Answer:
<u>Given :</u>
length offshore = CS=√(1+X^2)
Cable charged = 5000√(1+X^2)
onshore length = 4-X
laying cost = 3000(4-X)
total cost:
C=5000√(1+X^2) +3000(4-X)
DC/DX
= [5000*(0.5)*2X/{√(1+X^2)}]-3000=0... for optimum
5000X=3000√(1+X^2)
25X^2=3+3X^2
22X^2=3
X=√(3/22)
= 0.3693 miles
So, it would be laid offshore to S in a manner that
BS=X=0.3693 miles
Onshore=4-0.3693
=3.6307 miles
1 imperial foot has approximately 30.48 metric centimeters, and for one present we need 3 ft, or namely 3(30.48) cm, how many can we get from 102 cm? 102 ÷ 3(30.48) ≈ 1.1155, so barely just one present.
Answer:
2.625
Step-by-step explanation:
2 + 5/8
2 + .625 = 2.625
5/8 = .625
Answer:
24
Step-by-step explanation:
So, Dersan's path is a right triangle. You have to find the last side. You can use the triples or 6^2+8^2=c^2 but I'll use a triple.
3,4,5.
The last side is 10.
6+8+10=24
hope this helped!
slope = (25-4)/(30-10)
slope = 21/20
slope =21/20
using point slope form
y-y1 = m(x-x1)
y-4 = 21/20 (x-10)
y = 21/20x -21/2 +4
y = 21/20 x -21/2 +8/2
y = 21/20x -13/2
let x=40
y = 21/20 (40) -13/2
y = 42-13/2
y = 35.5 games
If we round the slope to 1
slope =1
using point slope form
y-y1 = m(x-x1)
y-4 = 1 (x-10)
y = 1x -10 +4
y = x -6
let x=40
y = 40-6
y = 34 games
