CPhill

(1/9)^3  =  27 * 9^n

9^(-3)  = 3^3 * 9^n

(3^2)^(-3)  = 3^3 * ( 3^2)^n

3^ (-6) = 3^3 * 3^(2n)

3 ^ (-6)  =  3^ ( 3 + 2n)

Solve for the exponents

-6  = 3 + 2n

-6 - 3  = 2n

-9 = 2n

n =  -9/2

Hard to see this but I guess you want to prove that the left side = the right side....if so....

cos (pi/2 - x)  =

cos (pi/2) cos (x)  + sin (pi/2) sin (x)  =

0 *  cos (x)  +  1 * sin (x)   =

sin x

Although  I don't know the answer to this.....here is a  very similar question answered by amy :

https://web2.0calc.com/questions/help-is-appreciated-thank-you

I think that you only need to change a few numbers and  then  graph the inequalities.....it should give  you a trianglular area that you can  calculate  to the total area

(a + b)  = 69          square both sides    a^2 + 2ab + b^2  =  4761  ⇒  2ab = 4761 - a^2 - b^2  ⇒

2ab = 4761 - (a^2 + b^2) ⇒

2ab = 4761  - ( 1944)

2ab =  2817

ab = 2817/2

Note that  a^3 +b^3    =    (a + b) ( a^2  - ab   + b^2)

So  we have 

a^3 + b^3  =  ( a + b ) ( [ a^2 + b^2 ] - ab )   =

(69) [ 1944 - 2817/2 ]  =  

73899 / 2

300 factors as     5^2 * 2^2 * 3

I see the possible identifiable sequences as

5 * 5 * 2 * 2 * 3 =    5!/(2! * 2!)  = 30

5 * 5 * 4 * 3 * 1 =    5! / 2! =  60

5 * 5  * 6 * 2 * 1 =  5! / 2! =  60

I get  150

Anyone else ???

3 - 5 - 7  by the triangle inequality works

Looks good, qjin  !!!

You are very good at this  counting stuff......I'm usually stuck on many  of these  !!!

2x^(1/2)  = 12 / x^(1/2)  - 5                multiplying through by x^(1/2)    we get

2x = 12 - 5x^(1/2)

2x - 12  =  - 5x^(1/2)           square both sides

4x^2 - 48x + 144  =  25x

4x^2 - 73x + 144  = 0        factor  as

(4 x - 9) ( x - 16)  = 0

Setting both factors to 0  and  solving for  x we get that

x = 9/4   or  x  =16

Which is true  ????

If x = 9/4  then x^(1/2)  = 3/2

2 (3/2) =  12 / (3/2) - 5

3  =  8 - 5       true

If x = 16  then x^(1/2)  = 4

2(4)  = 12 / 4 - 5

8 = 3 - 5    not  true

The answer is   x =   9/4

We get the extraneous answer  of x =16  because of squaring......this often happens   !!!!

1/x + 1/y  =  1/3     ⇒   (x + y) / xy  =   1/3   ⇒  (x + y)  = (1/3)xy  ⇒ 3(x + y) = xy

And

xy + ( x + y) = 4

3xy + 3(x + y) = 12

3xy + xy = 12

4xy = 12

xy  = 3

Since (x +y) / xy  =  1/3   this implies that  (x + y)   =1

So

x^2y  + y^2x   =

xy ( x + y)   =

3 ( 1)   = 3