Answer:
1) ∫ x² e^(x) dx
4) ∫ x cos(x) dx
Step-by-step explanation:
To solve this problem, eliminate the choices that can be solved by substitution.
In the second problem, we can say u = x², and du = 2x dx.
∫ x cos(x²) dx = ∫ ½ cos(u) du
In the third problem, we can say u = x², and du = 2x dx.
∫ x e^(x²) dx = ∫ ½ e^(u) du
Answer:
$32,000
Step-by-step explanation:
Yearly investment = $3,000
Number of years = 20
Investments earned an average of 8 percent a year.
Total invested amount = $3,000 × 20 = $60,000
Current value of investment = $92,000.
We need to find the total earnings on the investment.
Earnings = Current value of investment - Total invested amount
Earnings = $92,000 - $60,000
Earnings = $32,000
Therefore, the total earnings is $32,000.
(88sin20)(t - 16t²) = 0
(8(0.3090169944)(t - 16t²) = 0
(2.472135955)(t - 16t²) = 0
2.472135955t - 39.55417528t² = 0
Answer:
case 2 with two workers is the optimal decision.
Step-by-step explanation:
Case 1—One worker:A= 3/hour Poisson, ¡x =5/hour exponential The average number of machines in the system isL = - 3. = 4 = lJr machines' ix-A 5 - 3 2 2Downtime cost is $25 X 1.5 = $37.50 per hour; repair cost is $4.00 per hour; and total cost per hour for 1worker is $37.50 + $4.00
= $41.50.Downtime (1.5 X $25) = $37.50 Labor (1 worker X $4) = 4.00
$41.50
Case 2—Two workers: K = 3, pl= 7L= r= = 0.75 machine1 p. -A 7 - 3Downtime (0.75 X $25) = S J 8.75Labor (2 workers X S4.00) = 8.00S26.75Case III—Three workers:A= 3, p= 8L= ——r = 5- ^= § = 0.60 machinepi -A 8 - 3 5Downtime (0.60 X $25) = $15.00 Labor (3 workers X $4) = 12.00 $27.00
Comparing the costs for one, two, three workers, we see that case 2 with two workers is the optimal decision.
