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The corect spelling is "you're."

Yes i am smart

a^2+b^2=c^2

Si, es verdad.

what if a man working 15 years , his company have loan over $14 trillion dollars and have to paid interest at 3.9% percent on every monthly ,he forget to paid that interest on all those 15 years ,after that all total money of loan and interest rate money ,add together , how much is that 15 years of loan and interest rate 's money?

Using the Pythagorean Triple, 5, 12, 13, y = 12.....however, in the 4th quadrant, y would be negative.....so we have

csc Θ = r / y   =  13/-12 = -13/12   and

tan Θ = y / x = -12 / 5

I agree, my answer assumed each ball was separately identifiable, whatever the colour.

Here's my take on this one......

We're just counting sets of things, and order witin sets doesn't matter, so this is a combination problem, rather than a permute problem.

The number of possible sets made by choosing any 3 b***s from 9 =C(9,3) = 84

And the number of possible sets made by choosing no black b***s is just C(6,3) = 20

So, the number of possible sets made by choosing at least one black ball = 84 - 60 = 64

{I'm not great at probability, so I really don't know if i'm correct, ot not....!!!}

no, come on you can do it

They made bikes?

0 either way

[32^(1/5)]^5  =

[2]^5  =

32   !!!

X-1      X+2

---- +  -----  = 7

  3          6

Getting a common denominator we have

(2(x -1) + x+2) / 6 = 7    multiply both sides by 6

2(x-1) + x + 2  = 42  simplify

2x - 2 + x + 2 = 42

3x = 42     divide both sides by 3

x = 14

Sorry Melody, I overlooked your question until now.

Yes, I remember the original problem. I remember too thinking this was an odd question to become a snark. After some research I found this turned into a snark question partly because of a computer program malfunction. It also became a fascinating research project on crowed behavior.

The original question was

“How much tax would be on 113.98 at 9%”

This is the original URL

$$\Text {Solution}\\\

\Text {Total number of ways to choose 3 b***s from 9}\\
\Text {nCr(9,3) = 84 (where the order is irrelevant).}\\

\Text {Six non-black b***s and three black b***s.} \\


\text {nCr(6,2)*ncr(3,1) = 45}\\
\text {Number of ways to select 2 of 6 non-black b***s and select 1 of 3 black b***s.} \\

\text nCr(6,1)*ncr(3,2) = 18 } \\
\hspace{5ex}\; \text {Number of ways to select 1 of 6 non-black b***s and select 2 of 3 black b***s.}\\

\text {nCr(6,0)*ncr(3,3) =1 } \\
\text {Number of ways to select 0 of 6 non-black b***s and select 3 of 3 black b***s.}$$

(45+18+1) = 64 Total number of ways to choose 1, 2 or 3 black b***s of 3 b***s from 9 b***s where 3 are black and 6 are non-black.

The question asks for Permutation, but this is really a Commutation problem not a Permutation problem. The Binomial distribution solution should apply.

If these b***s were unique such as with numbers along with the colors, then Alan’s solution would apply. This means the drawing order of the black b***s along with the order of the non-black b***s is unique in each instance, and there would be many more arrangements.

~DQD~

cos(sin (pi)) = cos(0)  = 1

And

tangent(1) = about 1.557     assuming radian measure

y = 8 is a horizontal line at, well, y =8

And since no single vertical line "cuts" this line in more than one place, it is a function......(this is known as the vertical line test)

sin(30)/1 = 6/x     cross-multiply

xsin 30 = 6          divide both sides by sin30

x = 6 /(sin 30)     and sin 30 = 1/2  ... so...

x = 6/(1/2)

x = 12

Call x the number of tickets they gave to employees. Then, 5x is the number they gave to the listeners.

Then

x + 5x = 294

6x = 294      divide both sides by 6

x = 49

And that's the number they gave to the employees....

Mod(17,5)

$${\mathtt{17}} \,{Mod}\, {\mathtt{5}} = {\mathtt{2}}$$

on the home screen just click on the button that says 2nd 

log72 (x2 - x) = 1

In exponential form, we have

72^1 = x^2 - x  rearrange

x^2 - x - 72 = 0     factor

(x - 9) (x + 8) = 0   and setting each factor to 0, we have that x = 9 and x = -8

And both solutions "work".....

4,999,999,980 idk