CPhill

If it decreased by 5%, then 741 is 95% of what it used to be.......so....

741 / .95 =  780    ....and that's what it was in 2011

OK...I see the "ice cream" thing at the bottom of TR's post......hey.....bring me back some vanilla, will ya'  ????

Mmmm...OK, BJ.......  I'll go with "hamburgers".....BTW....what was the question, again ???????

For the first one, we're just putting (x - 1)   into  x^2+x....so we have

(x - 1)^2 + (x - 1)   =   (x^2 - 2x + 1) + (x - 1)  =   x^2 - x

For the second, we're just multiplying f(x) by 2, so ......    2 [ x^2 + x ] = 2x^2 + 2x

Both legs = 26/√2

If, for some k > 1, 2^k - 1 is prime then 2^k-1(2^k - 1) is a perfect number

For instance, the first perfect number, 6   =  2^(2 - 1) (2^2  - 1)  =  2^1 ( 3)   = 6

So far, only even perfect numbers have been found......no one knows if there are any odd ones...if there are, the above formula would not applly......[it produces only even numbers ]

20√7 * 7√8  =     140√56  = 140√[4 * 14 ] =  140√4*√14  = 280√14  = about 1047.66 sq units

This is asking to solve this system.....   6x+2y=2    5x-7y=6      ....divide the first equation by 2.... this gives 3x + y = 1    rearrange       y = 1 - 3x    .......and subbing this into the second equation, we have ......

5x - 7(1 - 3x) = 6     simplify

5x - 7 + 21x  = 6       add  7 to both sides and simplify the left hand side

26x = 13       divide both sides by 26

x = 13/26 = 1/2 and   y =  1 - 3(1/2)   =   1 - 3/2  = -1/2

If we're taking about segment XY, its perpendicular bisector intersects it at the point (1, 3)

If we're talking about  segment XZ, using the mid-point formula, its perpendicular bisector intersects it at the point (7/2, 3)

And for segment YZ, the point of intersection would be (7/2, 1)

A 100% increase doubles it

A 200% increase triples it

A 300% increase quadruples it.......