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1co/2g * 6g/5ch =  6 co/10ch

10ch/6co *6co = 10 chickens

Where k = chickens

g = goats

c = cows:

 c  = 2g

6g = 5k

6c = xk

First, we start by finding the number of cows that equal chickens.

We have that 6g/2g = 3, so 3c = 6g = 5k!

If 3c=5k and

  6c = xk

then x = 10

so 10 cows

:))))

What is the number of solutions to h + 3t = 8?

What is the probability that the frog goes over '8'? What happens if it gets three 3s?

Can you use modular arithmetic?

h+3t = 8

h = 2 (mod 3)

Let h be the number of heads.

Let t be the number of tails.

h = 3n + 2

so, t = 2-n

Ordered pairs of solutions:

(2, 2) ~ (5, 1) ~ (8, 0)

Non solutions that are greater than 8 but no extra leap over:

(0, 3) ~ (1, 3) ... count the rest, being mindful not to go an extra leap after you have reached >8.

Then answer is 3/ (non solutions + 3 )

The minimum value is 7/2.

The largest possible value of a is 8.

The probability is 7/20.

BX = sqrt(6).

The number of paths is 66.

The sum of the four numbers is 197.

We can form isosceles triangles with our angles.

We know that RQ is 8, and SQ is 2. 

However, we can't solve this problem without finding another isosceles triangle! This calls for a diagram! 

We can see now that ∆QTR is isosceles, with QT = QR = 8. 

But now we know that QT = 8, so ST = 8 - QS = 8 - 2 = 6!

 By AA similarity, ∆QTR is similar to ∆RST! By similar triangle ratios, RT/8 = 6/RT, and RT^2 = 48.

Thus, RT = 4 √3.

Also, by the base angles being equal, by AA similarity we have ∆PQR is similar to ∆QRT, and thus we have PQ/8 = 8/4√3 , so PQ = 64/4√3. 

Rationalizing the denominators, we have PQ = $\frac{16√3 }{3}$

If the angle WXA = 22.5º then  AP = 5.991618724

I'm not a guest. 

oops! that was wrong!

The possible values of cos(θ) are 0, 1/sqrt(2), and 1.

1. The maximum distance from A to P is 4 + sqrt(7).

1 - Without repeating the letters, you have: 26 nCr 4 =14,950 alphagrams.

2 - If each letter can be repeated up to 4 times, then you have: [26 + 4 - 1] nCr 4 =23,751 alphagrams.

Here is another way:

Since Carlos, working for 3.5 hours, saved her only 1/2 an hour, that means she does as much work in 1/2 an hour as Carlos does in 3.5 hours. Or Carlos rate is: 3.5 / 0.5 =7 times slower than that of Paula. And since she can paint the whole room in 6 hours, it would, therefore, take Carlos: 6  x  7 = 42 hours to finish the same room!!!!.

Agree with guest...but isn't it really  [-2,5)    ?

Note that without Carlos 'helping' Paula would finish the room painting at 3 PM by herself....so Carlos is not a lot of help

   after painting ofr 2 hours  there is  4 hours left for Paula   to finish the job

      but it takes 3.5 hours with both of them

       1/4    + 1/ C = 1/ 3.5

           solve for C = 28    but this is for 2/3 of the job

                Carlos alone would be     28 * 3/2 = 42 hours to paint the room....he must have a very small brush?

This is a bowl shaped parabola ....

minumum will occur at x = - b/2a 

   after you compute this value, use it in the equation to compute the minimum value of the function......

In interval notation, -2 <= x < 5 is {-2, 5).

AC = 7 units

I count 70 such permutations as follows:

(1233, 1323, 1332, 2133, 2331, 3123, 3132, 3213, 3231, 3312, 3321, 11223, 11232, 11322, 12123, 12132, 12213, 12231, 12312, 12321, 13122, 13212, 13221, 21123, 21132, 21213, 21231, 21312, 21321, 22113, 22131, 22311, 23112, 23121, 23211, 31122, 31212, 31221, 32112, 32121, 32211, 111123, 111132, 111213, 111231, 111312, 111321, 112113, 112131, 112311, 113112, 113121, 113211, 121113, 121131, 121311, 123111, 131112, 131121, 131211, 132111, 211113, 211131, 211311, 213111, 311112, 311121, 311211, 312111, 321111)=70 permutations.

area = 20 units squared

Since you don't want the answer given to you, then the only hint I can give you is that "n" must be a multiple of 9 !!. Good luck.