Answer:
293.38 pounds
Step-by-step explanation:
We are given that
Distance between poles=35 feet
Weight of cable=10.4 per linear foot
We have to find the weight of the cable.
Differentiate w.r.t
Let 
![s=\frac{2}{0.0225}\times\frac{2}{3}[t^{\frac{3}{2}}]^{17.5}_{0}](https://tex.z-dn.net/?f=s%3D%5Cfrac%7B2%7D%7B0.0225%7D%5Ctimes%5Cfrac%7B2%7D%7B3%7D%5Bt%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%5D%5E%7B17.5%7D_%7B0%7D)
![s=2\times \frac{2}{3\times0.0225}[(1+0.0255x)^{\frac{3}{2}]^{17.5}_{0}](https://tex.z-dn.net/?f=s%3D2%5Ctimes%20%5Cfrac%7B2%7D%7B3%5Ctimes0.0225%7D%5B%281%2B0.0255x%29%5E%7B%5Cfrac%7B3%7D%7B2%7D%5D%5E%7B17.5%7D_%7B0%7D)
s=28.21
Weight of cable=
pound
A bearing of 34° corresponds to corresponding angle of θ=90-34=56°
The (x,y) values for the position of the ship after completing its first heading are:
x=(10.4cos 56)
y=(10.4sin56)
The trigonometric angle for θ=90-90=0
The (x,y) values for the postion of the ship after completing the second bearing is:
x=(10.4 cos56)+(4.6cos0)≈10.4 mi
y=(10.4sin56)+(4.6sin0)≈8.6mi
the distance from the port will therefore be:
d=√(10.4²+8.6²)≈13.5 miles
It trigonometric angles is:
θ=arctan(y/x)
θ=arctan(8.6/10.4)
θ≈39.6°
Thus the bearing angle is:
90-39.6=50.4°
Answer:
x= 10
I hope i helped, best of luck!
It y=4z-3/6 BOOMM done for ya
Answer:
$30
Step-by-step explanation:
Calculation for the tickets that must be purchased for Carnival T and Carnival Q to be the same
Based on the information given let x be the number of ticket to be purchased .
Carnival T entrance fee= $7.00
Ride= $0.50 per ticket
Carnival Q entree fee =12.00
Ride= $0.25 per ticket
Tickets=$7.00 + ($0.50* x) = $12.00 + ($0.25* x)
.25 x = 5.00
Hence:
x=$.25+$5.00
×=$30
Therefore the amount of tickets that must be purchased in order for the total cost at Carnival T and Carnival Q to be the same will be $30
