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what is 9,10 and 11:zach

How to solve z+6/z-9=11/-10

z = 79/20-sqrt(3841)/20 and  z = 79/20+sqrt(3841)/20

DEEZ NUTS

(1+sin a)(1-sin a)=cos^2a    expand the left side

1 - sin^'2a   = cos²a       add sin²a  to each side

1 = sin²a + cos²a       and this is always true

21=deez nuts

thx

Just divide 8,500/12=$708.33

the square root of 9999999999 is 100000

$84000(.075)  = $6300

thanks

$ [ 2500 + (.04)95000]  = $6300

three thousand eight hundred

hi sddffgf

Assuming that we have

[x-6] / [x-5]=[x-5] / [x-4]        ..... cross-multiply

[x - 6] [ x - 4]  = [x - 5] [x - 5]     simplify

x^2 - 10x + 24  = x^2 - 10x + 25        subtract  [ x^2 - 10x ]  from both sides

24  = 25    and this is impossible.....so.....no solutions exist........!!!!

x = { }        [the empty set]

-5+7=2

-5+7=2

2^5/6,  4^2/3,  8^1/2,  32^1/3,

2^5/6, 4^2/3, 8^1/2,32^1/3 

This isn't as hard as it seems if we can write them in a little different mannner.....

8^1/2      =   [2³]^1/2   = 2^3/2     =  2^9/6

4^2/3   =   [2²]^2/3  = 2^4/3    =  2^8/6

32^1/3  =  [2⁵]^1/3  = 2^5/3     = 2 ^10/6

2^5/6           =   [2]^5/6     =  2^5/6    

And  since     10/6 > 9/6 > 8/6  > 5/6

Then     2 ^10/6  > 2^9/6 > 2^8/6  > 2^5/6   

So....in ascending order, we have

2^5/6 ,  4^2/3,  8^1/2, 32^1/3

bye

Here's a good explanation :     http://www.purplemath.com/modules/quadform.htm

Remember that    a^-n = 1 / aⁿ    ..... so we have ......

(3 / 4)^-3   =  1 / (3 / 4)³   =   1 / ( 27 / 64)   =  [1 / 1] / [ 27 / 64]  =

[1 / 1 ] * [ 64 /  27 ]  =

64 / 27 

Notice another shortcut to our answer, here

(a / b)^-n =  (b /a)ⁿ    ......so.....

(3 / 4) ^-3    =  (4 /3)³ =  64 / 27

evaluate (3/4)^-3 ???=64/27=2 10/27

-5 + 7 = 2

Think of (or draw out) a number line if you're stuck on adding negative and positive numbers.

(72,000,000)(0.0002)  =

[ 7.2 x 10^7]  [ 2 x 10^-4] =

14.4 x 10^3  =

1.44 x 10^4