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sqrt(-18) =sqrt(-1 x 9 x 2)=3i*sqrt(2)

55 is the answer

5X11 or,

5+5 eleven times

Hit the "2nd" button....you will see the "asin"  function...

Can someone actually help?

8 / (2/3)   =

[8/1]  * [3/2]  =

24 / 2  =

12  desserts

2x+3+x-1=3 (2x-3)     simplify

3x + 2  = 6x - 9       add 9 to, subtract 3x from both sdes

11   =  3x    divide both sides by 3

11/3  = x

Then what about part b?

We can solve this if we assume that the cone is inverted......

Volume of water in half sphere  =   (2/5)(2/3)pi (6)^3  =  288pi/5

So.... the radius of the water  surface =

288pi/5  = pi * !2 * r^2 / 3

288/5  = 4 r^2

288 / [ 5 * 4]  = r^2

72 / 5  = r^2

6 √ [2/5]  = r  ≈  3.79 cm

The 8th term of tetranacci sequence are the odds out 2^10 chances. Since the 8th term of this sequence is 56, therefore the odds of at least 8 Heads are: 56/(2^10)=0.0546875 ~ 5.5%.

We have

P(White ball is drrawn first) +  P (black ball is drawn first)  = 10/21

5/ [5 + k]  * [ k]/ [ 4+k]  +  [ k]/[5 + k] * [ 5]/ [4 + k]  = 10/21

[5k  + 5k]  / [ (5 + k) (4 + k)] = 10/21

[10k] / [ k^2 + 9k + 20 ]  = 10/21

k / [ k^2 + 9k + 20 ]  =1/21     cross−multiply

[ k^2 + 9k + 20 ]  = 21k

k^2 − 12k + 20   = 0    factor

(k − 10) (k − 2)  = 0       set each factor to  0 and  k =2  or  k=10

So ..... k = 2

Proof :   5/ [7]  * [ 2]/ [ 6]  +  [ 2]/[7] * [ 5]/ [6]   = 10/42  + 10/42  = 20/42  = 10/21

The Law of Cosines can be used in either  SSS  or  SAS  situations.

If you know the three sides of a triangle, you can use the Law of Cosines to find any of the angles.

If you know two sides and the included angle of a triangle, you can use the Law of Cosines to find the third side.

P (Heads  ≥ 8 flips)  =  P( Tails ≤ 2 flips )  =

C(10,0)(1/2)^10   + C(10,1) (1/2)^10  + C(10,2)(1/2)^10  =  7 / 128

6!  * 4! * 5! * 3!   = 12,441,600  ways

Answered  here :  http://web2.0calc.com/questions/three-geometry-questions

Hexagon inscribed in a circle : 

http://www.mathopenref.com/constinhexagon.html

3. The coordinate of the rectangle are (-8,2) (0,4) (1,0) (-7,-2). What is the perimeter of the rectangle to the nearest tenth of a unit? 

(-8,2) = A

(0,4) =B

(1,0)= C

(-7,-2) = D

AB  =  √ [(0+ 8)^2 + (4 -2)^2  ]  = √ [64 + 4 ] = √ 68  = 2√ 17

BC  =  √ [(1- 0)^2 + (4 -0)^2  ]  = √ [1 + 16 ] = √ √ 17

Perimeter  = 2 ( AB + BC)  =  2( 2√ 17  + √ 17 )  = 2 (3√ 17)  = 6√ 17  ≈  24.7  units

2. What is the area of the triangle whose vertices are (4,-6) (9,-6), and (6,-2)

A  = (4,-6)

B =  (9,-6)

C  = (6,-2)

Call AB  the base  = 5 units

The height  = 4 units

Area  = (1/2) B * H  =  (1/2)5 * 4   =  10 units^2

Sorry

That is a very interesting formula!!.

Guess #1 =2.1

[10/2.1^2 + 2*2.1] / 3 =2.1558578987150415721844293272865........1st. iteration.

[10/2.155...^2 + 2*2.155...] / 3 =2.1544356293689498231197972784037......2nd iteration.

[10/2.154...^2 + 2*1.154.....] / 3 =2.1544346900322932740260161820501.....3rd. iteration.

So, let us see: 2.1544346900322932740260161820501^3 =10.000000000005702919684002568331,

accurate to 11 decimal places in 3 iterations!!!!!!. Pretty good!. It seems to converge quadratically.

(nC4) = 5(nC3)     we have

n! / [ (n − 4)! 4! ]   =  5  n! / [ (n − 3)! 3! ]    implies that

5 [ [ (n − 4)! 4! ]  =   [ (n − 3)! 3! ]

5 * 4!  * ( n −4)!  = (n − 3)! 3!      rearrange and note that     →    [5 * 4!   = 5!]

5! / 3!   = (n − 3)!  / ( n − 4)!

120/6  = n −  3

20  = n − 3

23  = n

Check

(23C4)   =  8855

5 (23C3)  =  8855