Answer:

Step-by-step explanation:
The Cylindrical coordinates are:
x = rcosθ, y = rsinθ and z = z
From the question, on the xy-plane;


where:
0 ≤ r ≤
and 0 ≤ θ ≤ 2π
∴












The combo is 4592 I think
Answer:
24 ft
Step-by-step explanation:
Use proportions! The old tent has sides of 10ft, and a base of 15ft. The new tent has sides of 16ft and a base of ____ ft.
So, 10/15 = 16/___. Cross multiple and you get 24ft!
Step-by-step explanation:
1. when x is 0
y will be
4x+2y=10
4(0)+2y=10
0+2y=10
collect like terms
2y=10-0
2y=10
divide both sides by 2
2y/2=10/2
y=5
so when x =0 y will equal to 5
2. when x is 1
y will be
4x+2y=10
4(1)+2y=10
4+2y=10
collect like terms
2y=10-4
2y=6
divide both sides by two
2y/2=6/2
y=3
so when x is 1, y is 3
3. when x is 2
4x+2y=10
4(2)+2y=10
8+2y=10
(CLT) 2y=10-8
2y =2
y= 1