All Answers 355934 Answers

Hint for number 2: Count the number of possible arrangements...and subtract by the number of arrangements that the letters are next to each other. For this, try to use a superblock for (A), (L), (S)...because they all repeat twice

The question isn't sufficiently specific.

For instance, if there are only three meals and each of the three choices can be eaten only once per day, there are  3 x 2 x 1  =  6  ways.

However, if there are three meals and each food can be eaten more than once per day, there are  3 x 3 x 3  =  27 ways.

If there are more than three meals, then each food must be eaten more than once per day and there will be a different answer.

Since each "word" must start with a vowel, there are  5  choices for the first letter.

Each of the following  2  letters can be any of the 26 letters of the alphabet.

Multiplying the choices together, we get  5 x 26 x 26 or  3380 possibilities.

x - 2y  = 0     --->     x  =  2y

Substituting:  since  x + z  =  5   and   x  =  2y     --->     2y + z  =  5

Combining  2y + z  =  5   and   y - 2z  =  0

   2y + z  =  5   --->   x 2   --->   4y + 2z  =  10

    y - 2z  =  0                   --->     y -  2z  =    0

   Adding down:                        5y          =  10     --->     y  =  2

Since  y  =  2     --->     x  =  2y     --->     x  =  2(2)     --->     x  =  4

Since  x  =  4     --->     x + z  =  5     --->     4 + z  =  5     --->     z  =  1

No, but I do know that would equal 10^x=12. I just don't know how to go from there

Solve for p:
(-125 i - 5) + 5 p = 10000

Add 5 + 125 i to both sides:

5 p + ((-125 i - 5) + 125 i + 5) = 125 i + 5 + 10000

(-125 i - 5) + (125 i + 5) = 0:

5 p = 10000 + 5 + 125 i

10000 + 5 + 125 i = (10000 + 5) + 125 i = 10005 + 125 i:

5 p = 125 i + 10005

Divide both sides of 5 p = 125 i + 10005 by 5:

(5 p)/5 = (125 i + 10005)/5
p = (125 i + 10005)/5

Let p be the probability of an even roll.

For option A

E[moves]=4p - 3(1-p) = 7p-3

For option B

E[moves]=5p-4(1-p) = 9p-4

Playing with the numbers it turns out that p=P[even roll]=3/4

I suggest you confirm this for yourself.

EA=21/4 - 12/4 = 9/4

EB = 27/4 - 16/4 = 11/4

EB offers the greater likelihood of moving closer to the finish line.

Look at the graph here :

Let O = (0, 0) 

By the Law of Cosines   we can find the height, h, of point F above the x axis

h^2  = 1^2 + 1^2  - 2(1)(1)cos120

h^2  = 2 - 2(1/2)

h^2 = 2 - 1

h = sqrt(3).....so F =  (.5, sqrt(3) )

And D  = (1, sqrt(3)/2 )

And let the vertex of the parabola  be (0, k)

Since the parabola passes through D and F  we have that

sqrt(3) = a( .5)^2 + k

sqrt(3)/2 = a(1 )^2 + k      subtract these and we have

sqrt(3)/2  = a (-3/4)

-4sqrt(3)/6 = a

-2sqrt(3)/3 = a

-2/sqrt(3)  = a

So....we can find k as

sqrt (3) = -2/sqrt(3)(.5)^2 + k

sqrt(3) + 2/ [4sqrt(3) ] = k

[12 + 2 ] / [ 4sqrt(3) ]  = k

7/[2sqrt(3) ] = k

To find the positive x coordinate of  "B"  we have

0 = -2/[sqrt(3)] (x)^2 + 7/ [2sqrt(3)]

-7/ [ 2sqrt(3) ] = - 2/ [ sqrt(3) ] (x)^2

7/4  =  x^2

sqrt(7)/2  = x 

So....by symmetry,    AB  =  2 *sqrt(7) / 2    =  sqrt (7) 

9 pieces was wrong

1) A rope is 120 feet long. Starting from one end, make a red mark every 3 feet, make a blue mark every 4 feet, and cut the rope at every mark. (Where there’s both a red mark and a blue mark, we only make one cut, not two.) How many pieces of rope do we have now?

To see the answer....note that....in the first 12 feet  we make cut at the  3, 4, 6, 8, 9 and 12 foot marks

So...every 12 feet....we have 6 pieces of rope

So....we have   (10) (6)  =  60 pieces in total 

Eh...I left out 60....the other multiple of 4, 5

So

(25 + 45) - 5  =  40

And 

100 - 40  =  60  !!!

Here's my best effort :

74.   cos ( θ + k π)  =

cosθ   cos kπ   -  sin θ  sin π  =

cos θ cos kπ

If  k is even  then     cos kπ  =  1 ...... so cos θ * (1)  =   (-1)^kcosθ   

If k is odd   then    cos kπ  =  -1 ...... so cos θ * (-1)  =   (-1)^kcosθ

xxxxxxxxxx

Thanks!

The correct answer was 60. I think you just forgot to add one since the number was inclusive. 

Oops...sorry....you wanted the ones that remain  !!!

That would be  100  - 41   = 59

3)  If all multiples of 4 and all multiples of 5 are removed from the set of integers from 1 through 100, how many integers remain?

We have 25  multiples  of 4   and  20 multiples of 5

However....we've  "double-counted"  the multiples of 4 and 5   =  20, 40, 80  and 100

So

[ 25 + 20 ] -  4   =     41

1) In a survey of 100  students who watch television, 21 watch American Idol, 39 watch Lost, and 8 watch both. How many of the students surveyed watch at least one of the two shows?

(21 - 8)  =  13  = Those that watch "American Idol" only

(39 - 8)  =  31 =  Those that watch "Lost" only

So....the number that watch at least one  is  

13 + 8 +  31  =

52

Thanks! :)

Mmmmm....must be a "ghost in the machine"......consider it "un-flagged"

Why is my post flagged?

Nice, Geno   !!!