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I'll try my best on this one, I'm not particularly good with this topic.

So the first Ace can appear anywhere, so 52 choices, times 4 since it could be any of the 4 aces.

The ace adjacent to it has 2 choices, left or right, and has 3 choices on which type of ace.

The probability of this happening is \(\frac{52\cdot4\cdot2\cdot3}{52\cdot52\cdot2\cdot51}\) which will end up being \(\frac{1}{221}\)

so the expected number of times is 0.

Wow, thank you so much! The application I'm doing this on says #2 is wrong ("You can use Vieta's formulas in reverse"), but I think I can figure it out by myself (somehow). Again, thanks a lot for spending all that time (lol)!

Ive googled this question. Electric Pavlov also answered this, and he got no answer as well. Check with your teacher, or the website, or whoever made this question.

you are x, your brother is y

\(x-5 = y+5\)

\(x+10 = 2(y-10)\)

now we solve

\(x+10 = 2y-20\)

\(x = 2y-30\)

\(x = y+10\)

\(2y-30 = y+10\)

\(y = 40\)

your brother has 40 dollars right now.

Alright, we set the largest of the 7 number to be x.

\(x-6+x-5+x-4+x-3+x-2+x-1+x = 214x\)

\(7x-21=214x\)

\(-21 = 207x\)

\(x = -21/207\)

that is not possible since they have to be integers.

hmmmm. Is this a trick question?

5n /( n + 2)  =  6

5n = 6 (n + 2)

5n = 6n + 12

-n = 12

n  = -12

NOTE: notice that \(\frac{2x+3}{2x+3} = 1\)

a. since both are equal to one, any number we plug in will just be 1.

b. \(x=-\dfrac32\)for this, the denominator will equal zero when you plug it in.

c. No other real numbers can make the denominator 0, therefor all others work.

2. When the set of natural numbers is listed in ascending order, what is the smallest prime number that occurs after a sequence of five consecutive positive integers all of which are non prime?

24,25,26,27,28

Smallest prime  = 29

So.

Since it is a bisector, DC will be 10. Since Angle C is 15 degrees, we can write DE as

\(10\cdot tan(15^\circ)\)

That is equal to

\(10\cdot \frac{\sqrt{3}-1}{\sqrt{3}+1}\)

which is equal to 

\(20-10\sqrt{3}\)

Since

\(BE = 2\sqrt{200-5\sqrt{3}}\)

Im probably wrong, so anyone please correct me.

BE will  equal CE

To find CE

angle CED = 90  -15 = 75

CD / sin 75 =  CE / sin 90

10 / sin 75  = CE  =  10 (sqrt 6 - sqrt 2)  ≈  10.35 = BE

2. When the set of natural numbers is listed in ascending order, what is the smallest prime number that occurs after a sequence of five consecutive positive integers all of which are non prime?

Thank you, that made a lot of sense now.

100!  / 60!  =   100 * 99 * 98 * .....* 62 * 61

This is definitely smaller than 100^40

100!  = 

(100*99) * (98 *97) * (96 * 95) * (94 * 93) * (92 * 91) * ......* (10 * 11)

Note that we  have   ( 100 - 10) / 2 + 1 =  46 terms in this partial evaluation of 100!....each is > 100

So

100! > 100^40

So

1001/ 60! < 100^40 < 100!

      A

      4            5

     B        D          C

                3

Since AD is a  bisector.....Let BD = x  and DC  =3 - x   and  we have that

BD/ AB  = DC / AC

x / 4 = (3 - x) / 5

5x = 4 (3 - x)

5x  = 12 - 4x

9x= 12

x =  4/3  = BD

[ ABD ] = (1/2) (4/3)(4)  =  16/6 = 8/3

[ABC ]  = (1/2)(4)(3)  = 6

[ADC ]   =  [ ABC ] -  [ ABD ]  = 6 - 8/3  =   10/3 =  3 + 1/3  =   3 (to the nearest integer )

Can you tell me how you solved it?

100 ! / 60!   < 100^40 < 100!

Because CX is an angle bisector, CA / CB = AX / BX

So

10 /10  =AX /BX

1 = AX /BX

BX = AX

Triangles BCX and ACX are under the same height and have equal bases.....therefore, the ratio of their areas is

1 : 1  =   1 / 1

                    A

               x        x + 2

          B                 C

                 x + 4

Law of Cosines

(x + 4)^2 = x^2 + (x +2)^2  - 2 (x * (x + 2)) cos (120°)

x^2 + 8x + 16 =  x^2  + x^2 + 4x + 4  - 2 (x^2 + 2x)(-1/2)

x^2 + 8x + 16  = 2x^2 + 4x + 4  +  x^2 + 2x

x^2 + 8x + 16  =   3x^2 + 6x + 4

2x^2 - 2x - 12  =   0

x^2 - x - 6  =  0

(x - 3) ( x + 2) =  0             

x - 3  = 0                    x + 2 =   0

x = 3                           x = -2   (reject)

This is AoPS homework.  Do not post AoPS homework.

2. I assume the quadratic referred to is that in (1)

We will have the form   x^2  +  mx + n

If so

a^2 = 25

b^2 = 9

Sum of the roots  =  - m  =   (a^2 + b^2 )   =  34

Product of the roots =  a^2 * b^2     = (25 * 9)  =   225

The quadratic is

x^2  - 34x  + 225

OK....how ' bout  .....

(3x + - 5) (4x + 7)  =  12x^2  + 1x + - 35