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I believe I remember how to do this, but.......someone will have to check me on it !!!.....

In any   m x n    grid, there are   C ( m + n , m) paths to the right and upward, assuming that we start at the lower left vertex of the grid and end up at the top right vertex....  or, alternatively,  C ( m + n, n)  paths......these should work out the same

So.....the first grid is a 2 x 3 grid, which means there are C ( 2 + 3, 2)   = C ( 5, 2)  = 10 paths from the lower left starting point and moving to the right and up to the intersection of the two grids .......this would be the same as C ( 2 + 3, 3)  = C (5, 3)  = 10

The next grid is a  4 x 3 grid.....which implies that there are  C(4 + 3 , 3)  = C(7, 3)   = 35 paths which we can take [ moving in the specified manner ] from the intersection of the two grids to the top right vertex of the upper grid

So....the total paths become  C(5,2) * C (7,3)   =  10 * 35   = 350 distinct paths moving to the right and upward from the lower left vertex of the bottom grid to the upper left vertex of the top grid.

Here's a good website that presents the logic behind this........http://betterexplained.com/articles/navigate-a-grid-using-combinations-and-permutations/

A=lw

(Area=length*width)------------------- so if 14 is both the length and the width it is 14^2

or 196

If f(x) = x^2 and g(x) = 3x – 5, find (f ° g)(x)

The notation is a fancy way of saying that we're "plugging"  the function "g"  into the function "f"  and then evaluating what we have......so.....

f(x)  = [x]^2       and we're substituting  3x - 5    for  x    ......so we have.....

f( 3x - 5)  =     [3x - 5]^2   =  [3x -5] [3x -5]  =   9x^2 - 30x + 25

This is often confusing.... [I know it was for me].....but think about this.....if I wanted to know what f(2) was, I would evaluate

f(2)  =   (2)^2   = 4

Here......we're just doing the same thing with a function, rather than a number..... 

It is always meter unless it is plural... so just 2/3 meter will do if you are to lazy to write of a.

In parenthesis (x/y)

for example.. ¼ = (1/4)

lol...

I gave the "Guest" five stars and you gave him four.....I guess that will trump the five you gave me  and the five I gave you, everytime......LOL!!!!!

However....if we add ours together, we beat him by one star.....!!!!!

21\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\Gamma \leftarrow\)

One pound is about .45 kg

I like how this answer was rated better than CPhill's even though he had a better, more explained answer!!!

1 Tonne=1,000 kilograms

474/1,000 =.474 tonnes

the average of 4 consecutive even integers is 17. what are the integers and what is the process for solving this problem?

n +(n+2) + (n+4) + (n+6)=17 x 4

4n+ 12 = 68 subtract 12 from both sides

4n=56 divide both sides by 4

n=56/4

n=14. So the numbers are:14, 16, 18, 20

Let 2N be the first even number

2N + 2 the next

2N + 4 the next

2N + 6  the last   .......so we have

Average  =   [sum of the numbers ] / [ how many numbers we have]

17  = [ 2N + ( 2N + 2) + ( 2N + 4) + (2N + 6)] /  4      simplify

17  =  [ 8N + 12]  / 4    multiply both sides by 4

68  = 8N + 12     subtract 12 from both sides

56  = 8N    divide both sides by 8

7 = N   

So the first even number =  2(N)  = 2(7)  = 14

Which means that the next three are  16, 18 and 20

14,16,18,and 20. These even numbers are conescutive and average to 17. All you have to do is look at the closest 2 numbers higher and 2 numbers lower than 17!

the mean is the average; the sum of all the numbersin a set divided by however many numbers there are.

i.e if a set were 1, 2, and 3, you would add them to get 6, then divide by the amount of numbers, getting 2 (6/3=2)

Mmmm....this is an interesting one !!!

The volume of the snow needed for the snowman is given by :

(4/3)pi [ 5^3 + 3^3 + 2^3 ]   = about  670 ft^3   of snow

The volume of snow in the yard is given by :

[  50 ft * 25 ft * (2/3) ft ] =  about 833.33 ft^3   of snow

Looks like there's more than enough......!!!!

Sorry, Max.....I believe you made an error....look at your second step....you have.....

l 2x + 6  l  + 16 = 0       (1)

If we claim that the answer is -11, then we have

l 2(-11) + 6 l + 16 = 0

l -22 + 6 l  + 16  = 0

l -16 l + 16   = 0

16 + 16   = 0

32  = 0     .......which is clearly false

We have no solution because the left side of (1)   will always be >  0

cool beans

PAR - TAY!!!!!!!!

Let half-life =t

47/48 = 2^(-30/t)

47/48 = 2^(-30/t) is equivalent to 2^(-30/t) = 47/48:
2^(-30/t) = 47/48

Take reciporicals of both sides:
2^(30/t) = 48/47

Take the logarithm base 2 of both sides:
30/t = (log(48/47))/(log(2))

Take the reciprocal of both sides:
t/30 = (log(2))/(log(48/47))

Multiply both sides by 30:
Answer: |  t = (30 log(2))/(log(48/47))=987.70 minutes-half-life of the element.

6(3-x)=3x    distribute the 6 across the terms in the parenteses on the left

18 - 6x   = 3x         add 6x to both sides

18  = 9x               divide both sides by 9

2  = x

y = x^2  - 1       add 1 to both sides

y + 1  = x^2       take the positive/negative square root of both sides

± √ [y - 1 ]  =    x

No such thing as -288 degrees C, it only goes down to -273 absolute zero

I know you