CPhill

So what do you want to do with this??

9[g-(3-g)+2]=-1+7[3g-(4+g)]

Let's work through this one....since it's a little tricky....

Working within the brackets, first, we have

9[g- 3 + g + 2] = -1 + 7[3g - 4 - g ]

9[2g - 1 ] = -1 + 7[2g - 4 ]

Distribute the 9 and the 7 across the terms in the brackets

18g - 9 = -1 + 14g - 28      simplify

18g - 9 = -29 + 14g

Add 29 to both sides and subtract 18g from both sides

20 = -4g

Divide both sides by -4

-20/4 = -5 = g

Assuming a 30-day month, we have

9855/30 = 328.5 months

Because........ 7  "8"  9........

Also......can February March???.....no......but, April May..............

We can set this up as a proportion....

1137.50 / 50 =  x / 81      where x is the cost for 81 games

multiply both side by 81

(81)(1137.50)/50 = x = $1842.75

To figure the per game cost, divide $1137.50 by 50 =  $22.75

So, to find our how many games you could attend for $728, just divide this by $22.75 = 32 games

(2b^4 c^-2)^5 (3b^-3 c^-4)^-2

Applying a rule of exponents that (a^m)ⁿ = a^m*n......we have.........

(2)^5 *( b^4)^5* (c^-2)^5 * (3)^-2 * (b^-3)^-2 * (c^-4)^-2

32 * b^20 * c^-10 * (1/9)  * b^6 * c^8

And using another rule that a^m * aⁿ  = a^m+n  ...... we have.......

(32)(1/9)(b^20)(b^6)(c^-10)(c^8) =

(32/9)*(b^26)(c^-2) =               ............... remember c^-2 = 1/c^2............

[32b^26 ] / [9c^2] 

First, subtract the x values and square this result...

So we have   -1- 4 = -5     and (-5)^2 = 25

Do the same thing for the y values

4 - 1 = 3     and    (3)^2 = 9

Add the results

25 + 9 = 34

Take the square root = √34

And that's the distance between W and T

We can only add (or subtract) matrices of the same dimension. We just add entries having the same position(s) in the matrices.

For instance

[3 4 ]   +   [ 5 6 ]    =  [ 8 10 ]

This idea can be extended to any number of matices of like dimension.

LOL!!!.......

4x * x^2 =

4x^3