CPhill

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.

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

So, 20 * 15% = 3 times

( 0.17 - 3.5j ) ( 1.12 - 0.91j )

I'm assuming you want to "expand" this??

Start by multiplying .17 * 1.12 = .1904

Then mutiply (-3.5j) * (1.12) = -3.92j

Then multiply (.17) * (-.91j) =  -.1547j

Finally , multiply (-3.5j) * (-.91j) = 3.185j²

So putting this all together, we have

3.185j² -3.92j - .1547j + .1904 =

3.185j² - 4.0747 + .1904j

Instead of "Where's Waldo??"....we could play "Where's Melody???"

Excellent answer, chilledz3non......"thumbs-up" from me...

Note something.....x = -84/33 = -28/11.......we could also leave the answer in this form....some people like the "exact" answer, rather than a repeating decimal....I'm actually indifferent, but I might be in the minority......

Note that there are 7 parts of the 105 pounds....so each "part" must be 15 pounds

Since Bill gets 4 of these parts....... 4 * 15 =  60 pounds

Let's take the second function, first...we have

y(x) = (x^2) / (2x)....if we simplify this "rational function," we get

y(x) = (1/2)x        Note that this is just a linear function with a "hole" at x =0 (because we can't divide by zero in the original function)

This function is constantly increasing on the intervals where it is defined (-inf,0) U (0,inf), just as your derivative indicated

As to the first function, I don't know of any way to find the intervals of increase or decrease, except by graphing......here's a graph (I hope it displays correctly)

If we DID take the derivative, we'd find that this function has "minimums" at x = -1 and x = 1.

The intervals of decrease are (-inf, -1) and (0, 1)

The intervals of increase are (-1, 0) and (1, inf)

I think this is correct....maybe some other forum members know a way to find the intervals of increase and decrease with respect to the second function without graphing or calculus......