12,800cm3s
This is a classic Related Rates problems. The idea behind Related Rates is that you have a geometric model that doesn’t change, even as the numbers do change.
For example, this shape will remain a sphere even as it changes size. The relationship between a where’s volume and it’s radius is
##V=4/3pir^3##
As long as this geometric relationship doesn’t change as the sphere grows, then we can derive this relationship implicitly, and find a new relationship between the rates of change.
is where we derive every variable in the formula, and in this case, we derive the formula with respect to time.
So we take the derivative of our sphere:
##V=4/3pir^3##
##(dV)/(dt)=4/3pi(3r^2)(dr)/dt##
##(dV)/(dt)=4pir^2(dr)/dt##
We were actually given ##(dr)/(dt)##. It’s ##4(cm)/s##.
We are interested in the moment when the diameter is 80 cm, which is when the radius will be 40 cm.
The rate of increase of the volume is ##(dV)/(dt)##, which is what we are looking for, so:
##(dV)/(dt)=4pir^2(dr)/dt##
##(dV)/(dt)=4pi(40cm)^2(4(cm)/s)##
##(dV)/(dt)=4pi(1600cm^2)(4(cm)/s)##
##(dV)/(dt)=4pi(1600cm^2)(4(cm)/s)##
##(dV)/(dt)=12,800(cm^3)/s##
And the units even work out correctly, since we should get a volume divided by time.
Hope this helps.
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Marketing study $100,000
New PDA cost $200,000
Fixed Cost $2.1 million
Estimated sales per year:
Year 1 – $75,000
Year 2 – $85,000
Year 3 – $80,000
Year 4 – $70,000
Year 5 – $65,000
Unit price of new product $240
Equipment $10.5 million
Value of equipment in 5 years – $1.1 million
Networking Capital 22% No initial outlay for NWC
Corporate tax 30%
Required return 10%
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sales of a substitute, such as a telephone, decrease.
sales of a substitute, such as a telephone, increase.
inventory of computers increases.
inventory of computer software increases.
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If the probability distribution for the random variable Xis given in thetable, what is the expected value of X? Show work. Xi 730 10 20 60p,- 0.40 0.30 0.20 0.10 Enter answer — with its sign — rounded to nearest hundredth (2 placesafter decimal)
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** Show work please
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If The Radius Of A Sphere Is Increasing At A Rate Of 4 Cm Per Second How Fast Is
/in Uncategorized /by developer12,800cm3s
This is a classic Related Rates problems. The idea behind Related Rates is that you have a geometric model that doesn’t change, even as the numbers do change.
For example, this shape will remain a sphere even as it changes size. The relationship between a where’s volume and it’s radius is
##V=4/3pir^3##
As long as this geometric relationship doesn’t change as the sphere grows, then we can derive this relationship implicitly, and find a new relationship between the rates of change.
is where we derive every variable in the formula, and in this case, we derive the formula with respect to time.
So we take the derivative of our sphere:
##V=4/3pir^3##
##(dV)/(dt)=4/3pi(3r^2)(dr)/dt##
##(dV)/(dt)=4pir^2(dr)/dt##
We were actually given ##(dr)/(dt)##. It’s ##4(cm)/s##.
We are interested in the moment when the diameter is 80 cm, which is when the radius will be 40 cm.
The rate of increase of the volume is ##(dV)/(dt)##, which is what we are looking for, so:
##(dV)/(dt)=4pir^2(dr)/dt##
##(dV)/(dt)=4pi(40cm)^2(4(cm)/s)##
##(dV)/(dt)=4pi(1600cm^2)(4(cm)/s)##
##(dV)/(dt)=4pi(1600cm^2)(4(cm)/s)##
##(dV)/(dt)=12,800(cm^3)/s##
And the units even work out correctly, since we should get a volume divided by time.
Hope this helps.
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If the reaction Fe2N (s) + 3/2 H2 (g) -> <- 2Fe(s) + NH3(g) comes to equilibrium at a totalpressure of 1 atm, analysis of the gas shows that at 700 K and 800 K pNH3/pH2 = 2.165 and1.083,…
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