Answer:
55 is the answer
Step-by-step explanation:
180-76-49= 55 !!!!!
Answer:
1.
<u>An extraneous solution is a root of a transformed equation that is not a root of the original equation as it was excluded from the domain of the original equation.</u>
It emerges from the process of solving the problem as a equation.
2.I begin like:
The vertical asymptotes will occur at those values of x for which the denominator is equal to zero:
for example:
x² − 4=0
x²= 4
doing square root on both side
x = ±2
Thus, the graph will have vertical asymptotes at x = 2 and x = −2.
To find the horizontal asymptote, the degree of the numerator is one and the degree of the denominator is two.
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
$x-2>\frac{4}{5}\quad:\quad x>\frac{14}{5}$
$\mathrm{Combine\:the\:intervals}$
$x<\frac{6}{5}\quad\mathrm{or}\quad\:x>\frac{14}{5}$
(t-2)(t+1)(t+1)
t^2+t-2t-2(t+1)
t^2-t-2(t+1)
t^3-t^2-2t+t^2-t-2
t^3+-3t-2
Answer:
Your score is 100.
Step-by-step explanation:
If you have -300 and add +300 you will end up with 0 because when a positive and a negative that are equal cancel each other out. So if you add +300 to -300 you get 0 but its 400 so 300 + 100 is 400 so you have +100 left. : )
