CPhill

coockies are packed  in 2 kinds of boxes : big and small. the number of coockies in a small box is half of the number of coockies in the big box. how many coockies there are in every box if 3 big boxes and 5 small boxes have 165 coockies together 

OK.....let's call the number of cookies in the small box "x."  And since the number of cookies in the big box is twice as many, let's call that "2x."

So ....  the number of small boxes times the number of cookies in each plus the number of big boxes times the number of cookies in each equals 165.  Mathematically, we have

5(x) + 3(2x) = 165   So we have

11x = 165  Divide by 11  and we get

x = 15    So........that's the number of cookies in the small box

And since the big boxes hold twice as many, there must be 30 in each one of them. 

And that's it 

What is 2 to the 2048th power?

2^(2^11)

In other words........BIG   !!!

This number is too large for the on-site calculator to evaluate (BTW.....it's not "infinity", despite what the calculator claims), but if you're interested, you can do that here:

(1)   286 centigrams = ??? Milligrams

(2)    1 centigram = 10 milligrams

Multiply both sides of (2) by 286 and you will have your answer  

Re: how do you find percentage profit

Well....that depends on what we're taking a percentage of.  Generally, we're interested in finding either:

1) Percentage of profit with repect to cost (usually called the "mark-up")    .....  OR  ......

2) Percentage of profit with respect to sales (usually called the "margin")

Let me give you a simple example to demonstrate the difference. Suppose we bought something for $8 (our cost) and sold it for $10 (our sales price).  Then, we made a profit of $2.

The percentage of profit with respect to cost = (profit)/((cost) = (2)/(8) = (1)/(4) = 25%

The percentage of profit with respect to sales = (profit)/((sales price) = (2)/(10) = (1)/(5) = 20%

(Of course, most businesses usually prefer the former measure. It makes them seem more "profitable.")

Re: (2a-b) (3a-2b) resultado

We can evaluate this in several ways. One way is to "disrtibute" (multiply) the first expression over both terms in the second. This gives us

[(2a-b)*(3a)]  + [(2a-b)*(-2b)]    which gives us

[6a² - 3ab] + [ -4ab + 2b²]  =

6a² - 3ab - 4ab + 2b²   and combining "like" terms, we have

6a² - 7ab + 2b²

We also could have distributed the second expression over the first.....the answer would have been the same!!!

Let's take another look at this one.... (If you want to)

Let's assume that we DO use a "weighted average" system to calculate the average

So we have something like this

[63x + 83y + 100z] / [x + y + z] = 70     where   x, y, z   are the "weights"

Multiplying by [x + y + z] on both sides gives us

63x + 83y + 100z = 70 ( x + y + z)   And after a little manipulation, we have

-7(x) + 13(y) + 30(z) = 0     Now, solving "for z," we get

z = (7x - 13y)/30

Note that, if I let x = 2 and y = 1, I get "1"  as a numerator and "z" = (1/30).......but, I need "z" to be an integer. Therefore, if I just mutiply everything in the nunerator by 30, then we get

z = 30 [7(2) - 13(1)]/ 30 = [7(60) -13(30)]/30 = [(420-390)/30] = (30/30) = 1

Then our "solution set" for (x, y, z) = (60, 30, 1)

And this is just a "3 space" vector. And any positive integer multiple of this vector could be could be a "solution" to the weighted values of x, y, and z.

So, the general soltion to this problem - if "70" is indeed the "answer" - is just given by: n(x, y, z) .....  where (x, y, z) = (60, 30, 1)  and n is a positive integer.

Re: 15% + 1/15 + 0.015        Change everything to a fraction  =

15/100 + 1/15 + 15/1000       Reduce where possible     =

3/20 +1/15 + 3/200               Get a common denominator    =

90/600 + 40/600 + 9/600       Add the numerators    =

139/600

V = LWH₂ - LWH₁=  (3.5)(2.5)(2.5) - (2.5)(1.5)(1.5)

= 21.875 - 5.625 = 16.25 in³ =   None of the above

Cornbread..........

Bertie is correct.......this would have to be a weighted average problem if "70" is correct.