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whats the distance between -12 and -5

They are both negative so it is only 7 jumps on the number line,

also

distance = big one - little one = -5- - 12 = -5+12 = 7

What's 2.5! ?

\(\begin{array}{rcll} 2.5! &=& \Gamma(3.5) \\ &=& \Gamma(2.5+1) \\ &=& 2.5\cdot\Gamma(2.5) \\ &=& 2.5\cdot\Gamma(1.5+1) \\ &=& 2.5\cdot 1.5\cdot\Gamma(1.5) \\ &=& 2.5\cdot 1.5\cdot\Gamma(0.5+1) \\ &=& 2.5\cdot 1.5\cdot 0.5\cdot\Gamma(0.5) \qquad \Gamma(0.5) = \sqrt{\pi}\\ &=& 2.5\cdot 1.5\cdot 0.5\cdot\sqrt{\pi}\\\\ &=& 1.875\cdot\sqrt{\pi}\\ &=& 1.875\cdot 1.77245385091\\ &=& 3.32335097045\\ \mathbf{2.5!} & \mathbf{=} & \mathbf{3.32335097045} \end{array}\)

{nl} Guest {nl} 11 mins ago     

What's 2.5! ?  

http://www.wolframalpha.com/input/?i=2.5%21

A 3.32038

B 3.32335

What's 0 ÷ 0 ?     undefined

A ∞

B indefinite

What's 0^0 ?     undefined

A 1

B indefinite    

What's 2.5! ? The answer is: B

A 3.32038

B 3.32335

What's 0 ÷ 0 ? The answer is: B. not "indefinite", but "undefined"

A ∞

B indefinite

What's 0^0 ? The answer is: A

A 1

B indefinite

How do I find the geometric mean when there is an even number of terms (eg. 7 and 8 7 5, there are 4 terms)

\(\begin{array}{rcll} \text{geometric mean } &=& \sqrt[n]{ x_1\cdot x_2\cdot x_3\cdot x_4 } \qquad x_1 = 7\qquad x_2 = 8\qquad x_3 = 7\qquad x_4 = 5\\ &=& \sqrt[4]{ 7\cdot 8\cdot 7\cdot 5 } \\ &=& \sqrt[4]{ 1960 } \\ \mathbf{ \text{geometric mean } } & \mathbf{=} & \mathbf{ 6.65371229032 } \end{array}\\\)

11/12 - 1/6

Thanks guest but I think you made a small mistake  :(

11/12 - 2/12 = 9/12 = 3/4

what do I find cot0=12.3

\(cot\theta = 12.3\\ tan\theta = 1\div 12.3\\ atan(1/12.3) = 4.648 \;degrees\\ \theta = 4.648 +180n\quad degrees \qquad n\in Z \)

Thanks guest

can (x^4) + (5x^2) +4 be simplified to (x^2) + (5x) + 2

Here is another way of showing you.

Let x=3

LHS= 81+45+4 = 130

RHS = 9 + 15 +2 = 36

They are not equal :(

-3=-9-5x

5x=-9+3

5x=-6 divide both sides by 5

x=-6/5

x=-1 1/5

Do you mean region or like the area of a panda? Area like a square.

No offense, but is this a joke?

No, it cannot. See the result below.

Simplify the following:
x^4+5 x^2+4

x^4+5 x^2+4 = (x^2)^2+5 x^2+4:
(x^2)^2+5 x^2+4

The factors of 4 that sum to 5 are 4 and 1. So, (x^2)^2+5 x^2+4 = (x^2+4) (x^2+1):
Answer: |  (x^2+4) (x^2+1)
 

 cool story bro

Yum?

v/5 -6=3 What's the number for v?

Start by adding 6 to both sides and then tell me what you have :/

0.05+1.15(rm-rRF)=0.12

1.15(rm-rRF)=0.12-0.05

1.15(rm-rRF)=0.07

r m - r R F=0.07/1.15

r m - r R F= 7 / 115

You cannot solve thies equation because there are 3 unknowns.....

So I do not know what you want us to do with it  ????

28

Beijing

idk bro

(1/6)2= 2/12

11/12 - 2/12= -9/12= -3/4

-3/4 is the reducuded form

Yee

1/14 that's what it is.

17 jumps

Hint:

factors are things that are multiplied together

and

terms are things that are added (or subtracted)  together.

Hi  MFL   :))

(xy^2)^3 X (x^-2)

\((xy^{2})^3 \times (x^{-2})\\ =(x^{1}y^{2})^3 \times (x^{-2})\\ =x^{1*3}y^{2*3} \times (x^{-2})\\ =x^{3}y^{6} \times x^{-2}\\ =x^{3+-2}y^{6} \\ =x^{1}y^{6} \\ =xy^{6} \\~\\ or\\~\\ (xy^{2})^3 \times (x^{-2})\\ =\frac{xy^2*xy^2*xy^2}{x^2}\\ etc\)