You have to complete the square to get it into vertex form. Do this by setting the function equal to zero and at the same time moving the constant over to the other side of the equals sign so you have this:

. Now we can complete the square on the polynomial by taking half the linear term, squaring it, and adding it to both sides. Our linear term is 4. Half of 4 is 2, and 2 squared is 4, so we add 4 to both sides:

and simplify to get

. In this process we have created a perfect square binomial on the left, which happens to be

. Now move the -6 back over by addition to get

. The vertex is found at (-2, 6), the third choice down.
Answer:
<h2>
AC = 36.01</h2>
Step-by-step explanation:
Given ΔABC and ΔADB, since both triangles are right angled triangles then the following are true.
From ΔADB, AB² = AD²+BD²
Given AB = 24 and AD = 16
BD² = AB² - AD²
BD² = 24²-16²
BD² = 576-256
BD² = 320
BD = 
BD = 17.9
from ΔABC, AC² = AB²+BC²
SInce AC = AD+DC and BC² = BD² + DC² (from ΔBDC )we will have;
(AD+DC)² = AB²+ (BD² + DC²)
Given AD = 16, AB = 24 and BD = 17.9, on substituting
(16+DC)² = 24²+17.9²+ DC²
256+32DC+DC² = 24²+17.9²+ DC²
256+32DC = 24²+17.9²
32DC = 24²+17.9² - 256
32DC = 640.41
DC = 
DC = 20.01
Remember that AC = AD+DC
AC = 16+20.01
AC = 36.01
It wants expanded form and the distributive property (something like 2(5+6b), so I would think it wants something like 7(200+50+6). This way, you're multiplying 7 by each expanded part of the number, which you would do anyway when solving like normal. Please correct me if I'm wrong!
Answer:
A
Step-by-step explanation:
I am not sure though. Sorry if it is wrong :(
$20
Step-by-step explanation:
I'm high a.f. but I'm certain it's $20