![\bf ~~~~~~\textit{parabola vertex form} \\\\ y=a(x- h)^2+ k~\hspace{7em}vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{f(x)}{y}=(x+1)^2-8\implies y=1[x-(\stackrel{h}{-1})]^2+(\stackrel{k}{-8})~\hfill \stackrel{vertex}{(-1,-8)}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Bparabola%20vertex%20form%7D%0A%5C%5C%5C%5C%0Ay%3Da%28x-%20h%29%5E2%2B%20k~%5Chspace%7B7em%7Dvertex~~%28%5Cstackrel%7B%7D%7B%20h%7D%2C%5Cstackrel%7B%7D%7B%20k%7D%29%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A%5Cstackrel%7Bf%28x%29%7D%7By%7D%3D%28x%2B1%29%5E2-8%5Cimplies%20y%3D1%5Bx-%28%5Cstackrel%7Bh%7D%7B-1%7D%29%5D%5E2%2B%28%5Cstackrel%7Bk%7D%7B-8%7D%29~%5Chfill%20%5Cstackrel%7Bvertex%7D%7B%28-1%2C-8%29%7D)
the equation is in x-terms, meaning is a vertical parabola, and therefore the axis of symmetry will be the x-coordinate of the vertex, namely x = -1.
Literally just divide are you slow
Answer:
(x-12)(x+7)
Step-by-step explanation:
x^2 - 5x - 84
You find the factors of 84:
1,84; 2,42; 3,28; 4,21; 6,14; 7,12
7 and 12 are 5 apart, so we use them:
(x - 12)(x + 7)
The 12 is negative because it's greater than 7, and the coefficient of x is negative 5.
Wellu needa big d1ck ya fa4ggot
Answer:
Given differential equation,
![\frac{dy}{dx}+5y=7](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%2B5y%3D7)
![\frac{dy}{dx}=7-5y](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%3D7-5y)
![\implies \frac{dy}{7-5y}=dx](https://tex.z-dn.net/?f=%5Cimplies%20%5Cfrac%7Bdy%7D%7B7-5y%7D%3Ddx)
Taking integration both sides,
![\int \frac{dy}{7-5y}=\int dx](https://tex.z-dn.net/?f=%5Cint%20%5Cfrac%7Bdy%7D%7B7-5y%7D%3D%5Cint%20dx)
Put 7 - 5y = u ⇒ -5 dy = du ⇒ dy = -du/5,
![-\frac{1}{5} \int \frac{du}{u} = \log x + C](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B5%7D%20%5Cint%20%5Cfrac%7Bdu%7D%7Bu%7D%20%3D%20%5Clog%20x%20%2B%20C)
![-\frac{1}{5} \log u = \log x + C](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B5%7D%20%5Clog%20u%20%3D%20%5Clog%20x%20%2B%20C)
![-\frac{1}{5}\log(7-5y) = \log x + C---(1)](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B5%7D%5Clog%287-5y%29%20%3D%20%5Clog%20x%20%2B%20C---%281%29)
Here, x = 0, y = 0
![\implies -\frac{1}{5} \log 7= C](https://tex.z-dn.net/?f=%5Cimplies%20-%5Cfrac%7B1%7D%7B5%7D%20%5Clog%207%3D%20C)
Hence, from equation (1),
![-\frac{1}{5}\log(7-5y)=\log x -\frac{1}{5}log 7](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B5%7D%5Clog%287-5y%29%3D%5Clog%20x%20-%5Cfrac%7B1%7D%7B5%7Dlog%207)
![\log(7-5y)=\log (\frac{x}{7^\frac{1}{5}})](https://tex.z-dn.net/?f=%5Clog%287-5y%29%3D%5Clog%20%28%5Cfrac%7Bx%7D%7B7%5E%5Cfrac%7B1%7D%7B5%7D%7D%29)
![7-5y=\frac{x}{7^\frac{1}{5}}](https://tex.z-dn.net/?f=7-5y%3D%5Cfrac%7Bx%7D%7B7%5E%5Cfrac%7B1%7D%7B5%7D%7D)
![7-\frac{x}{7^\frac{1}{5}}=5y](https://tex.z-dn.net/?f=7-%5Cfrac%7Bx%7D%7B7%5E%5Cfrac%7B1%7D%7B5%7D%7D%3D5y)
![\implies y=\frac{1}{5}(7-\frac{x}{7^\frac{1}{5}})](https://tex.z-dn.net/?f=%5Cimplies%20y%3D%5Cfrac%7B1%7D%7B5%7D%287-%5Cfrac%7Bx%7D%7B7%5E%5Cfrac%7B1%7D%7B5%7D%7D%29)