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2^3 x 3 x 5 = 120 =(1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120) >>Total = 16 divisors.

2^(x + y)  * 3^(x - y)  * 6^(2x + 2y)   =  72

2^(x + y) *  3^(x - y) * ( 6^2)^(x + y)  =  72

2^(x + y) * 3(x - y) * 36(^(x + y)  =  72

( 2 * 36)^(x + y)  * 3^(x - y)  =  72

(72)^(x + y)  * 3^(x - y)  =  72

Since

72^(1) * 3^(0)  =   72

This will be true when

x +  y   =   1

x -  y  =     0              add these

2x   =  1

x   =1/2

And  

1/2 + y   = 1

y  = 1/2

(x,y)   =  (1/2, 1/2)

Oh yea! I remember the 2 roots thing from quadratic equations but I didnt know that one was a minus number. I remember some them had two plus numbers. How do you know which one to use if they are both plus numbers?

Divisors of 10^10 =2^10 x 5^10 = 11 x 11 =121

Divisors of 12^12 =2^24 x 3^12 = 25 x 13 =325

Divisors of 15^15 =3^15 x 5^15  = 16 x 16 =256

121 + 325 + 256 =702 divisors of all 3 numbers. However, I counted 34 as duplicates. If we remove these 34 duplicates, then we have: 702 - 34 =668 divisors of at least one of your 3 numbers (10^10, 12^12, 15^15).

Thanks, EP......here's a general way of solving something like this

Assuming that  the road is uphill one way and downhill the other way....we have that....

Avg rate    =    Total Distance  / Total Time

Call the total  distance  =  2D

The total time is   ( D/ 30)  + (D /20) 

So...we have that

Avg rate  =     ( 2D)                               2D                              2* 600

                  _____________  =       ______________  =       ______     =  24 mph

                     D/30 + D/24                D [ 30 + 20] /600                50

CORRECTED   !!!!

We can use  the triangle  inequality to solve this

Let   S  be the unknown  side

So....

S + 5  >  7           S + 7  >   5             7  +   5   >   S

S  >  2                  S  >   -2                   12  >    S

                                                              S   <  12

Taking the most restrictive positive interval for S......  

S  = ( 2 , 12)

We have   9  integers  that  will satisfy this   (3, 4, 5, 6, 7, 8 , 9 , 10 , 11)

So....the probability  =     9  / 15    =    3   /  5

I'll try:

THere is a 95% chance that you DO order a sandwich or soup

.95 x .8 x .7 = 53.2 %      (I usually get probability questions incorrect !)

225 / 36     has two   roots  : (-15/6)     and  (15/6)   

If we square both of these.....we get   225 /36

But.....since    "a"  is positive....we need   15 / 6    =   5   / 2

Does that make sense  ???

Determine the set of all real x satisfying (x^2+3x-1)^2<9
Enter your answer in interval notation.

Take both roots and we have that

x^2 + 3x - 1   =    3                     x^2   + 3x - 1  = - 3  

x^2  + 3x  - 4   = 0                     x^2  +  3x  + 2   =   0

(x + 4) (x - 1)  = 0                     (x  + 2)  ( x + 1)   =  0

Setting each factor to 0 and solving for x  gives the answers that  x  = { -4, - 2,- 1, 1 }

So....we have the possible answers  

(-inf, -4)    ( -4, -2)   ( -2, -1)   (-1, 1)    or  ( 1 inf )

Testing a point in each interval in the original inequality we see that the  solution intervals are

(-4, -2)  U   (-1, 1)

Really cool! Thanks CPhill!  Man Im glad I found this math site

What do you mean by positive root.

ab  =  11.25       ⇒  b  =  11.25 / a       (1)

ac  = 20   ⇒  c  = 20  / a        (2)

bc  = 36         (3)

Sub (1)  and (2)   into (3)

(11.25 /a ) (20 / a)  = 36

225/ a^2  =  36       rearrange as

a^2  =  225/36       take the positive root

a  =  15 / 6    =   5 / 2

So....using (2).....

c =  20 / ( 5/2)   =   8

I'm not sure if I understand your question fully!. But, I will give it  a try.

My undestanding is that you want to know how many pemutations are there where 2 (even number) is in the 2nd position from the left, and 4 (even number) is in the 4th position from the left and 6 (even number) in the 6th position from the left. If that is the case, then:

7! =5,040 permutations. Each of the 7 numbers appears:5,040 / 7 =720 times in EACH of the 7 positions, from left to right. So, 2 will appear in the 2nd position from the left 720 times. So will 4 appear in the 4th position from the left 720 times as well. And so will 6 appear in the 6th position from the left 720 times.

Therefore, all three even numbers(2, 4, 6) will appear in their repective positions in: 3 x 720 =2,160 permutations in total (if I understood your question!).

Lay out a  rectangle with  a width of 5  and a height of  3

As shown here :  https://www.desmos.com/calculator/qnelrpyogs

The ant will cross 7  distinct squares

So....in a  rectangle with a width of 20  and a height of 12......the ant will cross 4 times as many squares  = 28

Sequences with one T  =

(TAR)  (TAG) (TAE)

(TRG) (TRE)

(TGE)

And each of these can be arranged in 6 ways....so....6 * 6  =   36       (!)

Sequences with two T's  =

(TTA) (TTR) (TTG) (TTE)

And each of these can be arranged in 3 ways ,,,,so   4 * 3  =  12     (2)

So...adding (1)  and (2)   we get  48 arrangements with at least one T

Cool!! Thankx EP!

Note that most 'Answerers' are less inclined to provide methods of finding an answer to 'guest' posters and may just post  an answer.....

    Ever thought of signing up as a registered user?    ~EP

b=11.25/a     (from first equation)

c=36/b = 36a/11.25

ac=20

a  * [36a / 11.25] = 20

solve for a =15/6

c= 36a/11.25 = 36(15/6) /11.25 = 8

It's 10 to the power of 10?

https://file.coffee/eAPmrxE_a.png

What are: 1010, 1212, 1515 ??? Are they supposed to be: 10, 12, 15?

Without the workings, this is fucking useless!