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BlackZzzverrR [31]
3 years ago
12

Write 8.379 as a mixed number

Mathematics
1 answer:
BaLLatris [955]3 years ago
5 0
8.379 as a mixed number is 8 and 379 over 1000.
379
8 --------
1000
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What is 0.0592 in scientific notation
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Answer:

5.92 times 10 to the 2nd power.

Step-by-step explanation:

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Find the value of f(-9)
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Answer: 2

Step-by-step explanation:

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7 0
2 years ago
the distance in feet that Karina swims in a race is represent by 8d - 8, where d is the distance for each lap.What is an express
sergij07 [2.7K]

Answer:    The equivalent expression for  is  

Step-by-step explanation:

The distance in feet that Karina swims in a race is represented by 4d - 4, where is d is the distance for each lap.

To get equivalent expression we factor the expression 4d-4

Greatest common factor is 4

Factor out 4 from the expression 4d-4

when we factor out 4, we divide each term by 4

The equivalent expression for  is

Step-by-step explanation:

4 0
3 years ago
A cylindrical tank has a base of diameter 12 ft and height 5 ft. The tank is full of water (of density 62.4 lb/ft3).(a) Write do
saw5 [17]

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

7 0
3 years ago
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