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Sorry, Asinus.....we can't just "cancel" exponents in that fashion.......

Your answer ends up being =  99999/100000 =  .99999

However..........

99999^100000 / 100000^100000 =

[99999/10000] * [ 99999/100000] * [99999/100000].........100,000 times

[99999/100000]^100000  ≈  .3679

(6x^2)^4=1,296x^8

Hello Guest!

Whats (99,999^100,000)/(100,000^100,000)?

99999^(10⁵) / 10⁵^(10⁵⁾     [ * (10^-5) / (10^-5)

= (10⁵ - 1) / 10⁵

Greetings asinus :- )    !

(99999^100000)(100000^100000) = (infinity)*(infinity)

Highly disturbed! Seek professional help.

How do I solve an inequality of -2.51 < or = to -2h/4

 -2.51 < or = to -2h/4

-2.51<=-2h/4

-2.51<=-h/2 Reverse

-h/2>=-2.51 multiply by -1

h/2>=2.51 multiply by 2

h  ≤  5.02

whatever......Zzzzzz  Zzzzzz .....

Do you have any MATH questions?

As always, Algebra wizard!!.

Note :

(a + 5)^2  - (a - 5)^2   factors as a difference of squares   =

[ (a + 5)  - (a - 5)]  [(a + 5) + (a - 5) ]  =

[ 10] [ 2a]  =

20a

Note that :

 sin^2(2x)   = [1 - cos(4x)]/ 2     ......so we have.....

[ 1 - cos4(x + h)] / 2] * [ 1/h]    -    [1 - cos(4x)] / 2 * [ 1/h ] =

[1 - cos(4x + cos4h)] / 2h   - [1 - cos(4x)] 2h  =

[cos(4x)  -  cos(4x + 4h) ] / 2h   =

[ cos(4x)  - [cos4xcos4h  - sin4xsin4h] ] / 2h  =

[ cos4x ][1 - cos4h] / 2h    +  sin4xsin4h/ 2h        [multiply top/bottom of both fractions by 2]

[2cos4x] [(1 - cos4h) / 4h ] +   [2sin4x] [sin4h/ 4h ]       let h →  0

[2cosx * 0]  +  [2sin4x * 1 ]  =

2sin4x  =

2sin(2x + 2x)  =

2[ sin2xcos2x + sin2xcos2x]  =

2 [ 2 * sin2xcos2x]  =

4[sin2x] [cos2x]

-------------------------------------------------------------------------

Note that :   [sin(2x)]^2  =  [sin2x ] * [sin2x]     (1)

And using the Product Rule, the derivative  of (1)  =

[2cos2x] [sin2x] + [sin2x][ 2 cos2x]  =

[2cos2x] [ sin2x + sin2x]  =

[2cos2x] [2sin2x] =

4[sin2x][cos2x]

Simplify the following:

(a+5)^2-(a-5)^2

(a+5) (a+5) = (a) (a) + (a) (5) + (5) (a) + (5) (5) = a^2+5 a+5 a+25 = a^2+10 a+25:

a^2+10 a+25-(a-5)^2

(a-5) (a-5) = (a) (a) + (a) (-5) + (-5) (a) + (-5) (-5) = a^2-5 a-5 a+25 = a^2-10 a+25:

25+10 a+a^2-a^2-10 a+25

-(a^2-10 a+25) = -a^2+10 a-25:

25+10 a+a^2+-a^2+10 a-25

Grouping like terms, 25+10 a+a^2-25+10 a-a^2 = (10 a+10 a)+(a^2-a^2)+(25-25):

(10 a+10 a)+(a^2-a^2)+(25-25)

10 a+10 a = 20 a:

20 a+(a^2-a^2)+(25-25)

a^2-a^2 = 0:

20 a+(25-25)

25-25 = 0:

Answer: |20a     IT IS TRUE!!.

This can be a kite ...

True, because:

Simplify the following:

(a+5)^2-(a-5)^2

(a+5) (a+5) = (a) (a) + (a) (5) + (5) (a) + (5) (5) = a^2+5 a+5 a+25 = a^2+10 a+25:

a^2+10 a+25-(a-5)^2

(a-5) (a-5) = (a) (a) + (a) (-5) + (-5) (a) + (-5) (-5) = a^2-5 a-5 a+25 = a^2-10 a+25:

25+10 a+a^2-a^2-10 a+25

-(a^2-10 a+25) = -a^2+10 a-25:

25+10 a+a^2+-a^2+10 a-25

Grouping like terms, 25+10 a+a^2-25+10 a-a^2 = (10 a+10 a)+(a^2-a^2)+(25-25):

(10 a+10 a)+(a^2-a^2)+(25-25)

10 a+10 a = 20 a:

20 a+(a^2-a^2)+(25-25)

a^2-a^2 = 0:

20 a+(25-25)

25-25 = 0:

Answer: |20a

(a + 5)²   =  (a + 5)(a + 5)  =  a(a) + (a)(5) + (5)(a) + (5)(5)  =  a² + 5a + 5a + 25  =  a² + 10a + 25

(a - 5)²   =  (a - 5)(a - 5)  =  a(a) + (a)(-5) + (-5)(a) + (-5)(-5)  =  a² - 5a - 5a + 25  =  a² - 10a + 25

(a +5)² - (a - 5)²  = [a² + 10a + 25] - [a² - 10a + 25]  =  a² + 10a + 25 - a² + 10a - 25  =  20a

Yes; it's correct!

Always true!.

\(6\frac{3}{25}\)

how does 1816 have 6 circles??

16!!!!

Just use ANY calculator and divide the TOP number by the BOTTOM number. Example: 1/8= Enter 1 first then / followed by 8 and then = key. The calculator should give you .125

yea :)

2.Because 9999 have 4 circles,8888 have 8 circles,1212 have 0 circles,1816 have 6 circles.So 1919=2,because 1919 have in the numbers.

I agree with Bree...

You know, we can tell what your talking about unless you say who...

Youre cristi?

If the recipe is doubled....each ingredient amount is doubled....so

2 * 3.25 =    6.5  cups of flour