by dividing 23 by 99 on a calculator.  Result:  0.23232323....

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1024 is the answer to this problem.

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If tan theta =24 over 45 find tan theta minus cos theta. <br> ​
juin [17]

Answer:

24/45 - 45/51

Step-by-step explanation:

Tangent can be expressed as opposite/adjacent. So, create a right triangle with one leg = 24 and the other = 45. The hypotenuse is 51 by the Pythagorean theorem. Cosine is adjacent/hypotenuse, so it would be 45/51.

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One garden has 5 times more raspberry bushes than another garden. After 22 bushes were transplanted from the first garden to the
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Suppose you invest $150 a month for 5 years into an account earning 7% compounded monthly. After 5 years, you leave the money, w
lisabon 2012 [21]

Answer:

About 0.3 billion dollars

Step-by-step explanation:

5 years = 60 months.

The 150 of the first month will be 150*1.07^60 in 5 years.

The 150 of the second month will be 150*1.07^59 in 5 years.

The 150 of the third month will be 150*1.07^58 in 5 years.

And so forth.

So we sum that up:

( sum_(n=1)^(60) 150×1.07^n)

And multiply with

× 1.07^(5×23)

to account for the increase in value in the following 23 years.

7 0
3 years ago
According to data from a medical​ association, the rate of change in the number of hospital outpatient​ visits, in​ millions, in
GaryK [48]

Answer:

a) f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

b) f(t=2015) = 264,034,317.7

Step-by-step explanation:

The rate of change in the number of hospital outpatient​ visits, in​ millions, is given by:

f'(t)=0.001155t(t-1980)^{0.5}

a) To find the function f(t) you integrate f(t):

\int \frac{df(t)}{dt}dt=f(t)=\int [0.001155t(t-1980)^{0.5}]dt

To solve the integral you use:

\int udv=uv-\int vdu\\\\u=t\\\\du=dt\\\\dv=(t-1980)^{1/2}dt\\\\v=\frac{2}{3}(t-1980)^{3/2}

Next, you replace in the integral:

\int t(t-1980)^{1/2}=t(\frac{2}{3}(t-1980)^{3/2})- \frac{2}{3}\int(t-1980)^{3/2}dt\\\\= \frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}+C

Then, the function f(t) is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+C'

The value of C' is deduced by the information of the exercise. For t=0 there were 264,034,000 outpatient​ visits.

Hence C' = 264,034,000

The function is:

f(t)=0.001155[\frac{2}{3}t(t-1980)^{3/2}-\frac{4}{15}(t-1980)^{5/2}]+264,034,000

b) For t = 2015 you have:

f(t=2015)=0.001155[\frac{2}{3}(2015)(2015-1980)^{1/2}-\frac{4}{15}(2015-1980)^{5/2}]+264,034,000\\\\f(t=2015)=264,034,317.7

3 0
3 years ago
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