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Home » r » What is the volume of the solid of revolution generated by revolving R about the x-axis?

What is the volume of the solid of revolution generated by revolving R about the x-axis?

Q. Let f(x) = e^(-3x/2)r />Let R be the region between the graph of f and the x-axis on the interval 0 < x < 1. Find the volume V of the solid of revolution generated by revolving R about the x-axis.

I've been trying all night to figure this problem out with no luck at all. Please help!

A. Use the method of disks.

Draw the region R and a typical disk. This disk has a radius of [e^(-3x/2) - 0] and a thickness of dx. It contribues an increment of volume dV given by

dV = Ïr²dx = Ï(e^(-3x/2))² dx
= Ï e^(-3x) dx

The total volume V is given by

1
â« Ï e^(-3x) dx
0

Original Question

What percentage does h & r block charge for a refund anticipation loan?
Q. I need money and am expecting a decent tax return. Does anyone know the approximate amount that H & R block charges for one of its refund anticipation loans? And how much do they charge to do the actual taxes? I know it's a ripoff but I really need the cash so I am considering it.

A. DO NOT USE H&R BLOCK FOR A RAL.....Efile online yourself using direct deposit to get the refund in 8-14 days. H&R BLOCK along with HSBC BANK are running a scam this year with the Refund Anticipation Loan -- telling consumers that they have been approved for the loan and will receive the money in 1-2 days. However, people that have been getting the loan for the past 3, 5, 8 years are being denied all of a sudden, with the bank stating that their criteria for the loan changes every year. If you fill today 01/29/2009 online with direct deposit, according to the IRS Refund Cycle Chart, you will see your refund deposited on 02/06/2009 -- just one week. Save yourself $300 and the headache of dealing with H&R Block and the Emerald Card, which is a way for H&R Block to "Hijack" your money, collect interest and cause a delay in your return.

Original Question

Prove that if R is a commutative ring with unity, then the invertible elements form a group with respect to th?
Q. Part 1) An element a in a commutative ring R with unity e is said to be invertible if there is an element b in R such that ab=e. Prove that if R is a commutative ring with unity, then the invertible elements form a group with respect to the multiplication of the ring.

Part 2) What is the group of invertible elements in ZZ4?

A. Write U = set of invertible elements in R.
1) Closure: a and b are in U then there exist a^-1 and b^-1 so that aa^-1 = bb^-1 = 1. Then (ab)(b^-1a^-1) = a(bb^-1)a^-1 = a(1)a^-1 = aa^-1 =1, so that ab is also in U.
2) Clearly U is associative since R is.
3) 1 is clearly in U.
4) Clearly, by the definition of U every element has an inverse.

----------------------------------------------------------------------------------------------------------------------
U(Z/4) = {1,3}. If a is invertible in Z/4 then ab = 1 mod 4 is solvable. So a can't be 2 otherwise 2a = 1 mod 4 implies 2 divides 1. If a = 1 then 1*1 = 1 mod 4, and if a = 3 then 3*3 = 9 = 1 mod 4.

A more general way to do this part is:
Using Bezout's theorem/Euclidean algorithm, gcd(a,4) = 1 implies ab = 1 +4d = 1 mod 4. The only numbers with gcd(a,4) =1 are a = 1 and 3.

Original Question




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Title : What is the volume of the solid of revolution generated by revolving R about the x-axis?
Description : Q. Let f(x) = e^(-3x/2) r />Let R be the r egion between the graph of f and the x-axis on the interval 0 < x < 1. Find the volume...

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