f(x)= x^2 -6x +21
0 =x^2-6x+21
subtract 21 from each side to prepare for completing the square
-21 = x^2 -6x
completing the square (b/2)^2
(-6/2) ^ =3^2 = 9
add 9 to each side
-21 +9 = x^2 - 6x + 9
-21 +9 =(x-3) ^2
-12 = (x-3)^2
take the square root of each side
+- sqrt(-12) = x-3
add 3 to each side, take out the negative inside the square root
3 +- i sqrt(12)=x
simplify the square root sqrt(12)= sqrt(4)sqrt(3) =2sqrt(3)
3 +- i 2sqrt (3) = x
Answer: 3+2isqrt(3)
3-2isqrt(3)
Answer:
![\begin{aligned}\bullet\ &f(1)=800;f(n)=f(n-1)+900, \text{for $n\ge 2$}\\ \bullet\ & f(n)=900n-100\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7D%5Cbullet%5C%20%26f%281%29%3D800%3Bf%28n%29%3Df%28n-1%29%2B900%2C%20%5Ctext%7Bfor%20%24n%5Cge%202%24%7D%5C%5C%20%5Cbullet%5C%20%26%20f%28n%29%3D900n-100%5Cend%7Baligned%7D)
Step-by-step explanation:
See attachment for the figure.
Using arithmetic sequence with a first term of 800 and a common difference of 900. The general form for such a sequence is given by,
an = a1 +d(n -1)
an = 800 +900(n -1) = 900n -100
If n is the function, this can be written as,
f(n) = 900n -100
When considered as a recursive relation, we find the first term is still 800:
f(1) = 800
and that each term is 900 more than the previous one:
f(n) = f(n-1) +900 . . . . for n ≥ 2
You need to consider that huge numbers of the different answer decisions are debasements of either of these structures, so you should look at them cautiously.
Ratio = 40 : 166 = 20 : 83
Answer: 20 : 83
Answer:
Both answers are C!!!
Step-by-step explanation:
Im so lucky, I guessed and got them both right on the usatestpreptest.
answer 1: petri dish A will have a density of 19.6
answer 2: Petri dish A by 3.3 bacteria/mm