So you need to come up with a perfect square that works for the x coefficients.
like.. (2x + 2)^2
(2x+2)(2x+2) = 4x^2 + 8x + 4
Compare this to the equation given. Our perfect square has +4 instead of +23. The difference is: 23 - 4 = 19
I'm going to assume the given equation equals zero..
So, If we add subtract 19 from both sides of the equation we get the perfect square.
4x^2 + 8x + 23 - 19 = 0 - 19
4x^2 + 8x + 4 = - 19
complete the square and move 19 over..
(2x+2)^2 + 19 = 0
factor the 2 out becomes 2^2 = 4
ANSWER: 4(x+1)^2 + 19 = 0
for a short cut, the standard equation
ax^2 + bx + c = 0 becomes a(x - h)^2 + k = 0
Where "a, b, c" are the same and ..
h = -b/(2a)
k = c - b^2/(4a)
Vertex = (h, k)
this will be a minimum point when "a" is positive upward facing parabola and a maximum point when "a" is negative downward facing parabola.
Arc length (L) = 83.7758 ft
33 plus 4 is 37 and 10plus 1 is 12 so add that and get 49
Answer:
Step-by-step explanation:
We want to differentiate the equation:
With respect to <em>t</em>, where <em>x, y, </em>and <em>z</em> are functions of <em>t. </em>
<em />
So:
![\displaystyle \frac{d}{dt}\left[x^2-y^3+z^4\right]=\frac{d}{dt}\left[1\right]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft%5Bx%5E2-y%5E3%2Bz%5E4%5Cright%5D%3D%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft%5B1%5Cright%5D)
Implicitly differentiate on the left. On the right, the derivative of a constant is simply zero. Hence:
Answer:
20x^50
Step-by-step explanation:
