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solong [7]
3 years ago
13

How do i round 186.282?

Mathematics
2 answers:
liq [111]3 years ago
8 0
Okay so if you want to know how to Round all you have to remember is 4 or more add 1 more 4 or less just ignore.

If your rounding 186.282 in the tenths you would look in your tenths place and look at the number in the left. ( the number in the tenths place is 8)

So is 2 greater or lesser than 4?

its lesser.


so ignore the rest of the two numbers there ( 8 & 2)

And your answer would be 186.2







adoni [48]3 years ago
6 0

Well, first you need to decide what place you want to round it TO.

Example:  Round it to the nearest hundredth:

The next larger hundredth is  186.29 .
The next smaller hundredth is  186.28 .
Now look at it.
186.282  is closer to  186.28  than it is to  186.29 .
So the nearest hundredth is  186.28 .

-- When  186.282  is rounded to the nearest hundredth, it becomes  186.28 .

Similarly . . .
-- When  186.282  is rounded to the nearest tenth, it becomes  186.3 .
-- When  186.282  is rounded to the nearest whole number, it becomes  186 .
-- When  186.282  is rounded to the nearest ten, it becomes  190 .
-- When  186.282  is rounded to the nearest hundred, it becomes  200 .
-- When  186.282  is rounded to the nearest thousand or anything larger,
                                                                                       it becomes  zero.

I'm curious . . . where did this number come from ?

It happens to be one thousandth of the speed of light, in miles per hour.

Did it come up in science class, or did a science geek use it for
one of the problems in math ?

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Answer:

a.  71884.8 π lb/ft-s²∫₀⁵(9 - y)dy

b.  23961.6 π lb/ft-s²∫₀⁵(5 - y)dy

c. 99840π lb/ft-s²∫₀⁶rdr

Step-by-step explanation:

.(a) Write down an integral for the work needed to pump all of the water to a point 4 feet above the tank.

The work done, W = ∫mgdy where m = mass of cylindrical tank = ρA([5 + 4] - y) where ρ = density of water = 62.4 lb/ft³, A = area of base of tank = πd²/4 where d = diameter of tank = 12 ft.( we add height of the tank + the height of point above the tank and subtract it from the vertical point above the base of the tank, y to get 5 + 4 - y) and g = acceleration due to gravity = 32 ft/s²

So,

W = ∫mgdy

W = ∫ρA([5 + 4] - y)gdy

W = ∫ρA(9 - y)gdy

W = ρgA∫(9 - y)dy

W = ρgπd²/4∫(9 - y)dy

we integrate W from  y from 0 to 5 which is the height of the tank

W = ρgπd²/4∫₀⁵(9 - y)dy

substituting the values of the other variables into the equation, we have

W = 62.4 lb/ft³π(12 ft)² (32 ft/s²)/4∫₀⁵(9 - y)dy

W = 71884.8 π lb/ft-s²∫₀⁵(9 - y)dy

.(b) Write down an integral for the fluid force on the side of the tank

Since force, F = ∫PdA where P = pressure = ρgh where h = (5 - y) since we are moving from h = 0 to h = 5. So, P = ρg(5 - y)

The differential area on the side of the tank is given by

dA = 2πrdy

So.  F = ∫PdA

F = ∫ρg(5 - y)2πrdy

Since we are integrating from y = 0 to y = 5, we have our integral as

F = ∫ρg2πr(5 - y)dy

F = ∫ρgπd(5 - y)dy    since d = 2r

substituting the values of the other variables into the equation, we have

F = ∫₀⁵62.4 lb/ft³π(12 ft) × 32 ft/s²(5 - y)dy

F = 23961.6 π lb/ft-s²∫₀⁵(5 - y)dy

.(c) How would your answer to part (a) change if the tank was on its side

The work done, W = ∫mgdr where m = mass of cylindrical tank = ρAh where ρ = density of water = 62.4 lb/ft³, A = curved surface area of cylindrical tank = 2πrh  where r = radius of tank, d = diameter of tank = 12 ft. and h =  height of the tank = 5 ft and g = acceleration due to gravity = 32 ft/s²

So,

W = ∫mgdr

W = ∫ρAhgdr

W = ∫ρ(2πrh)hgdr

W = ∫2ρπrh²gdr

W = 2ρπh²g∫rdr

we integrate from r = 0 to r = d/2 where d = diameter of cylindrical tank = 12 ft/2 = 6 ft

So,

W = 2ρπh²g∫₀⁶rdr

substituting the values of the other variables into the equation, we have

W = 2 × 62.4 lb/ft³π(5 ft)² × 32 ft/s²∫₀⁶rdr

W = 99840π lb/ft-s²∫₀⁶rdr

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