Answer:
3000 text messages can be sent without breaking the budget.
Step-by-step explanation:
Since each month your cell phone company charges $ 50 for your plan plus 3 cents for each text you send, and you have $ 140 budgeted for cell phone expenses for the month, to make a determination about the number of texts you can send each month the following calculation must be performed:
(140 - 50) / 0.03 = X
90 / 0.03 = X
3000 = X
Therefore, 3000 text messages can be sent without breaking the budget.
The get (x-1)(x-√2)(x+√2)(x-√3)(x+√3).
Expanding, you get x^5-x^4-5x^3+5x^2+6x-6.
Part a)
Answer: 5*sqrt(2pi)/pi
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Work Shown:
r = sqrt(A/pi)
r = sqrt(50/pi)
r = sqrt(50)/sqrt(pi)
r = (sqrt(50)*sqrt(pi))/(sqrt(pi)*sqrt(pi))
r = sqrt(50pi)/pi
r = sqrt(25*2pi)/pi
r = sqrt(25)*sqrt(2pi)/pi
r = 5*sqrt(2pi)/pi
Note: the denominator is technically not able to be rationalized because of the pi there. There is no value we can multiply pi by so that we end up with a rational value. We could try 1/pi, but that will eventually lead back to having pi in the denominator. I think your teacher may have made a typo when s/he wrote "rationalize all denominators"
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Part b)
Answer: 3*sqrt(3pi)/pi
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Work Shown:
r = sqrt(A/pi)
r = sqrt(27/pi)
r = sqrt(27)/sqrt(pi)
r = (sqrt(27)*sqrt(pi))/(sqrt(pi)*sqrt(pi))
r = sqrt(27pi)/pi
r = sqrt(9*3pi)/pi
r = sqrt(9)*sqrt(3pi)/pi
r = 3*sqrt(3pi)/pi
Note: the same issue comes up as before in part a)
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Part c)
Answer: sqrt(19pi)/pi
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Work Shown:
r = sqrt(A/pi)
r = sqrt(19/pi)
r = sqrt(19)/sqrt(pi)
r = (sqrt(19)*sqrt(pi))/(sqrt(pi)*sqrt(pi))
r = sqrt(19pi)/pi
Answer: It will take 4.4 hours to fill the pool.
Step-by-step explanation: One pump can fill a swimming pool in 8 hours and another pump can fill it in 10 hours. If both pumps are opened at the same time, how many hours will it take to fill the pool?
Aprox. 4.4 hours.
Hope this Helps!
First, notice that the series you wrote is
∑ (-3)(n) for n ∈ [0,5]
There is a general formula for the sum of a basic geometric series that you should know or learn.
Here's a hint: How can you simplify the product of (1-x) and a summation, for n∈ [0,N], of all the n'th powers of x ((1-x)·∑ xn for n∈ [0,N])?
You can see you have to perform a distributive multiplication here, and if you write out the first three or four terms, and the last three or four terms, you should see a lot of cancellation going on. Pay special attention to what happens at n=0, n=N.
(x0+x1+x2+...+xN-2+xN-1+xN)-(x1+x2+...+xN-1+xN+xN+1)
Notice how all the values in between end up cancelling?
You can use your result to find an appropriate formula. In your case, x is 3. N is 5.
If your answer is correct, your formula will work for summing up any number of terms of any geometric series (any value of x and any value N). It is easy to compare your answer using your formula to the answer found manually by actually adding up all the terms.
