We have that
<span>(c-4)/(c-2)=(c-2)/(c+2) - 1/(2-c)
</span>- 1/(2-c)=-1/-(c-2)=1/(c-2)
(c-4)/(c-2)=(c-2)/(c+2)+ 1/(c-2)------- > (c-4)/(c-2)-1/(c-2)=(c-2)/(c+2)
(c-4-1)/(c-2)=(c-2)/(c+2)---------------- > (c-5)/(c-2)=(c-2)/(c+2)
(c-5)/(c-2)=(c-2)/(c+2)------------- > remember (before simplifying) for the solution that c can not be 2 or -2
(c-5)*(c+2)=(c-2)*(c-2)------------------ > c²+2c-5c-10=c²-4c+4
-3c-10=-4c+4----------------------------- > -3c+4c=4+10----------- > c=14
the solution is c=14
the domain of the function is (-∞,-2) U (-2,2) U (2,∞) or
<span>all real numbers except c=-2 and c=2</span>
Answer:its 0.58
Step-by-step explanation:
Answer:
0.75 mg
Step-by-step explanation:
From the question given above the following data were obtained:
Original amount (N₀) = 1.5 mg
Half-life (t₁/₂) = 6 years
Time (t) = 6 years
Amount remaining (N) =.?
Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:
Half-life (t₁/₂) = 6 years
Time (t) = 6 years
Number of half-lives (n) =?
n = t / t₁/₂
n = 6/6
n = 1
Finally, we shall determine the amount of the sample remaining after 6 years (i.e 1 half-life) as follow:
Original amount (N₀) = 1.5 mg
Half-life (t₁/₂) = 6 years
Number of half-lives (n) = 1
Amount remaining (N) =.?
N = 1/2ⁿ × N₀
N = 1/2¹ × 1.5
N = 1/2 × 1.5
N = 0.5 × 1.5
N = 0.75 mg
Thus, 0.75 mg of the sample is remaining.
