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do (2+1/3).

^ That is how it will show up.

Yes.

37 is a prime number because its only whole factors are 1 and 37.

Find the smallest odd prime number. No restrictions on the operation. use these numbers 2,3,5,7,11

3 / [(11+2) - (7+5)]

3/[13 - 12]

3 / 1

3 is the smallest odd prime number.

yes

life doesn't happen to you it happens for you, how do i know this...i dont im just making sound. thats the good thing.

Since  _nC₂  =  n·(n - 1) / (2·1):

_nC₂  =  56     --->     n·(n - 1) / 2  =  56

                     --->                n² - n  =  112

                     --->         n² - n - 112  =  0

                            But this can't be factored!                     

There is no answer!

The voltage required for a circuit is found by , where is the power, in watts, and is the resistance, in ohms. What is the approximate difference in the voltages required to light a 45-watt bulb and a 90-watt bulb, both of which have a resistance of 120 ohms?

The meaning of life is 4x^2+5(5x)10y=b 

1524157875019052100

😯5+25=30

to live to the life fullest

How do you go from (sqrt 2)^17 to 256 * sqrt 2 !!!?????

Guest #5

          Thank you very much for those detailed instructions. We are allowed to use financial calculators, and I followed your instructions and got exactly the same numbers as you did.

So, I thank you and Melody for solving this problem that was nagging me for a couple of days, which is the reason I posted the question. Now and in future, I should know how to handle problems of this sort.

Which equations could represent Relationship B? Select each correct answer. y=1.4x y=2.2x y=3x y=1/2x

.

pie is life

Don't worry about it I figured it out.

How does (sqrt 2)^17 = 256 * sqrt 2

Simplify the following:
(sqrt(2))^17

Multiply exponents. (sqrt(2))^17 = 2^(17/2):
2^(17/2)

2^(17/2) = 2^(16/2+1/2) = 2^(16/2)×sqrt(2):
2^(16/2) sqrt(2)

16/2 = (2×8)/2 = 8:
2^8 sqrt(2)

2^8 = (2^4)^2 = ((2^2)^2)^2:
((2^2)^2)^2 sqrt(2)

2^2 = 4:
(4^2)^2 sqrt(2)

4^2 = 16:
16^2 sqrt(2)
Answer: |256 sqrt(2)

Stupidest question ever am i right?

Here is another approach to solving the same problem:

1) Find the PV of the four $25,000 that he is withdrawing at the beginning of each year. I assume you know the formula that is used for that. I get a figure of $92,933.25 @ 5% comp.monthly.

2) The above amount is 10 years in the future, so we have to find its PV as of today, using the common PV formula, which I assume you know. By doing that, I get a PV of $56,425.45 @ 5% comp.monthly.

3) Now this number that we just got, $56,425.45, is the important figure for the parents, because they now have the PV of ALL their monthly payments for 13 years or 156 months.

4) Now, we use the annuity formula to find the monthly payment. By doing that, we get a monthly payment of $492.63, which is the same number that Melody calculated.

The only thing that you have to be careful about is the number of years or months that the parents will continue to deposit the money. It is 13 not 14 years, because by the end of the 13th year, they should have saved enough money for their son to be able to withdraw the last payment of $25,000 at the beginning of the 14th year. Good luck to you.

-2x+3y=6 

x= -1.5

y = 1

-2x+3y=6 

x= -1.5

y = 1