P would equal negative three.
Answer:
Option A. √(x + 1)
Step-by-step explanation:
Data obtained from the question include:
f(x) = √(x² – 1)
g(x) = √(x – 1)
(f/g) (x) =..?
(x² – 1) => difference of two square
(x² – 1) => (x – 1)(x + 1)
f(x) = √(x² – 1)
f(x) = √(x – 1)(x + 1)
(f/g) (x) = f(x) /g(x)
f(x) = √(x – 1)(x + 1)
g(x) = √(x – 1)
(f/g) (x) = √(x – 1)(x + 1) / √(x – 1)
(f/g) (x) = √[(x – 1)(x + 1) / (x – 1)]
(f/g) (x) = √(x + 1)
It’s a diagram ur welcome
Answer:
x<6/5, x>14/5
Step-by-step explanation:
Steps
$5\left|x-2\right|+4>8$
$\mathrm{Subtract\:}4\mathrm{\:from\:both\:sides}$
$5\left|x-2\right|+4-4>8-4$
$\mathrm{Simplify}$
$5\left|x-2\right|>4$
$\mathrm{Divide\:both\:sides\:by\:}5$
$\frac{5\left|x-2\right|}{5}>\frac{4}{5}$
$\mathrm{Simplify}$
$\left|x-2\right|>\frac{4}{5}$
$\mathrm{Apply\:absolute\:rule}:\quad\mathrm{If}\:|u|\:>\:a,\:a>0\:\mathrm{then}\:u\:<\:-a\:\quad\mathrm{or}\quad\:u\:>\:a$
$x-2<-\frac{4}{5}\quad\mathrm{or}\quad\:x-2>\frac{4}{5}$
Show Steps
$x-2<-\frac{4}{5}\quad:\quad x<\frac{6}{5}$
Show Steps
$x-2>\frac{4}{5}\quad:\quad x>\frac{14}{5}$
$\mathrm{Combine\:the\:intervals}$
$x<\frac{6}{5}\quad\mathrm{or}\quad\:x>\frac{14}{5}$
Hi there’s some math apps for this
