X(30;0)
y(0;50)
substitue 0 for y and solve for x
substitute 0 in for x and solve for y
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
$x-2>\frac{4}{5}\quad:\quad x>\frac{14}{5}$
$\mathrm{Combine\:the\:intervals}$
$x<\frac{6}{5}\quad\mathrm{or}\quad\:x>\frac{14}{5}$
Answer:
1. y = sin(x²+C)
2. see below
Step-by-step explanation:
1. 
separation of variables
integration both sides
you should get :
remember constant of integration!!
2. y = sin(x²+C)
3 = sin(0+C)
y(0) = 3 does not have a solution because our sin graph is not shifted vertically or multiplied by a factor whose absolute value is greater than 1, so our range is [-1,1] and 3 is not part of this range
Thats when point =0. equation of a line mx +b where m is the slope
and b is the intercept
