6m3 - 16m2 + 15m - 40<span> Simplify —————————————————————
2m2 + 5
</span>Checking for a perfect cube :
<span> 4.1 </span> <span> 6m3 - 16m2 + 15m - 40</span> is not a perfect cube
Trying to factor by pulling out :
<span> 4.2 </span> Factoring: <span> 6m3 - 16m2 + 15m - 40</span>
Thoughtfully split the expression at hand into groups, each group having two terms :
Group 1: 15m - 40
Group 2: <span> -16m2 + 6m3</span>
Pull out from each group separately :
Group 1: (3m - 8) • (5)
Group 2: <span> (3m - 8) • (2m2)</span>
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Add up the two groups :
<span> (3m - 8) • </span><span> (2m2 + 5)</span>
<span>Which is the desired factorization</span>
<span>3m-8 is the answer</span>
Answer: Yes, it means 3 times the square root of 2. First, find the square root of 2. Then multiply that value by 3. You can also type it into a calculator, and it will give you the exact value.
Answer:
Step-by-step explanation:
<u>The side of the square is w/2, then its area is:</u>
<u>The shaded area is the difference of areas of the rectangle and the square:</u>
<u>This can be simplified:</u>
- 56w - w² = 336
- w² - 56w + 336 = 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.