The solution set of IR - 31 = 10 is
1 answer:
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9514 1404 393
Answer:
- 459 mg
- about 20 hours
Step-by-step explanation:
The decay factor is 1 -25% = 0.75 per hour, so the exponential equation can be written ...
r(t) = 1450·0.75^t . . . . . milligrams remaining after t hours
__
a) After 4 hours, the amount remaining is ...
r(4) = 1450·0.75^4 ≈ 458.79 . . . mg
About 459 mg will remain after 4 hours.
__
b) To find the time it takes before the amount remaining is less than 5 mg, we need to solve ...
r(t) < 5
1450·0.75^t < 5 . . . . use the function definition
0.75^t < 5/1450 . . . . divide by 1450
t·log(0.75) < log(1/290) . . . . . take logarithms (reduce fraction)
t > log(1/290)/log(0.75) . . . . . divide by the (negative) coefficient of t
t > 19.708
It will take about 20 hours for the amount of the drug remaining to be less than 5 mg.
Answer:
(0, 3 ), (5, 0), (10, -3), (15, -6)
Step-by-step explanation:
3x − 5y = 15
-5y = 3x + 15
y =
Point 1: (0, 3 )
y = -3/5(0) + 3
y = 0 + 3
y = 3
Validate:
3 = -3/5(0) + 3
3 = 3
Point 2: (5, 0)
y = -3/5(5) + 3
y = -3 + 3
y = 0
Validate:
0 = -3/5(5) + 3
0 = 0
Point 3: (10, -3)
y = -3/5(10) + 3
y = -6 + 3
y = -3
Validate:
-3 = -3/5(10) + 3
-3 = -3
Point 4: (15, -6)
y = -3/5(15) + 3
y = -9 + 3
y = -6
Validate:
-6 = -3/5(15) + 3
-6 = -6