Answer:
167.27 mg.
Step-by-step explanation:
We have been given that the half-life of Radium-226 is 1590 years and a sample contains 400 mg.
We will use half life formula to solve our given problem.
, where N(t)= Final amount after t years, 
= Original amount, t/2= half life in years.
Now let us substitute our given values in half-life formula.
Therefore, the remaining amount of Radium-226 after 2000 years will be 167.27 mg.
The way you do this is 5322÷5 which equals 1064.4
0.6 is your answer. Have a nice day :^)
Answer:
6 units
(Use the distance formula and plug in everything)
(a) Yes all six trig functions exist for this point in quadrant III. The only time you'll run into problems is when either x = 0 or y = 0, due to division by zero errors. For instance, if x = 0, then tan(t) = sin(t)/cos(t) will have cos(t) = 0, as x = cos(t). you cannot have zero in the denominator. Since neither coordinate is zero, we don't have such problems.
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(b) The following functions are positive in quadrant III:
tangent, cotangent
The following functions are negative in quadrant III
cosine, sine, secant, cosecant
A short explanation is that x = cos(t) and y = sin(t). The x and y coordinates are negative in quadrant III, so both sine and cosine are negative. Their reciprocal functions secant and cosecant are negative here as well. Combining sine and cosine to get tan = sin/cos, we see that the negatives cancel which is why tangent is positive here. Cotangent is also positive for similar reasons.
