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Write an equation: \(4200 > t(70 - t)\)

Now, just solve for t...

\({11 \over 30} \times {10 \over 29} \times {12 \over 28} \times {11 \over 27} \times {10 \over 26} \times {5 \choose 2} = \text{____}\)

Note that for the product to be divisible by 4, there need to be at least 2 even numbers. 

The probability of rolling only odd numbers is \({5 \over 9}^4 = {1025 \over 6561}\)

The probability of rolling exactly 1 even number is \(4 \times 5^3 \times {4 \choose 3} = 2000\)

So, the probability is \(1 - ({1025 \over 6561} + {2000 \over 6561}) = \text{_____}\)

Hi Cphill,

so in other words my aproach was correct, the answer in this case is 16?

See https://web2.0calc.com/questions/algebra_24481

Use the 3D plane! :D

P ( G(x)) =  4 + 2 / sqrt [ 4-8x + 2]   =  4 + 2 /sqrt [ 6 - 8x]

6 - 8x  ≥  0

6 ≥ 8x

x ≤  6/8   

x ≤   3/4

Then

a ≤  3/4

Interior angle of a regular hexagon  =   (6-2) (180) / 6  =  (2/3)(180) =  120°  =  α

Area =  (h/2)  ( sum of bases)

144  = (12/2) ( 16 + x)

144 = 6 ( 16 + x)

144/6=  16 + x

24  = 16 + x

x = 8  = top base

Right edge  =  sqrt   [ (16 - 8)^2  + 12^2 ] = sqrt [ 8^2 + 12^2 ]  = sqrt [ 208] = 4sqrt (13)

Perimeter =  12 + 16 + 8 + 4sqrt (13)  =   

[36 + 4sqrt (13)] cm

Hello, Thank you for the help, 

But that answer seems to be wrong. 

Costruct  a circle with a center of  (8,3)  and a  radius of  sqrt 58

The equation of this  circle   is

(x - 8)^2  + ( y  - 3)^2 =  58

Our point must be on this  circle and is 6 units from  the x axis (y coord of the point  =  6)

So

(x - 8)^2  + (6 -3)^2    =  58

(x - 8)^2  + 9 = 58

(x - 8)^2  =  49

x - 8  =  -sqrt (49)

x =  8 - 7    = 1

n = 1^2  + 6^2 =    37

(a)   5 / (5 + 1 + 2)  =  5 / 8

(b)    Before the race Lothar had  (2/8) = 1/4 of the total

When he lost (1/2)  he now had 1/8 of the total   (he lost 1/8 of the total )

Zane originally had  (1/8) of the total

After the race he now has (1/8) + 1/8 =  2/8 =  1/4 of the total

So

(1/4) total  =  (1/8) total + 3

(1/4 - 1/8) total =  3

(1/8)total  = 3

total =  24 

Zane had   (1/4) (24)  = 6   at the end

Lothar had ( 1/8) (24)  = 3 at the end

Johnny had (5/8) (24) =  15 at the end

Case 1: Superball

The probability is 1/10, because you choose 1 out of 10.

Case 2: 2 whites

The probability is 1/C(10,2) = 1/45, because you choose 2 out of 10.

Case 3: 3 whites

The probability is 1/C(10,3) = 1/120, because you choose 3 out of 10.

Adding these, you get 1/10 + 1/45 + 1/120 = 47/360.

It's absolutely ridiculous to not only solve for 20! and 12! but to also take the prime factorization!

2 , 3 , 5 , 7 , 11 , 13 , 17 , 19 , 23 , 29 , 31 , 37 , 41 , 43 , 47 , 53 , 59 , 61 , 67 , 71 , 73 , 79 , 83 , 89 , 97 , 101 , 103 , 107 , 109 , 113 , 127 , 131 , 137 , 139 , 149 , 151 , 157 , 163 , 167 , 173 , 179 , 181 , 191 , 193 , 197 , 199 , 211 , 223 , 227 , 229 , 233 , 239 , 241 , 251 , 257 , 263 , 269 , 271 , 277 , 281 , 283 , 293 , >>Total =  62 >>Sum = 8275

n==600

LCM{600, 100} ==600

GCD{600, 100} ==100 + 500 ==600

The  interior  angles sum to   180 (9 -2)  =  1260°

So

100 + (100 + d) + (100 +2d)  + (100 + 3d)  +(100 + 4d) + (100 + 5d) + (100 + 6d) + (100 + 7d) +

(100 + 8d)  =  1260

900 + 36d =  1260

36d =  360

d =  10

Largest angle =     100 + 8 (10) =  180°

a + ab^2 =  40b       (1)

a - ab^2  = 32b         add these equations

2a  =  72b

a  = 36b    (2)

Sub   (2) into (1)

36b + (36b)b^2  =  40b

36b^3   =  4b

36b^3  - 4b  =  0

4b (  9b^2  - 1)  = 0

4b ( 3b -1)  (3b + 1)  = 0

4b = 0         3b  -1   = 0         3b + 1  = 0

b = 0            b  =1/3              b = -1/3

a = 36 (0)  =  0

a = 36 (1/3)   =  12

a = 36 (-1/3)   = -12

x - 2y  = 0      →  x   = 2y      (1)

y  - 2z =   0    → y  = 2z →  z = y/2    (2) 

x + z  = 5    (3)

Sub (1)  , (2)  into  (3)

2y  + y/2  = 5

2.5 y =   5

y  = 5 / 2.5  =   2

x = 2(2)  =  4

z = 2/2  =  1

(x, y, z)  = (4, 2 , 1 )

The solutions (a,b) are (12,3) and (-12,-3), so the possible values of a are 12 and -12.

The solution is (-inf,2 + sqrt(7)).

a^2 = 12

b = 27.

x = log(2)/4.

Since 12 ==2^2  x  3, we can see that the largest distinct factor of 12==3, and since:

20!  +  12!= 2^10 * 3^5 * 5^2 * 7 * 11 * 23 * 31 * 653 * 10909

Therefore, the largest exponent of 3 above==5 - the largest integer n, and:

[20!  +  12!]  mod  12^5 ==0