The base case is the claim that
which reduces to
which is true.
Assume that the inequality holds for <em>n</em> = <em>k </em>; that
We want to show if this is true, then the equality also holds for <em>n</em> = <em>k</em> + 1 ; that
By the induction hypothesis,
Now compare this to the upper bound we seek:
because
in turn because
So first, you have to get two of the same variables to cancel out. Let's do this for x. In order for the x's to cancel out, we could multiply the bottom problem by 2.
(2) 3x-6y=24
After multiplying all the numbers by 2, you get the equation 6x-12y=48
The set of equations is now
-6x+2y=12
6x-12y=48
Now you can add them. The x variables cancel out, so you are left with the y variable.
2y+-12y=-10y and 12+48=60
Then you would divide 60 by -10 to get y=-6.
You would plug the answer for y into one of the original equations, lets do the top one. -6x+2y=12 becomes -6x+2(-6)=12
You'd multiply the 2 and -6 to get -12 so the equation is
-6x-12=12
The negative 12 turn positive and you add to both sides to get the -6x alone.
-6x-12=12
+12=12
-6x=24
Then divide 24 by -6
X=4
(-4,-6) is your final answer.
Its D, incenter is the incircle of a triangle or other figure.
Hello :
by moivre theorem :
[2(cos15*+isin15*)]^3 = 2^3(cos(3×45*)+isin(3×45*))
=8 (cos45*+isin45*)
= 8 ( √2/2 +i √2/2)
= 4√2 + 4√2 i ..... ( <span>in standard form a+bi)
</span>a = b = 4√2
I think it would be A) The coefficient of the x^2 term is zero.
