Dec 20, 2014 - <span>The </span>volume V<span> of a right circular </span>cylinder<span> of </span>height h<span> and </span>radius r<span> is </span>V=pr^2h<span>. If the </span>height<span>is twice the </span>radius<span>, express the </span>volume V<span> as a function of r ... Hi Tammie,. OK so we have the </span>volume<span>formula. </span>V=(pi)<span>r2h</span><span>. and we know that </span>h<span> = 2r. so we can replace </span>h<span> with 2r and we get. </span>V<span> = (</span>pi)<span>r2</span>(2r).V=2(pi)r3<span>.</span>
the equation is 105=8.75×h
When at height>0, that is when it is below the water
find zeroes
0=-16x^2+18x
0=x(-16x+18)
solve
0=x
0=-16x+18
16x=18
x=18/16
x=9/8
from x=0 to x=9/8
9/8 seconds or1.125 seconds
Answer:
x > 3
Step-by-step explanation:
8x - 6 > 12 + 2x....,just move the values with variable x to one side and those without to the other side
; 8x - 2x > 12 + 6
; 6x > 18...., then divide by 6,both sides of the inequality
; x > 3
Answer: 72
Step-by-step explanation:
6 * 2 = 12
12 / 2 = 6
6 * 2 = 12
12 / 2 = 6
6 * 10 = 60
60 / 2 = 30
6 * 10 = 60
60 / 2 = 30
30 + 30 + 6 + 6 = 72