(6+6+6)(36+1) = (18)(37) = 666. ... If you don't want an answer posted, there's a simple solution ... don't post the question ! And if it still mysterously somehow appears, then for heaven's sake, don't offer points as a reward for an answer. ... Finally ... devil's numbers belong in the Old Wives' Hooey category, not in the Mathematics one. They're not real.
10x^6y^3 + 20x^3y^2 / 5x^3y =
5x^3y (2x^3y^2 + 4y) / 5x^3y=
<span>2x^3y^2 + 4y
</span>
First you find what the expressions 10x^6y^3 and 20x^3y^2 have in common, and that is 5x^3y, which is the same as the denominator, so you can easily divide them and get 1.
<span>Hope you understand!
</span>
Answer:
.
Step-by-step explanation:
Let the 
-coordinate of 
be 
. For 
to be on the graph of the function 
, the 
-coordinate of 
would need to be 
. Therefore, the coordinate of 
would be 
.
The Euclidean Distance between and 
is:
.
The goal is to find the a that minimizes this distance. However, 
is non-negative for all real 
. Hence, the 
that minimizes the square of this expression, 
, would also minimize 
.
Differentiate with respect to :
![\displaystyle \frac{d}{dt}\left[t^2 - 17 \, t + 81\right] = 2\, t - 17](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft%5Bt%5E2%20-%2017%20%5C%2C%20t%20%2B%2081%5Cright%5D%20%3D%202%5C%2C%20t%20-%2017)
.
![\displaystyle \frac{d^{2}}{dt^{2}}\left[t^2 - 17 \, t + 81\right] = 2](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%5E%7B2%7D%7D%7Bdt%5E%7B2%7D%7D%5Cleft%5Bt%5E2%20-%2017%20%5C%2C%20t%20%2B%2081%5Cright%5D%20%3D%202)
.
Set the first derivative, 
, to 
and solve for :
.
.
Notice that the second derivative is greater than for this . Hence, would indeed minimize . This value would also minimize , the distance between and .
Therefore, the point would be closest to when the -coordinate of is .
The answer is:
17 + 6c + d
You can combine the 23 with the -6 and the 8d with the -7d
