PLEASE ANSWER ASAPFind the value of x in each case:
1 answer:
You might be interested in
(tanθ + cotθ)² = sec²θ + csc²θ
<u>Expand left side</u>: tan²θ + 2tanθcotθ + cot²θ
<u>Evaluate middle term</u>: 2tanθcotθ = = 2
⇒ tan²θ + 2+ cot²θ
= tan²θ + 1 + 1 + cot²θ
<u>Apply trig identity:</u> tan²θ + 1 = sec²θ
⇒ sec²θ + 1 + cot²θ
<u>Apply trig identity:</u> 1 + cot²θ = csc²θ
⇒ sec²θ + csc²θ
Left side equals Right side so equation is verified
Answer:
<em>908360.67 lb-ft</em>
<em></em>
Step-by-step explanation:
height of tank= 6 ft
diameter of the tank = 4 ft
density of water p = 62.4 lbs/ft
A is the cross sectional area of the tank
A =
where r = diameter/2 = 4/2 = 2 ft
A = 3.142 x = 12.568 ft^2
work done = force x distance through which force is moved
work = F x d
Force due to the water = pgAh
where g = acceleration due to gravity = 32.174 ft/s^2
Force = 62.4 x 32.174 x 12.568 x 6 = 151393.44 lb
work done = force x distance moved
work = 151393.44 x 6 = <em>908360.67 lb-ft</em>