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Hello folks: Regarding the above example, I entered it in WolframAlpha engine in this manner and it gave the correct answer:

=-250*[(1-(1+.5/100)^-360)/(.5/100)]+F*(1+(.5/100))^-360-10000=0, solve for F

Just copy and paste the above and plug in WolframAlpha engine and you will see the result. Thanks and good luck.

1. 1,470

2. 1,476

3. 928,0

Example? Sure:

I want to save some money for my retirement. I already have $10,000 and I wish to invest it in a fund. In addition to this, I plan to deposit a monthly payment of $250 for the next 30 years. Will assume the fund pays interest at 6% compounded monthly. How much money will I have saved 30 years hence? In other words, what is the FV of my $10,000 and 30 X 12=360 payments of $250 each going to be?

Now, if you enter correctly all these values in the formula, you should get a FV of $311,354.51.i,e.,

PV=$10,000

P=$250

i=6/12=.5

N=30 X 12=360

FV=?

P.S. If you have somebody who is good at programming, can easily program this equation into his/her computer and test it for bugs, and if everything is working fine, then you can distribute the program to anybody who wants it. Best of luck.

Thankyou but you need to give examples of how this could be used :)

1.4

2/3 OF 27 IS EQUAL TO 6/5 OF WHAT NUMBER.?

2/3 X 27=N X 6/5

N=(2/3 X 27)/ 6/5=54/3 X5/6=270/18

N=15

Hallo

Кроме того один заполняет сначала 5-литровый ковш (1-ый из шага), льется от содержания так много в 3-литровом ковше, к этому полон (2-ой из шага), освобождает впоследствии 3-литровый ковш (3-ий из шага) и льет остающиеся два литра 5-литрового ковша в 3-литровом ковше (4-ый из шага).

Теперь один заполняет еще раз 5-литровый ковш (5-ый из шага) и льется этого снова так много в 3-литровом ковше, к этому полон (6-ой из шага). Поскольку 3-литровый ковш уже содержал два литра воды, только литр соответствуют в нем. Это факт, что в 5-литровом ковше все еще четыре литра содержатся - задача почти решена. Поскольку, однако, только (точно) четыре литра должны быть выкопаны от моря (озеро), следует наклонять наконец все еще содержание 3 литров ковша назад в море (озеро) (7-ой из)

2/3 of 27 = 18

So

18 = (6/5)N    multiply both sides by 5/6

18(5)/6 = (18/6) x 5   =  3 x 5   = 15  = your number

Solve for z:
8/z-7 = 9

Bring 8/z-7 together using the common denominator z:
(8-7 z)/z = 9

Multiply both sides by z:
8-7 z = 9 z

Subtract 9 z+8 from both sides:
-16 z = -8

Divide both sides by -16:
Answer: | 
| z = 1/2

Please help to solve: 1 barrel up to 3 liters of water
2 barrel up to 5 liters of water
in the third barrel of a maximum of 20 liters of water (in the second barrel of water is 20:20. In other barrels of water present)
you need to make a second barrel so it was 4 liters (maximum can be filled with barrels full)

Thanks, Mr. Genius......I let you keep your brownie......your answer was far more difficult to write out than mine was !!!....besides.....I knew how to write it out, but I let WolframAlpha do the 'heavy lifting".....hey, I'm lazy  !!!!

Thanks MrGenius for anticipating my answer (I recognize that anybody is allowed to give cookies to any one...)...

#39: for the graph of the funtion y=1/x^2 a tangent formed through (a,1/a^2)

A. Express using "a" the equation of the tangent

B. Express using "a" the unions with the coordinate systems (x=0),(y=0)

C. find "a" for whom the tangent creates "Isosceles triangle" with the coordinate systems

A. dy/dx (1/x^2) =  -2/x^3

Equation of the tangent at "a"

y - 1/ a^2 = (-2/a^3)(x - a)

y  - 1/a^2  = -2x/a^3 + 2/a^2

y =  -2x/a^3 + 3/a^2

y = [3a - 2x] / a^3

B.    y intercept →   x = 0    →  y = 3/a^2   →  (0, 3/a^2 )

       x intercept → y = 0  →  x = 3a/2   →  ( 3a/2 , 0 )

C.  Notice that many values of  'a' will produce isosceles triangles.......for instance......it we specify that the length of both tangent lines between the x and y axis = 5, we have

9/a^4 + 9a^2/4 = 25

9 + (9/4)a^6  = 25a^4

(9/4)a^6 - 25a^4 + 9  = 0

a≈ .78575  and a ≈ -.78575

Look at the graph, here......https://www.desmos.com/calculator/0twc2fqneu

Notice that two sides of the triangle formed by the points of the intersection of the tangent lines with both the x and y axis have lengths  ≈ 5

What an excellent answer. :)

Thanks answering guest. :))

or aleph null plus aleph x

Exactly.  Thanks CPhill for having the courage to rewrite the whole number (You deserved a brownie for this)

This was the correct answer in the American short scale. In the old British long scale, it would have been written like this :

«Thirty-four quadriliard, two hundred and eighty-six quadrillion, two hundred and twenty-seven trilliard, five hundred and seventy-four trillion, three hundred and eighty-seven billiard, fourty-five billion, four hundred and eighty-one milliard, six hundred and ninety-one million, three hundred and eighty-six thousand eight hundred and seventy-five».

Solve for c:
(2 (c-18))/3 = 7

Multiply both sides of (2 (c-18))/3 = 7 by 3/2:
((3×2 (c-18))/(2))/(3) = 3/2×7

3/2×7 = (3×7)/2:
(3×2 (c-18))/(2×3) = (3×7)/2

(3×2 (c-18))/(2×3) = (2×3)/(2×3)×(c-18) = c-18:
c-18 = (3×7)/2

3×7  =  21:
Answer: | 
| c-18 = 21/2

c=28 1/2

Bolt is an alloy of iron, carbon and other elements. But you can determine it yourself by getting yourself a blot; weigh it accurately in grams. In your kitchen, you should have a measuring cup graded in ml. Fill it with water to an accurate level, then immerse the bolt in it and see how much water it displaces. That would be the volume of your bolt. Then divide the weight by the volume, which will give you the density of your bolt. it should be somewhere between 7-8 g/ml(cubic cm).

\(\mbox{This problem is an example of a binomial distribution. here }n=6, p=0.73 \\ \\ \\ P[\mbox{k of 6 like carrots}]=\begin{pmatrix}n \\ k \end{pmatrix}p^k (1-p)^{n-k} \\ \\ \\ \mbox{The probability that at least 5 like carrots is }P[5]+P[6] = \\ \\ \\ \begin{pmatrix}6 \\ 5\end{pmatrix}(0.73)^5(0.27)^1 + \begin{pmatrix}6 \\ 6\end{pmatrix}(0.73)^6(0.27)^0 = \\ \\ \\ (6)(0.2073)(0.27) + (1)(0.1513)(1) = 0.487126\)