Given a polynomial p(x) with leading coefficient
and solutions
, we have

That's why we have to find the solutions first. We complete the square by adding 3 to both sides:

From here, we continue by taking the square root of both sides:

So, the two solutions are

Which yields the factorization

let's firstly convert the mixed fractions to improper fractions and then get their difference.
![\stackrel{mixed}{8\frac{7}{8}}\implies \cfrac{8\cdot 8+7}{8}\implies \stackrel{improper}{\cfrac{71}{8}} ~\hfill \stackrel{mixed}{6\frac{3}{4}}\implies \cfrac{6\cdot 4+3}{4}\implies \stackrel{improper}{\cfrac{27}{4}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{71}{8}-\cfrac{27}{4}\implies \cfrac{1(71)~~ -~~2(27)}{\underset{\textit{using this LCD}}{8}}\implies \cfrac{71-54}{8}\implies \cfrac{17}{8}\implies 2\frac{1}{8}](https://tex.z-dn.net/?f=%5Cstackrel%7Bmixed%7D%7B8%5Cfrac%7B7%7D%7B8%7D%7D%5Cimplies%20%5Ccfrac%7B8%5Ccdot%208%2B7%7D%7B8%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B71%7D%7B8%7D%7D%20~%5Chfill%20%5Cstackrel%7Bmixed%7D%7B6%5Cfrac%7B3%7D%7B4%7D%7D%5Cimplies%20%5Ccfrac%7B6%5Ccdot%204%2B3%7D%7B4%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B27%7D%7B4%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B71%7D%7B8%7D-%5Ccfrac%7B27%7D%7B4%7D%5Cimplies%20%5Ccfrac%7B1%2871%29~~%20-~~2%2827%29%7D%7B%5Cunderset%7B%5Ctextit%7Busing%20this%20LCD%7D%7D%7B8%7D%7D%5Cimplies%20%5Ccfrac%7B71-54%7D%7B8%7D%5Cimplies%20%5Ccfrac%7B17%7D%7B8%7D%5Cimplies%202%5Cfrac%7B1%7D%7B8%7D)
Answer:
Suppose the universal set is U = all whole numbers from 1 to 9. If A = {1, 2, 4}, then Ac = {3, 5, 6, 7, 8, 9}. As we saw earlier with the expression Ac ...
Let U be the set of whole numbers from 5-10 inclusive. Set A={5, 7, 9 ...
Answer:
Ok so I've had braces and it doesn't really hurt getting put in but it does hurt when they tighten them but you have nothing to worry about :)
Step-by-step explanation:
The number of defective modems in the inventory is 20%⋅ 30 + 8%⋅ 50 = 10 (out of 80).
Note that the number of defectives in the inventory is fixed, i.e., we are not told that there
is 1
8 probability that a modem in the inventory is defective, but rather that exactly 1
8
of
all modems are defective. The probability that exactly two modems in a random sample
of five are defective is = 0.102