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3 of a kind ==6 *[5 C 3 * 5^2]==1,500
4 of a kind ==6 *[5 C 4 * 5^1]==150
5 of a kind ==6 *[5 C 5 * 5^0]==6

Therefore, the probability of at least 3 showing the same number is: [1,500 + 150 + 6] ==1,656 / 6^5 ==23 / 108

The distance between \((-1,5)\) and \((-6,-2)\)i is 5 units across and 7 units down.

This means that the midpoint is 2.5 units to the right and 3.5 units above \((-6,-2)\). 

This means that the coordinates are: \(\color{brown}\boxed{(-3.5,1.5)}\)

Just as proyaop said :)

Add the values of the coordinates of the two points, then divide it by 2. Thus, the midpoint of the segment would be ((-1 - 6)/2, (5 - 2)/2)

Thus, the coordinates of the midpoint of the line segment is (-3.5, 1.5).

:)

Let AC  =  AB

tan (angle BEA) =  AB / EA  =  AB / [1/3) AB)   =   1/ (1/3)  = 3

arctan 3 ≈ 71.6°

Angle  BED ≈  180  - 71.6  ≈  108.4°

Triangle CHB is similiar to triangle GFB by angle angle similarity (shared angle B and right angle).

Thus, the ratio of CH to GF is 18:3 or 6:1. Since CH is 24, then GF is 4. 

If GF is 4, then the area of triangle GFB is 6. The area of triangle ADE is also 6. To calculate the area of the pentagon, just calculate the area of the entire trianglce ABC and subtract it by ([ADE] + [GFB]). The area of triangle ABC is 36 x 24 / 2 = 432 units^2.

Thus, the area of pentagon CDEFG is 420 units^2.

=   4*8  * n    =   32 n

Thank you CPhill!

Using sin law

(first note that angle C = 180 - 45-30 = 105 degrees)

sin A / sqrt 6 =  sin 105 / AB

AB = sqrt 6   sin 105 / sin 45

P ( 3 show the same number )  =   C(5,3) (1/6)^3 ( 5/6) (4/6)   =  25/ 972 = 200/7776

P( 4 show the same number)  = C(5,4) (1/6)^4 ( 5/6)   = 25/7776

P( 5 show the same number ) =  C(5,5) ( 1/6)^5  = 1/7776

Total probability  = (200 + 25 + 1) / 7776  =   226 /7776 = 113/ 3888

900 / 13 ==69 multiples of 13 between 100 and 999

Of the 69 numbers above, only the following 7 end in 3:

(143, 273, 403, 533, 663, 793, 923) >>Total = 7

But, there are 90 integers between 100 and 999 that end in 3 such as: 103, 113, 123........973, 983, 993.

Therefore, the probability that it is divisible by 13==7 / 90

Yep.....good to BE back !!!!

10 #/ft   *    1 ft / 12 inches    *    4 inches    *    16 oz / #    =    53.333 oz              (as proyaop found !)

Yay!!! You're back!!!

Nice work  guys  !!!!

See this:

If a chain weighs 10 pounds per foot, and a foot is 12 inches, and 16 ounces is a pound, then 4 inches is 10/3 pounds. 

Since 16 ounces are in a pound, there are 16 x 10/3 ounces in 4 inches of chain.

4 inches of chain weigh 160/3 ounces or 53.3333 ounces.

Since triangle ADC is an isosceles triangle, then angle DAC = angle ACD = 24 degrees. Thus, angle ADB = angle ACD + angle DAC by exterior angle theorem, so angle ADB = 24 + 24 = 48 degrees. Since triangle ABD is an isosceles triangle, then angle ABD = angle ADB = 48 degrees.

Angle ABC = angle ABD.

Angle ABC = 48 degrees.

Set  the discriminant =  2c + 3

We have

5^2 - 4c  = 2c + 3

25 - 3  = 6c

22= 6c

22 / 6  = c  =  11 / 3

I am assuming that we have a triangle ABC even though not posted in the question. If that is so, then the first step is to identify a few extra triangles we can draw in. 

I drew an altitude from point C down to AB, and called the point of intersection D. 

Angle A is 45 degrees, so triangle ADC is a right isosceles triangle with hypotenuse AC. 

Angle B is 30 degrees, so triangle BCD is a 30-60-90 triangle with hypotenuse BC.

If those are the cases, then DC is \({\sqrt{6}\over2}\). We also know that AD = DC, so AD = \({\sqrt{6}\over2}\). BD is \(\sqrt{3}\) x DC, so BD is \(3\sqrt{2}\over2\).

What we also do know is that AB = BD + AD = \({\sqrt{6}\over2} + {3\sqrt{2}\over2}\).

Thus, \(AB = {\sqrt{6} + 3\sqrt{2}\over2}\).

PRYAP

h = 8

8 = -4.9t^2 + 14 t - .4

0 = -4.9t^2 + 14t - 8.4     Use quadratic formula to solve for t     the difference in the two t's is the time above 8 m

                                            a = - 4.9       b = 14       c = -8.4

\(t = {-14 \pm \sqrt{14^2-4(-4.9)(-8.4)} \over 2(-4.9)}\)                reults in t = 2   and  .8571      difference = 1.143 sec        <==== time = or above 8 m

We will have exactly real one solution  whenever the discriminant   =  0 

We have the form  ax^2  + bx + c = 0

So....the  discriminant is

b^2  - 4ac

4^2 - 4 (3)(c)  = 0

16 - 12c  =  0 

16  =12c

16/12  = c

4/3  = c

I’d set x+5=y, so x=y−5 and so we know that y should be a divisor of 60 greater than 5. Hence we can only have:

y∈{6,10,12,15,20,30,60}

On the other hand, we need y(y−5) to be a multiple of 60. We can make a table:

y         y−5     y(y−5)

6          1            6

10        5           50

12        7          84

15       10        150

20      15        300

30      25       750

60      55       3,300

and we see that only two cases are possible, namely y=20 or y=60.

Now that we know that, if N is the other number, we need:

60N=y^2(y−5)

For y=20 we get N=100

For y=60 we get N=3300

Since we need the smallest positive integers, we therfore have:

x ==15, and the other number=100

GCD(60, 100)= 20

LCM(60, 100)==300