All Answers 358132 Answers

ayyyy my boy, lol jk. kinda bad but good at the same time

351 possible patterns. 

awsome you?? 

Sample mean  =  [ sum of the data values ] / number of values.....so we have

[ 8 + 8 + 8.5 +  8 + 9 + 8 + 9 + 9 + 8.5 + 8 ]  /  10   =

8.4

Take the square root of 555:

Sqrt(555) = 23 - perfect squares under 555

Cube root(555) = 555^1/3 =8 - perfect cubes under 555.

Note: 1 and 64 are shared as both perfect squares and perfect cubes.

Notice that r is the reciprocal of the golden ratio.

It always looks easy in retrospect! I looked at it in other, less productive ways first before I came up with the above.

Not positive

NaCl is an IONIC compound and readily dissociates in water  But stable as the solid 'salt'

H2O is a covalent polar compound but is relatively stable

Argon is a noble gas....noble gases (neon, krypton etc)  are very UN-reactive because they are stable

Clorine atoms are very reactive...chlorine is almost never found in a pure state, but rather as a compound with other elements

    So I will say Chlorine atom is the least stable config.

Thanks Alan, it always looks easy when someone else shows you how :)

It was sort of like that other circle one that you solved the other day. :) 

I suppose I mainly wanted to graph it but I know my graph is not quite right anyway.  I saw my error tonight.

Maybe I will fix it ...

Here's my take on this:

I guess what is wanted is that a = -1, b = 5 and c = 2.  However, this is a bit silly, as r can be expressed in an infinite number of ways simply by multiplying numerator and denominator by any positive number, p.  Then we would have a = -p, b = 5p² and c = 2p.

Thank you so much!

\(P[\text{win 1 game}] = \dfrac 1 3 \dfrac 1 2 = \dfrac 1 6\\ P[\text{win before 4th turn}] = P[\text{win game 1}]+P[\text{lose 1, win 2}]+P[\text{lose 1,2, win 3}]\)

\(P[\text{win game 1}] = \dfrac 1 6\\ P[\text{lose 1, win 2}] = \dfrac 5 6 \dfrac 1 6 = \dfrac{5}{36}\\ P[\text{lose 1,2, win 3}] = \left(\dfrac 5 6\right)^2 \dfrac 1 6 = \dfrac{25}{216}\\ P[\text{win before 4th turn}] = \dfrac 1 6 + \dfrac{5}{36} + \dfrac{25}{216} = \dfrac{91}{216}\)

x+9>3x

9>2x

x<4.5

x+9>x+4  always true

3x+x+4>x+9

3x>5

x>1.6

so

1.6

m=1.6

n=4.5

You can finish it.

Let the point where the top of the circle meets the base of the cone =    A

Let the radius of the cone = AB = 12

Let the center of the sphere = C

Let the radius of the sphere meet the side of the cone at D

Let the bottom "tip" of the cone = E

So......CD  meets the side of the cone at a right angle

And  BA  =  BD  = 12   [  tangents  drawn from the same point ]

So....Triangle CDE is a right triangle

The slant height of the cone, BE =   √ [ 24^2 + 12^2]  = √720 =  √ [ 144 * 5]  = 12√5

So DE  = BE - BD =    12√5 - 12

And  CE =  24 - r

And CD = r

So.....by the Pythagorean Theorem, we have that

CE^2 = CD^2 + DE^2

(24 - r)^2  = r^2 + (12√5 - 12)^2       simplify

r^2 - 48r + 576  = r^2 + 144(√5 - 1)^2

-48r + 576   =  144 (5 - 2√5 + 1)

-48r + 576   =  144(6 - 2√5)

-48r + 576  = 864 - 288√5

48r  =  288√5 - 288            

48r =  288 (√5 - 1)

r = 6 (√5 - 1)   =  6√5 - 6

So     a  +  c   =    6 + 5  =   11

Let the digits be x and y

Let the original number = 10x + y

So

x + y =  6        ⇒    y =  6 - x    (1)

And we know that

3 (10x + y)  + 6   =  (10y + x)     

30x + 3y + 6  =  10y + x      (2)       sub   (1) into (2)

30x + 3( 6 - x)  + 6 = 10(6  - x) + x

30x + 18 - 3x + 6    = 60 - 10x + x

27x + 24  =   60 - 9x

36x =  36   

 x = 1

And y = 6 - x  =  6 - 1    =  5

The original number is

10x + y   =      10(1) + 5  =   15