Answer:
51 meters
Step-by-step explanation:
Steve is turning half his backyard into a chicken fan. His backyard is a 24 m x 45 m rectangle. He wants to put a chicken wire fence that stretches diagonally from one corner to the opposite corner. How many meters of fencing will Steve need?
We are to find the meters of fencing for the diagonal.
We solve the question using Pythagoras Theorem
= c² = a² + b²
Where
c = Diagonal
a = Width
b =Length
Diagonal² = Width² + Length ²
Hence:
Diagonal ² = 45² + 24²
Diagonal = √45² + 24²
Diagonal = √(2601)
Diagonal = 51 m
Therefore, the meters of fencing for the diagonal that Steve would be needing = 51 meters
Answer:
15cm
Step-by-step explanation:
First, a square's diagonal is basically the hypotenuse of a 45-45-90 triangle. a 45-45-90 triangle has a really special relationship, where the side length is x, and the diagonal is x 
. So, the side length is 15.
Consider number 17. You know that
17=9+8.
Then for negative numbers you have the same rule (with respect to sign -):
-17=(-9)+(-8).
Since (-9)+9=9+(-9)=9-9=0, you have that
-17+9=(-9)+(-8)+9=(-9+9)+(-8)=0+(-8)=-8.
Answer: -17=(-9)+(-8).
We can rewrite the expression under the radical as
then taking the fourth root, we get
![\sqrt[4]{\left(\dfrac32a^2b^3c^4\right)^4}=\left|\dfrac32a^2b^3c^4\right|](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%7B%5Cleft%28%5Cdfrac32a%5E2b%5E3c%5E4%5Cright%29%5E4%7D%3D%5Cleft%7C%5Cdfrac32a%5E2b%5E3c%5E4%5Cright%7C)
Why the absolute value? It's for the same reason that
since both 
and 
return the same number , and 
captures both possibilities. From here, we have
The absolute values disappear on all but the 
term because all of 
, 
and 
are positive, while 
could potentially be negative. So we end up with
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.
