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(-2x+11y)2

\((-2x+11y)^2\\ =(-2x)^2\;+\;2*(-2x)*(11y)\:+\;(11y)^2\\ =4x^2\;-\;44xy\:+\;121y^2\)

Compute. i+i^2+i^3+\cdots+i^{258}+i^{259}.

\(Compute. i+i^2+i^3+\cdots+i^{258}+i^{259}\\ i^1=i\\ i^2=-1\\ i^3=-i\\ i^4=1\\ i^5=i\\ \mbox{and so the pattern continues, so}\\ i^2+i^4+\dots +i^{256}=0\\ i^1+i^3+\dots +i^{255}=0\\ \mbox{so this leaves}\\ =0+i^{257}+i^{258}+i^{259}\\ =i^{4*64+1}+i^{4*64+2}+i^{4*64+3}\\ =i-1-i\\ =-1 \)

Solve for x over the real numbers:
sinh(2 log(x)) = 0

Take the inverse hyperbolic sine of both sides:
2 log(x) = 0

Divide both sides by 2:
log(x) = 0

Cancel logarithms by taking exp of both sides:
Answer: |x = 1

Simplify the following:
sqrt(75/2)-sqrt(3/2)

sqrt(3/2) = (sqrt(3))/(sqrt(2)):
sqrt(75/2)-(sqrt(3))/(sqrt(2))

Rationalize the denominator. (sqrt(3))/(sqrt(2)) = (sqrt(3))/(sqrt(2))×(sqrt(2))/(sqrt(2)) = (sqrt(3) sqrt(2))/(2):
sqrt(75/2)-(sqrt(3) sqrt(2))/(2)

sqrt(3) sqrt(2) = sqrt(3×2):
sqrt(75/2)-(sqrt(3×2))/(2)

3×2 = 6:
sqrt(75/2)-( sqrt(6 ) )/(2)

sqrt(75/2) = (sqrt(75))/(sqrt(2)):
(sqrt(75))/(sqrt(2))-(sqrt(6))/(2)

sqrt(75) = sqrt(3×5^2) = 5 sqrt(3):
(5 sqrt(3))/(sqrt(2))-(sqrt(6))/(2)

Rationalize the denominator. (5 sqrt(3))/(sqrt(2)) = (5 sqrt(3))/(sqrt(2))×(sqrt(2))/(sqrt(2)) = (5 sqrt(3) sqrt(2))/(2):
(5 sqrt(3) sqrt(2))/(2)-(sqrt(6))/(2)

sqrt(3) sqrt(2) = sqrt(3×2):
(5 sqrt(3×2))/(2)-(sqrt(6))/(2)

3×2 = 6:
(5 sqrt(6 ) )/(2)-(sqrt(6))/(2)

(5 sqrt(6))/(2)-(sqrt(6))/(2) = (5 sqrt(6)-sqrt(6))/(2):
(5 sqrt(6)-sqrt(6))/(2)

5 sqrt(6)-sqrt(6) = 4 sqrt(6):
(4 sqrt(6))/(2)

4/2 = (2×2)/2 = 2:
Answer: | 2 sqrt(6)

If you can't simplify it, then please say so.

 Did you forget something? What numbers are you supposed to use to make up the answer that you want, or 14946581?

3/16 were apple   = .1875  = 18.75%

This is a famous series that has been known for a long time. It converges extremely slowly (evaluating the first 10,000 terms gives 3 accurate digits)!!. It converges to ln(2)=0.69314718055......etc.

∑[ i^n], for n=1 to 259= -1

See if this helps :

https://www.wolframalpha.com/input/?i=1%2F%5B2n-1%5D-1%2F%5B2n%5D++sum+1+to+infinity

CORRECTION: I said PV instead of FV in this particular case. Sorry about that.

These types of deposits are very common in the business world, where small and large businesses deposit regular or irregular amounts of money into mutual funds or many other investment accounts.

The purpose, in this example, is to find the interest rate that these funds earned in 2015. This interest rate is called IRR, or Internal Rate of Return.

Many softwares are available to calculate such returns, including many spreadsheets such as Excel to calculate these returns. I happened to have software programmed into my computer to do these calculation very rapidly.

All these softwares do the same thing, which is to find the PV of all these payments which will equate them to the final market value of the fund, which in this case is $112,500. So, they use iteration to find a couple of interest rates, and then use interpolation between them to arrive at actual IRR. This is what Wolfram/Alpha does.

Now, entering these amounts into my computer, it comes out with a daily rate of=.00057492740959, which is essentially the same as Melody found using Wolfram/Alpha engine. The custom, in the business world, is to raise this number to the power of 365 days to find the Effective IRR.

So, 1.00057492740959^365=23.34%, which is basically the same as Melody found.

Thanks Omar,

You are right, I copied the question incorrectly but i think the steps are all good though :)

Hello Melody,

your steps are perfectly done.

there is a little mistake you've done 

\(\frac {200\times 25\; + \;5\times 800}{ 5 \times 20 \;+\; 25 \times 4 }\\ \)

you did 200*2+5*800

but the tutorial is nicely done.

the real answer to his problem btw is 45

What a nice thank you       I am really pleased that my answer was helpful. :)

The following deposits were made into a balanced mutual fund in 2015. The number of days is the length of time each deposit earned interest:
$30,000 on deposit for 300 days.
$10,000 on deposit for 257 days.
$25,000 on deposit for 179 days.
$15,000 on deposit for 143 days
$20,000 on deposit for 107 days
These deposits totaling $100,000 were worth $112,500 on Dec,31, 2015. What was the overall yield on this mutual fund for 2015?. Thank you for any help.

I have assumed that the interested is calculated and added daily.

I have let R = 1+ the daily interest rate.

\(112,500=30000R^{300}+10000R^{257}+25000R^{179}+15000R^{143}+20000R^{107}\\ 11.25=3R^{300}+1R^{257}+2.5R^{179}+1.5R^{143}+2R^{107}\\ \)

Now I have no idea how to do this so I entered it into Wolfram|Alpha

\((r+2i)(r-2i)\)

Thank you very much for the detailed answer, i will try to mimik it and memorize the technique into action.

Update: It worked as a charm with your advice, thank you.

How do i calculate 200 x 25 + 5x 800
                               5 x 20 + 25 x 4 

\(\frac {200\times 2\; + \;5\times 800}{ 5 \times 20 \;+\; 25 \times 4 }\\ =\frac {200\times 2\; + \;5\times 800}{ 5 \times 20 \;+\; 25 \times 4 }\\ \qquad 200*2=2*2\;and\; add\; the\; two\; zeros = 400\\ \qquad 5*800=5*8\;and\; add\; the\; two\; zeros = 4000\\ \qquad 5*20=100\\ \qquad 25*4=100\\ \qquad \mbox{Put it all together and you get...}\\~\\ =\frac {400\; + \;4000}{ 100 \;+\; 100 }\\ =\frac {4400}{ 200 }\\ =\frac {4400\div 100}{ 200\div 100 }\\ =\frac {44}{ 2 }\\ =\frac {44\div 2}{ 2 \div 2}\\ =\frac {22}{ 1 }\\ =22 \)

If there is a step that you do not understand then just ask :)

as well as you can do factoring if you want 

because you didn't explain what do you want

you can factor it by 2

so 2(x+3*2xy-y)=2(x+6xy-y)

2x+3y*4x-2y

2x+12xy-2y

That is all the simplifying that you can do. 

2x-y-3z=7.           (1)

3x+y+z=10.         (2)

5x+y+2z=15.       (3)

Add equations (1) and (2) term by term:   5x - 2z = 17.       (4)

Add equations (1)and (3) term by term:    7x - z = 22.         (5)

From (5) we have:   z = 7x - 22.     (6)

Put this in (4):  5x - 2(7x-22) = 17

                        5x - 14x + 44 = 17

                        -9x = -27

                           x = 3.              (7)

Put (7) into (6):   z = -1.             (8)

Put (7) and (8) into (1):   6 - y + 3 = 7

                                       y = 2

.

Hi Qaz,

I would do these with matrices but is there are still different methods.

Which method have ou been taught. If i know it I will try to do it the way you have been taught ://