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Note there are 6 ways to choose the first person in line, 5 ways to choose the second person in line, etc. So, the product of all these is just

6 x 5 x 4 x 3 x 2 x 1 = 6! = 720 ways

The 6! is read as "six factorial."  It represents the product of all the positive integers less than or equal to 6.

A (regular)  hexagon is comprised of 6 equal equilateral triangles. The total area is given by:

6*(1/2)*(s²)(√(3)/2) =   (1/2)(s²)(3√(3))   where s is the side length....note, since the triangles are equilateral, the sides are all the same length.

And the area is 425 ft²

So we have:

425 = (1/2)(s²)(3√(3))      multiply both sides by 2  

850 = (s²)(3√(3))              divide both sides by (3√(3)) 

850/(3√(3))  = (s²)            take the square root of both sides

s ≈  12.79 cm

In simple terms, it's the reciprocal of a number such that, when we multiply the number by that reciprocal, we get 1.

For instance, the multiplicative inverse of 3 is 1/3 because  3 * (1/3)  = 1.

And the multiplicative inverse for a fraction a/b is b/a because (a/b)*(b/a) = 1

So, in general, the multiplicative inverse of a number, n, is just 1/n .

Note that 0 doesn't have a multiplicative inverse because 1/0 isn't defined.

[5(2) -1] * [ 3(2) + 1 ] =

[9] * [7] = 63

I agree.......

Fire those publishers !!!!

BTW.....67 is a prime number...the only two possible positive integers that could multiply to this are 1 and 67.

So, [f*g](x) = f(x) * g(x)  =

[2x + 5] * [x^2 - 3] =

(2x)*(x^2) = 2x^3

(5)*(x^2) = 5x^2

(2x) * (-3) = -6x

(5)*(-3) = -15

So, putting this all together, we have

2x^3 + 5x^2 - 6x - 15   = [f*g](x)

I don't see any phrases, but the effect is to triple the volume.

To see this, divide the "new" dimensions by the "old."

(7 * 12 * 9) / (7 * !2 * 3)     =    3 times as much volume !!!

I'm assuming you want a formula for the area.

It's just base x height......like this......

If 76 was your only previous grade, then your new average would be

(76 + 80 + 100) /3 = 85.3

However, if 76 represents an average of some other grades, then, this would be a hard one to answer.

Consider that your grades could have been 90, 76, and 62. This would be an average of 76.

When we add 80 and 100, the average is now (90 + 76 + 62 + 80 +100)/5 = 408/5 = 81.6

However, suppose your grades were 78 and 74. This is also an average of 76. Now add 80 and 100, and we get   (78 + 74 + 80 + 100)/4 = 332/4 = 83

So, in short, we would need more information about your previous grades to give a definitive answer in the second situation.

Old total = 76*38

New total = 76*38+100+80

New grade = (76*38+100+80)/40 = 76.7

9.81*10⁶

5X+23=25

    -23  -23

5x=2

/5  /5

you'll get your answer

/ means divide by

the 100 was a test and the 80 was a homework. tests are more than everything else. I had 38 grades in before.

9.81x104 is how you would

This would actually depend on how your teacher grades and how many grades were input before those two.

I say this because some teachers will have quizzes be worth more to your average then say... homework

You should - theoretically - win 15% of the time.

So, 20 * 15% = 3 times

Thank you Alan!

Yes, this is much clearer now. Thanks a lot man! 

Only 1 out, that's not too bad given my memory !

c is a constant which is just 1/2 in your case.  

However, let me do it with just your numbers:

f = (5x - 4)^1/2

Let z = 5x - 4     ...(1)     so 

f = z^1/2

Differentiate f with respect to z

df/dz = (1/2)*z^(1/2)-1 or df/dz = (1/2)*z^-1/2    ...(2)

We want df/dx not df/dz, so we use the chain rule df/dx = df/dz*dz/dx   ...(3)

We have df/dz, so we need dz/dx

From (1) above we find dz/dx = 5  ...(4)

Put (2) and (4) into (3) to get

df/dx = df/dz*dz/dx = (1/2)*z^-1/2*5 = 5/(2*z^1/2)

Replace z using (1) to get

df/dx = 5/(2*(5x-4)^1/2)

Is this any clearer?

I think Ramanujan's number is 1729 Bertie.  It's the smallest integer that can be expressed as a sum of two cubes in two different ways:

1729 = 1³ + 12³

1729 = 9³ + 10³

$${\mathtt{ExpectedValue}} = {\mathtt{0}}{\mathtt{\,\times\,}}{\mathtt{0.962\: \!3}}{\mathtt{\,\small\textbf+\,}}{\mathtt{3}}{\mathtt{\,\times\,}}{\mathtt{0.035\: \!7}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1\,000}}{\mathtt{\,\times\,}}{\mathtt{0.000\: \!036\: \!9}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1\,000\,000}}{\mathtt{\,\times\,}}{\mathtt{0.000\: \!000\: \!143}} \Rightarrow {\mathtt{ExpectedValue}} = {\mathtt{0.287}}$$

Expected value = 0.287 dollars = 28.7cents.  

To the nearest whole number of cents: Expected value = 29

Let x = log₇(4)

This means that

7^x = 4

Take the natural log of both sides (ln means log to base e)

ln(7^x) = ln(4)

By a property of logarithms (see the Formulary at the top of this page) we have

x*ln(7) = ln(4)

so x = ln(4)/ln(7) or, substituting for x from the first line above):

log₇(4) = ln(4)/ln(7)

This is a big help, thanks. However, I am still having difficulty understanding as to how and why I can get the "z" equation, for instance what happens to "c"?

$$$$\frac{-1-x^2}{(1+x*y)^2+(x-y)^2}$$\\
Expand the denominator $$(1+x*y)^2+(x-y)^2=1+2x*y+x^2*y^2+x^2-2x*y+y^2$$\\
Collect terms in the denominator
$$1+x^2+y^2+x^2*y^2 = 1+x^2+y^2*(1+x^2)=(1+x^2)*(1+y^2)$$\\
The original numerator can be written as $$-(1+x^2)$$\\
Putting the numerator and denominator together again we have
$$\frac{-(1+x^2)}{(1+x^2)*(1+y^2)}= \frac{-1}{1+y^2}$$