CPhill

The light we see now from the star left that star 900 years ago......I'll let you figure out the year...

We have

100N =N^2 / 4   multiply both sides by 4

400N = N^2       rearrange

N^2 - 400N = 0   factor

N(N - 400) = 0     and setting both factors to 0, we have that  N = 0 or N = 400

Yeah....

Any value between 6 and 7 would work....

So we have

6 < x < 7

None.....to see why.....note that the only possible solution would have to lie between 7 and 8 or between 8 and 9......but, if r lies between 7 and 8, the answer will be less than 15.5, while, if r lies between 8 and 9, the answer will be greater than 15.5....

[I believe this is correct...!!!]

The circumference of a circle is given by pi*d...where d is the diameter...so

53 * ( pi * 837.0069km )  will give you your answer....

Thanks, geno....I'll recommend this one for Melody's Daily Wrap...!!!!

Just playing around with this in WolframAlpha, I noticed that...

floor(x)*(x) = 91 returns the solution  -91/10

but

floor(x)*(x) = 101  returns the solution 101/10

and

floor(x)*(x) = 122  returns the solution  122/11

but

floor(x)*(x) = 143  returns the solution  -143/12

I wonder if there's some rationale for the fact that, depnding on the number on the RHS of the equation, the parity of the solutions seems to "oscillate" ???  .... anybody ????

It is called the "radicand."

√[(-33--18)^2+(5--6)^2] =

√[-15)^2 + (11)^2] =

√(225 + 121) =

√346 = about 18.6

a*b=a-b+a-b. How much larger is 12*3 than 10*5?

Using the relation as defined, we have

12 - 3 + 12 - 3  = 18    and    10 - 5 + 10 - 5  = 10

So 12*3 is "greater" than 10*5  by 8......