CPhill

888x   +  889y  =   890

891x   +  892y  =   893

Subtract  the  first equation from the second

3x  + 3y   =  3

x +  y   =  1       ⇒   y =  1 - x

So....using the first equation

888x  +  889 (1 - x)  =  890

888x  +  889  -  889x  =  890

-x  =  1

x  =  - 1

So.....   y  =  1 - (-1)  =    2

So

x  -  y   =    -1  -  2   =    -3

Thanks, hectictar....

Note that  in the "formula"  provided by MIRB, 3, 4, 5 , 6, 9 and 10  all  divide 180,  while ( n - 2) will always be an integer

So...we only need to check the values  n  = 7  and n  = 8

( 7 - 2) * 180 / 7  =   (5/7)*180  ≈  128.57°

(8 - 2) * 180  / 8   =   (3/4)* 180  =  135°

So.....a regular heptagon [ 7 sides ]  will not have integer-valued interior angle measures

7^3x  = 400    take the log of both sides

log 7 ^3x  =  log 400      and we can write

3x  *  log 7  =  log 400          divide both sides by 3log 7

x  =  log (400)  / [ 3log (7)]  ≈ 1.026  =  1.03

ln  (4)  + ln (3x)  = 2     we can write

ln  ( 4 * 3x)  = 2

ln (12x )   =  2

This says that

e^2  =  12x            divide both sides by 12

e^2  / 12 =   x   ≈  .615  =  .62

x^2  + 7x  +  20  =  0

The sum of the roots  =  -7/1  =  -7

The product of the roots is  20/1  = 20

So.....this implies that in  Ax^2 + Bx + 1  = 0

-B/A  =  -7    ⇒  B/A  = 7 ⇒  B =  7A

And

1/A  = 20  ⇒  A =  1/20

So..... B  = 7(1/20)  = 7/20

So.....  A + B   =    1/20  + 7/20  =  8/20   =  2/5

Happy (belated) Birthday, Geno.....!!!

In the form  ax^2  +  bx  + c  =  0

The sum  of  the  roots will be  = -b/a

And the product of the roots will be  = c/a

Sum  of the roots   =   2m

Product of the roots  =  m^2  + n^2

So

2m  =  6/2   ⇒   m  =  3/2

And

m^2  + n^2  =  5/2    ⇒   (9/4) + n^2  =  5/2  ⇒  n^2  = 10/4 - 9/4  =  1/4  ⇒  n  =  1/2

So

m * n   =   (3/2)(1/2)  =  3/4

Notice the pattern

7^3  =  343

7^4  = 2401

7^5  =  16807

7^6  =  117649

7^7 = 823543

7^8  =  5764801

7^9 = 40353607

7^10  = 282475249

So...it appears that  for n ≥ 0   where n is an integer

7^(4n + 3)   has 4 as a tens digit

7^(4n + 4)  has 0  as a tens digit

7^(4n + 5)  has 0 as a tens digit

7^(4n + 6 )  has 4 as a tens digit

So

7^2005   =   7^ ( 4(500) + 5)  ⇒   has  0  as a tens digit

a)  sin θ + cos θ   =   0          square both sides

sin^2 θ  +  2sin θ  cos θ   +  cos^2 θ  =    0

1  +  2sinθcosθ   =  0

2sinθcosθ    =   - 1

sin(2θ)  =  -1

let 2θ  = x

So

sin x  =  - 1        and this occurs at  x =  270°  and at  x  = 630°

So

2θ  =  270°            and       2θ  =    630°

So

θ  =  135°        and   θ  =   315°

We can equate the area as follows

(1/2)base * height =  A  ⇒  65 *  height  =  A      (1)  

120 * (1/2)side  = A   ⇒  60 * side  =  A    (2)

Set (1)  = (2)      and solve for the side

60 * side  =  65 * height

side  = (65/60) * height

side  =  (13/12)* height  =   (13/12)*h

The side  is the hypotenuse of a right triangle with the height and 1/2 base being the legs

So.......Using the Pythagorean Theorem, we have that

sqrt  [  side^2  -  height^2 ]   =  1/2 base

Substitute

sqrt  [  ( (13/12)h)^2  - h^2]  =  1/2 base        

sqrt  [  (169/ 144)h^2 -  (144/144)h^2 ]  =  1/2 base

sqrt [  (169 - 144) h^2    / 144 ]   =  1/2 base

sqrt [   25h^2 / 144 ]  =  1/2 base

(h / 12)  sqrt (25)  =  1/2 base

(5/12)h  =  1/2 base

(5/12)h  =  65

h = (65)(12) / 5  

h  = 13 * 12  =   156  m

So....Area  = 

(1/2)base * height

(1/2)130m * 156m

65m * 156m  =   

10140 m^2

Thanks, Mr. Owl.......