Y = 1 + i
<span>(1 + i)^3 - 3 * (1 + i)^2 + k - 1 = -i </span>
<span>(1 + 3i + 3i^2 + i^3) - 3 * (1 + 2i + i^2) + k - 1 = -i </span>
<span>1 + 3i - 3 - i - 3 - 6i + 3 + k - 1 = -i </span>
<span>1 - 3 - 3 + 3 - 1 + 3i - i - 6i + k = -i </span>
<span>-3 - 4i + k = -i </span>
<span>k = 4i - i + 3 </span>
<span>k = 3i + 3 </span>
<span>k = 3 * (1 + i) </span>
<span>k = 3y</span>
Answer:
Step-by-step explanation:
x < -12, so x+12 < 0
| x-(-12) | = | x+12 | = -x-12
Answer:

Step-by-step explanation:
we know that
The formula to calculate continuously compounded interest is equal to
where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest in decimal
t is Number of Time Periods
e is the mathematical constant number
we have
substitute in the formula above
solve for t
simplify
Apply ln both sides
Applying property of exponents
Remember that ln(e) =1

9t-7t-t^2
2t-t^2
Hope this helps :)