4.58 x 10^-5
hope I helped
Answer:
perimeter goes with the one that has 5x^2
i dont know the second im sorry
Step-by-step explanation:
Answer: none
Step-by-step explanation:
(A)
(16÷32/10) ×2 + 0.2×(90)
Using bodmas principle ; solve bracket
(16×10/32)×2 + (2/10×90)
10+18 =28
(B)
{(16÷32/10) × (2+2/10)} ×90
Open brackets
{(16×10/32) × (22/10)} ×90
(5×11/5) ×90
11×90 = 990
(C)
16÷{(32/10×2) + (2/10×8)} +82
Open brackets, solve division first, dolled by addition
16÷(32/5 + 8/5) +82
16÷(40/5) +82
16÷8 +82
2+82= 84
(D)
[16÷(32/10 ×2) + 0.2× (90)]
16÷ (32/5) + 2/10 ×90
Solve division
16×5/32 + 18
5/2 + 18
L.c.m of denominator (2&1) =2
(5+36) / 2 = 41/2
=20.5
Answer:
The answer is 4.35, rounded two decimals to hundredths place.
<em>some text here to make the answer slightly longer for some good reasons. yep.</em>
Answer:
The work done to pump all of the kerosene from the tank to an outlet is 
Step-by-step explanation:
The work is defined by:
(1)
The force here will be the product between the volume and the kerosene weighing, so we have :
This force will be in-lbs.
Where R is the radius (3 feet)
Then using (1), we have:
Here 8-y is a distance at some point of the tank. Now, to get the work done from the base to the top of the tank we will need to take integral from 0 to 8 feet.
Therefore, the work done to pump all of the kerosene from the tank to an outlet is
I hope it helps you!
