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A = 1,  B = 4,  C = 2 All in base 6

So that: AB_6  +  C_6 =C0_6 =14_6  +  2_6 = 20_6

A + B + C = 1 + 4 + 2 =11_6

The probability is: 1 / [10C3 * 10] =1 / 1200

There are:  832  842  852  932  942  952  Total =  6 such numbers

[4 nCr 1] * 5^3 =[4 x 125] =500 / 6^4 =125 / 324 = 38.58%

OR:

[4 nCr 1] * (1/6)^1 * (5/6)^3 =38.58%

24 -30   =  7 days in Nov   + 19 days in Dec = 26 days

26 days * 6 show/day = ......... shows

side   x   side = area

side² = 221     solve for 'side'

To have only one real root, the discriminant of the quadratic formula (b^2 - 4ac) for this equation must = 0

  8^2 - 4 (a)(6) = 0

64 - 24 a = 0

         24a = 64

           a =   64/24    Simplify/reduce as needed.....

:D thank you

4= 12 (1 - .5)^n                  n = # half lives

1/3 = .5ⁿ

n= 1.585

1.585 * 1620 =2567.64 years

I found some happy banana phones but you will have to peal them yourself I am afraid.  

The probability is 1/C(20,3) = 1/1140.

Should have been one of the commandments.

Though shalt not divide by zero.

A movie is shown 6 times every day from November 24 through December 19. How many times was the movie shown in all?

Hello Guest!

                                                                19 days                            26

The film plays 7 days in November plus 24 days in December, i.e. 31 days.

Thanks ElectricPavlov!

  !

Oh, gorblimey! Pawsing to read Omi’s post definitely rings a bell and it’s doggone funny.... 

Given a trapezium with two parallel sides of lengths m and n, where m and n are integers, prove that it is possible to divide the trapezium into several congruent triangles.

4C2  *  (1/6)^2  *  (5/6)^2

Let

f(x) = 2x^2 - 3 if x <= 2

gx) = ax - 7 if x > 2
Find a if the graph of y = f(x) is continuous

Hello Guest!

$\color{BrickRed}f(x)=2x^2-3\\ \color{blue}f'(x)=4x=a\\ \color{BrickRed}g(x)=ax-7\\ 4x = a\ inserted\ in\ g (x)\\ g(x)=4x^2-7\\ \color{blue} f(x)=g(x)$

$2x^2-3=4x^2-7\\ 2x^2=4\\ \color{blue}x=\pm \sqrt{2}$

$y=2\cdot 2-3$

$y=1$

$g(x)=ax-7\\ 1=a\cdot \sqrt{2}-7\\ a=4\cdot \sqrt{2}$

$a=5.65685\\ g(x)=5.65685x-7$

\(y\in \{f(x)=2x^2-3\}\ |-∞  LaTeX is crazy today!

y $\in$ {f(x)=$2x^2$-3} | ($-∞$ < x $\le \sqrt{2}$ )

y $\in$ {g(x)=5.65685y-7} | ($\sqrt{2}$ < x $\le 2$)

y $\in$ {g(x)=5.65685x-7}    | ($2$ < x < $∞$)

  !

a)   Surface area is:  A = (54 * 2) + (36 * 24) = 972 cm² = 0.0972 m²

b)   Possible dimensions of the sides are:     9, 12, 15     (9+12+15=36) (9² + 12² = 15²)

so for 1/2 b*h=54
b*h-=108

what is the base and what is the height?

 

 Put right side over common denominator

(Ax-5A + 3x-6 ) / (x^2-7x+10)         now equate the L and R  numerators

Ax + 3x  - 5A -6   =  x + B      and equate the components

Ax + 3x = x

A = -4   

   then:

   -5A-6 =  B

   -5(-4) - 6 =B

      14 = B

a + b + c = 36  cm

      surface area =   36 * 24 = ........ cm²     then divide by 10 000 to get m²

 1/2 b * h  =  54 

        b * h = 108

                           9 x 12  is a possible dimension

                           4 x 27      6 x 18    etc

Simplify the following:
(i + 1)^16

(1 + i)^16 = ((i + 1)^8)^2 = (((i + 1)^4)^2)^2 = ((((i + 1)^2)^2)^2)^2:
((((i + 1)^2)^2)^2)^2

(i + 1)^2 = 1 + i + i - 1 = 2 i:
(((2 i)^2)^2)^2

(2 i)^2 = -4:
((-4)^2)^2

(-4)^2 = 16:
16^2

 =256