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The answer is 6.....to see this....let C = (0, 0)  and let D  = (9, 0)

[These are arbitrary...we can pick any two points that are 9 units apart ]

Then CD  = 9

The first transformation doesn't do anything to the length....it just shifts  C  to the right 3 units and down 2 units  ...so.... C'  =  (3, - 2)

And it does the same to D.....so we have D' =  (12, - 2)

The second transformation  is applied to both of these points.....so 

C''  = (2/3)(3, - 2)  =   (2, - 4/3)

And D''  = (2/3)(12, - 2)  =  ( 8, - 4/3)

And the distance between these points is

sqrt [  (8 - 2)^2 + ( -4/3  -  - 4/3)^2  ]  =

sqrt [ 6^2  + 0^2 ]  =

sqrt [ 36 ]  =  

6     .....just as we suspected !!!!!!

You're welcome.  I hope you fully understand :)

Draw an equilateral triangle of side length 2

now split it down the middle to form two congruent right angled triangles.

The angles on these 2 triangles are 90,60, and 30 degrees

the side lengths are  2,1 and sqrt3   (gained using Pythagoras's theorum)

The the hypotenuse of a similar triangle is 2root6 then that is a dilation of root6

2*sqrt6 = 2sqrt6

1*sqrt6 = sqrt6

sqrt3*sqrt6 = sqrt18 = 3sqrt2

Perimeter is

    \(2\sqrt6+\sqrt6+3\sqrt2\\ =3\sqrt6+3\sqrt2\;\;cm\)

wow! thanks melody!

We can equate the area of the the triangle on both sides of an equation

(1/2)*  (3 + 5) * sqrt [ (x + 2)^2 - 3^2 ]   =  (1/2) (x + 2) * sqrt  [ ( 8^2 - 2^2 ]

8 * sqrt [ x^2 + 4x + 4 - 9]  = (x +2 ) sqrt (60) 

8 * sqrt [ x^2 + 4x - 5]  = (x+ 2) sqrt (60)          square both sides

64 (x^2 + 4x  - 5 ]  =  (x^2 + 4x + 4 ) * 60

16(x^2 + 4x - 5)  = 15(x^2 + 4x + 4)

16x^2 + 64x - 80  =  15x^2 + 60x + 60

x^2 + 4x - 140  = 0       factor

(x - 10) ( x + 14)  = 0

Setting both factors to 0 and solving for x gives that x  = -14 (reject)  or  

x = 10

If the probability that a baby born in a certain hospital will speak in the next day is 1/4, what is the probability that at least 2 babies out of a cluster of 5 babies will speak tomorrow?

The prob that LESS than 2 will speak is

 5C0 * (0.25)^0 * (0.75)^5 +  5C1 * (0.25)^1 * (0.75)^4

= (0.75)^5 +  5 * 0.25 * (0.75)^4

(0.75)^5+5*0.25*(0.75)^4 = 81/128

so the prob that 2 or more will speak is    1- 81/128 = 47/128

I would guess 6 too but I don't really understand this question.

Here's my best attempt.....

Consider the sets  where one of the Americans wins gold...we have 3 possibilities for the American gold medal

And  of the other 5 sprinters, we want  to choose any of the other 5 non-Americans to receive the silver and any 4 of the non-Americans to recieve the bronze

So...this gives  3 * 5 * 4  = 60 possible ways

And we will have the same number of sets for an American winning silver  and the same number of sets for an American winning bronze

So far... this gives 60 * 3  =  180 ways

To this..we must add the sets where no American wins any medals....we have 5 ways to award the gold medal, 4 ways to award the silver and 3 ways to award the bronze = 5 * 4 * 3 = 60 ways

So  180 + 60  =   240 ways where, at most, one American wins a medal

No problem, here anytime! 

8P3=8!/5!=8*7*6=336.

It isn't. I want to know how to type it. I tried using LaTeX, though it doesnt seem to work....

The answer was really \(\frac{1}{663}\) 

IDK why really...

Sorry, but your answer was incorrect...

Luckily, I used a more efficient version of your method to get an answer of 118...

If the units digit is a 0, then the only two cases are that the 0 is the repeated digit, in which case the tens digit must be 0 and the hundreds digit can be 1 through 9, or the first two digits are the repeated ones, of which there are also 9 choices. Now, if the units digit is any even digit of the set \(\{ 2, 4, 6, 8 \}\), then the situation is different. If the repeated digit is to be the units digit, it can be repeated in the hundreds digit, in which case there are 9 choices of tens digit, or it can be repeated in the tens digit, in which case there are 8 choices of hundreds digit. Or, if the first two digits are the repeated ones, they can be any of 1 through 9 excluding the even units digit, for a total of 8. So, for each of 2, 4, 6 and 8, there are \(9 + 8 + 8 = 25\). So, the total number is \(2 \cdot 9 + 4 \cdot 25 = \boxed{118}\).

Thanks for your effort though!

Thanks so much, this helps!! 

The focus lies above the directrix....so this parabola turns upward

The vertex  lies midway between these and is given by  ( 4, (2 - 3) / 2 )  = (4, -1/2)

The focal distance,p, is the distance between the vertex and the focus....since the parabola turns upward this is positive and is given by  2 - (-1/2)  =  2 + 1/2  = 5/2

So....we have this form :

4p (y - k)  = (x - h)^2    where  the vertex is (h, k)  = (4, -1/2)  and p = 5/2

Filling in what we know, we have

4 (5/2) (y -  -1/2)  = (x - 4 ) ^2

10 ( y +1/2)  =  ( x - 4 )^2         divide both sides by 10

y +1/2  = (1/10) (x - 4)^2        subtract 1/2 from both sides

y  = (1/10) (x - 4)^2  - 1/2       this is in vertex form

Thanks so much! 

Thanks! I appreciate it! 

Yes, LaughingFace, you would be correct. \(WT=5\). This is because every segment extending from point W to a side is equidistant, by the properties of an incenter.

To be fair, the original question is ambiguous because rectangles do not have lengths; sides do. When the question says "Rectangle ABCD has a length of 20cm," what is that supposed to mean? I noted this as a disclaimer in my previous answer. I attempted to transform obscure information to something known, but I guess I was unsuccessful. I am sorry that I was unable to assist you. RainbowPanda, if there is any explanation on how to arrive at the answer, then I would like to see it. 

I have already solved 61; it is in my previous answer. I just did not label it. \(RQ\approx 17.9\)

Here are the permutations of 3-digit numbers beginning with "1". You have a total of 100 permutations and it appears that 10 EVEN combinations have a repeated digit. Since, there: 9 x 10 x 5 =450 EVEN 3-digit numbers between 100 - 999, and since there are 9 numbers beginning with "1", then I would generalize that there are: 10 x 9 = 90 such EVEN numbers with repeated digits.

{1, 0, 0} | {1, 0, 1} | {1, 0, 2} | {1, 0, 3} | {1, 0, 4} | {1, 0, 5} | {1, 0, 6} | {1, 0, 7} | {1, 0, 8} | {1, 0, 9} | {1, 1, 0} | {1, 1, 1} | {1, 1, 2} | {1, 1, 3} | {1, 1, 4} | {1, 1, 5} | {1, 1, 6} | {1, 1, 7} | {1, 1, 8} | {1, 1, 9} | {1, 2, 0} | {1, 2, 1} | {1, 2, 2} | {1, 2, 3} | {1, 2, 4} | {1, 2, 5} | {1, 2, 6} | {1, 2, 7} | {1, 2, 8} | {1, 2, 9} | {1, 3, 0} | {1, 3, 1} | {1, 3, 2} | {1, 3, 3} | {1, 3, 4} | {1, 3, 5} | {1, 3, 6} | {1, 3, 7} | {1, 3, 8} | {1, 3, 9} | {1, 4, 0} | {1, 4, 1} | {1, 4, 2} | {1, 4, 3} | {1, 4, 4} | {1, 4, 5} | {1, 4, 6} | {1, 4, 7} | {1, 4, 8} | {1, 4, 9} | {1, 5, 0} | {1, 5, 1} | {1, 5, 2} | {1, 5, 3} | {1, 5, 4} | {1, 5, 5} | {1, 5, 6} | {1, 5, 7} | {1, 5, 8} | {1, 5, 9} | {1, 6, 0} | {1, 6, 1} | {1, 6, 2} | {1, 6, 3} | {1, 6, 4} | {1, 6, 5} | {1, 6, 6} | {1, 6, 7} | {1, 6, 8} | {1, 6, 9} | {1, 7, 0} | {1, 7, 1} | {1, 7, 2} | {1, 7, 3} | {1, 7, 4} | {1, 7, 5} | {1, 7, 6} | {1, 7, 7} | {1, 7, 8} | {1, 7, 9} | {1, 8, 0} | {1, 8, 1} | {1, 8, 2} | {1, 8, 3} | {1, 8, 4} | {1, 8, 5} | {1, 8, 6} | {1, 8, 7} | {1, 8, 8} | {1, 8, 9} | {1, 9, 0} | {1, 9, 1} | {1, 9, 2} | {1, 9, 3} | {1, 9, 4} | {1, 9, 5} | {1, 9, 6} | {1, 9, 7} | {1, 9, 8} | {1, 9, 9}.

The "npr" button at the bottom left ....

Try watching this.

93.

P ( Having a pet  l Having a sibling)  = 

15 students have a sibling...of these 7 have a pet

So....the probability is just 7/15 ≈ 46.7%

94.

Note.......P(A or B)  = P(A) + P(B) - P(A and B)

P(ace  or 2)  = P(ace) + P(2)  - P(ace and 2)   =

                         4/52  +  4/52  -  1/52  =

                                     7 / 52

95.  P (5 or diamond)  = P(5) + P(diamond)  - P(5 and diamond) =

                                        4/52  +  13/52    -         1/52   =

                                              16/52  =

                                               4/13

96.   We have 26 black cards....2 of them are queens....so  2/26  = 1/13

97. We have 4 kings....only one is a diamond....so  1/4

98. Don't know the answer to this one, LF   ...sorry  !!!

Oh okay thank you for that explanation. The picture does help.

HMMM I have been thinking about it for awhile now and I think I am starting to get it!!

Sorry!