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Strictly through a little experimentation, I get  that  the  greatest length of EF  is ≈  19.98

However.....I'd like to  see  if someone knows  the  method of  generating a definite solution and  how they  get the  result....

Call the point we are looking for  (a, 0)

Since we are equating distances,  we can equate  the distances^2

So

( a - -2)^2  +(0 - 0)^2  = ( a - 0)^2  + (4 - 0)^2     simplify

(a + 2)^2   =  a^2  +  16

a^2 + 4a + 4  = a^2 + 16

4a + 4 =  16

4a  = 12

a =   3 =   the x coordinate we are  looking for

Let AP= AO

Angle PAO = 90

So angle APO =Angle AOP = 45

Then since PC is an angle bisector, then Angle APC  = 22.5

And Angle AOP  = Angle  AOD  = 45

The angle ABD  is an inscribed angle intercepting an arc of 45 ....so its measure = 22.5 = angle OBA

And in triangle AOB, OA  = OB....so Angle OBA  = Angle  OAB  =  22.5

And angle CAP =  Angle PAO + Angle OAB =   90  + 22.5 =  112.5

So in triangle APC, angle PCA  = 180  - 112.5 - 22.5  =  180 - 135  =   45 (degrees)

Thank you, CPhill !

4x   -3y  +  12  = 0     rearrange as

3y = 4x  + 12

y  = (4/3)x  +  4   (1)

A perpendicular   through   (10, 9)   will  have a negative reciprocal slope  =  -3/4

The equation of this line is 

y= (-3/4) ( x  -10) + 9

y = (-3/4)x  + 30/4 + 9

y =  (-3/4)x + 33/2     (2)

Find  the x coordinate of the intersection of (1)  and (2)

(4/3)x + 4 =  (-3/4)x  + 33/2

(4/3 + 3/4)x  =  33/2  - 4

( 25/12)x  = 25/2

x = ( 25/2) ( 12/25)  =  12/2  = 6

And y  = (4/3)(6) + 4 =  12

(6, 12)

This is the  midpoint  between    (9,10)  amd  the reflection point

So we have

(10 + x)/2  = 6                   (9 + y)  / 2  =  12

10 + x  = 12                         9 + y  =  24

x = 2                                    y = 15

The reflected point  is  (2,15)

Here's a graph  :  https://www.desmos.com/calculator/l4wolqqhos

(B)  

Since angle ADB  is obtuse....it is the greatest angle in triangle ADB

So BD <  AB

We can't say anything about the  relationship between CD and AB....we  would have to  know something about the ,measures of angles CAD  and DAB

For instance   angle   ADB  is obtuse   so   BD  <  AB

But here   CD is clearly >  AB

First one 

Angle ADC  = 180 - 120  =  60°

So angle CAD  = 180 - 60  - 65  =  55°

The  length of the sides   are   in line with  their opposite angles

Greatest angle  opp  greatest side

Next greatest angle opp next greatest  side

Least angle opp  shortest side

So

AD >  AC >  CD    (1)

And note that  angle ADB   =  180  -  120  - 35     = 25°

So   AB > BD  > AD     (2)

So   (1) and (2) imply that

AB > BD > AD > AC  >  CD

So  CD  is the  shortest side       (clearly, BC  is greater than CD)

I mean i dont know much about it myself since geo is my worst subject

But It should be that the smallest side of a triangle is opposite to the smallest angle in a triangle

Oh, sorry, seems like the LateX didn't work, so it might have some missing parts.

2)     Lets start with the red ball because it is easier. You have 1/10 chances to pick the same number that is drawn randomly. Thus your base probability is .1 Now going to the white balls, you need 2 numbers picked as the numbers drawn. Let's say we pick numbers 1-3. In order to win, the numbers 1 and 2, 2 and 3, 1 and 3, or 1 2 and 3 must be drawn. In the case only 2 are drawn, the probability for each is 0.075. Since there are three cases that this could happen, you multiply this probability by 3, becoming 1/4. However, there is also the chance that you get all three. The probability of this is 3/10*2/9*1/8.

.225 + .1 + 0.008333  = 33.333...% or 1/3

Don't want to tackle numero 1 for now...

I can start on the first question...

The shortest side would be the one across from the smallest angle since the larger the angle, the opposite side gets bigger in comparison to the others. (just look at a right triangle)

right, messed up there

I think its 2002, but someone else posted this problem some time ago.

Heres another way to do it MooMooooMooM:

We know that the numero of apes must have factors of 4 and 5
20 is the smallest number that has these factors,

the smallest number of zebras is 5.

Not sure but this is what I have...

1≤a

b

a2<100a2<100 so a<10a<10.

ab<100ab<100 so ab≤99ab≤99 so b≤99ab≤99a so b≤⌊99a⌋b≤⌊99a⌋.

b+1≤c≤100b+1≤c≤100.

∑9a=1(∑⌊99a⌋b=a+1(∑100c=ab+11))=∑a=19(∑b=a+1⌊99a⌋(∑c=ab+11001))=.

∑9a=1(∑⌊99a⌋b=a+1[100−ab])∑a=19(∑b=a+1⌊99a⌋[100−ab])

∑99b=2(100−b)+∑49b=3(100−2b)+∑33b=4(100−3b)+∑24b=5(100−4b)+∑19b=6(100−5b)+∑16b=7(100−6b)+∑14b=8(100−7b)+∑12b=9(100−8b)+∑11b=10(100−9b)=∑b=299(100−b)+∑b=349(100−2b)+∑b=433(100−3b)+∑b=524(100−4b)+∑b=619(100−5b)+∑b=716(100−6b)+∑b=814(100−7b)+∑b=912(100−8b)+∑b=1011(100−9b)=

∑98k=1k+∑94k=2;+2k+∑88k=1;+3k+∑80k=4;+4k+∑70k=5;+5k+∑58k=4;+6k+∑44k=2;+7k+∑28k=4;+8k+∑10k=1;+9k=∑k=198k+∑k=2;+294k+∑k=1;+388k+∑k=4;+480k+∑k=5;+570k+∑k=4;+658k+∑k=2;+744k+∑k=4;+828k+∑k=1;+910k=

∑98k=1k+∑47k=12k+∑29k=0;(3k+1)+∑20k=14k+∑14k=15k+∑9k=06k+4+∑7k=07k+2+∑308k+4+∑1k=09k+1=∑k=198k+∑k=1472k+∑k=0;29(3k+1)+∑k=1204k+∑k=1145k+∑k=096k+4+∑k=077k+2+∑038k+4+∑k=019k+1=

98∗992+2∗47∗482+329∗302+30+4∗20∗212+514∗152+69∗102+4∗10+7∗7∗82+2∗8+83∗42+4∗4+9∗1∗22+2∗1=98∗992+2∗47∗482+329∗302+30+4∗20∗212+514∗152+69∗102+4∗10+7∗7∗82+2∗8+83∗42+4∗4+9∗1∗22+2∗1=

49∗99+47∗48+3∗29∗15+30+2∗20∗21+5∗7∗15+3∗9∗5+40+7∗7∗4+16+4∗3∗4+16+9+2=49∗99+47∗48+3∗29∗15+30+2∗20∗21+5∗7∗15+3∗9∗5+40+7∗7∗4+16+4∗3∗4+16+9+2=

49∗99+47∗48+29∗45+30+20∗42+35∗15+27∗5+40+49∗4+32+48+11=10269

(Okay, I don't know how to simplify)

Say my age back then is x

So my moms back then was 7x

In 8 years, we add 15.

So 7x+15 and x+15

7x+15 is two times x+15

So 7x+15 equals 2x+30

5x=15

x=3

So back then your mother was 21.

Checking work

21+15=2(3+15)

"The square root of that is a*b*c*d=30."    Did you mean to say the cube root? 

The shaded area of a rug is 50 square units.

1)  since you have to choose three numbers and one superball to win the lottery, your chances will be:

chances of getting the 3 white numbers = 1/6*1/10 x 1/9 x 1/8 = 1/4320

chance of getting the superball = 1/10

chances of winning the jackpot = 1/4320 x 1/10 = 143200

2)

On average, 1 out of 10 times you will draw the AwesomeBall (and this wins all by itself so we don't care what the other balls are) for a winning probability of 0.10.

Now, for the 9 out of ten times that you do not draw the AwesomeBall:

We will have to get a losing blue ball, I'll call that B (and the probability of getting that losing ball is 0.90).

We can win by getting two or more of the white balls. I'll classify a winning white ball (whose probability is 1/10 = 010)

as W and a losing white ball as L (whose probability is 9/10 = 0.90).

So, we can still win if we have this:

B x W x W x W   =   (0.90) x (0.10) x (0.10) x (0.10)   =  0.0009

B x W x W x L    =   (0.90) x (0.10) x (0.90) x (0.10)   =  0.0081

B x W x L x W    =   ...

B x L x W x W    =   ...

Since these are independent, add the above 5 results together:  317/2500

The area of the region is 14.

There were 20 blue marbles originally.