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malfutka [58]
3 years ago
6

At the end of each of the following five years, Eason is going to withdraw $30,000 from an investment account that pays interest

at a rate of 8% compounded annually. How much must there be in the investment account today in order for his account to reduce to a balance of zero after the last withdrawal? Show all your calculations.
Mathematics
1 answer:
Sophie [7]3 years ago
5 0

Answer:

3,000×0.8.×5

12000+3000 = 15000

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Answer:

The first five terms in Pattern B are 3, 7, 11, 15, 19.

Step-by-step explanation:

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Set up the integral that uses the method of cylindrical shells to find the volume V of the solid obtained by rotating the region
Ganezh [65]

Answer:

first exercise

V = 16 π

second exercise

V= 68π/15

Step-by-step explanation:

Initially, we have to plot the graph x = (y − 5)2 rotating around y = 3 and the limitation x = 4

<em>vide</em> picture 1

The rotation of x = (y − 5)2 intersecting the plane xy results in two graphs, which are represented by the graphs red and blue. The blue is function x = (y − 5)2. The red is the rotated cross section around y=3 of the previous graph. Naturally, the distance of "y" values of the rotated equation is the diameter of the rotation around y=3 and, by consequence,  this new red equation is defined by x = (y − 1)2.

Now, we have two equations.

x = (y − 5)2

xm = (y − 1)2 (Rotated graph in red on the figure)

The volume limited by the two functions in the 1 → 5 interval on y axis represents a volume which has to be excluded from the volume of the 5 → 7 on y axis interval integration.

Having said that, we have two volumes to calculate, the volume to be excluded (Ve) and the volume of the interval 5 → 7 called as V. The difference of V - Ve is equal to the total volume Vt.

(1) Vt = V - Ve

Before start the calculation, we have to take in consideration that the volume of a cylindrical shell is defined by:

(2) V=\int\limits^{y_{1} }_{y_{2}}{2*pi*y*f(y)} \, dy

f(y) represents the radius of the infinitesimal cylinder.

Replacing (2) in (1), we have

V= \int\limits^{5 }_{{3}}{2*pi*y*(y-5)^{2} } \, dy - \int\limits^{7 }_{{5}}{2*pi*y*(y-5)^{2} } \, dy

V = 16 π

----

Second part

Initially, we have to plot the graph y = x2 and x = y2, the area intersected by both is rotated around y = −7. On the second image you can find the representation.

<em>vide</em> picture 2

As the previous exercise, the exclusion zone volume and the volume to be considered will be defined by the interval from x=0 and y=0, to the intersection of this two equations, when x=1 and y = 1.

The interval integration of equation y = x2 will define the exclusion zone. By the other hand, the same interval on the equation x=y2 will be considered.

Before start the calculation, we have to take in consideration that the volume of a cylindrical shell is defined by:

(3) V=\int\limits^{x_{1} }_{x_{2}}{2*pi*x*f(x)} \, dx

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V= \int\limits^{1 }_{{0}}{2*pi*x*(\sqrt{x} + 7) } \, dy - \int\limits^{1 }_{{0}}{2*pi*x*(x^{2}+7) } \, dy

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Rina8888 [55]
V=72 because you need to divide both sides by 2 which makes it v=72
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Answer:

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hope this helps

have a good day :)

Step-by-step explanation:

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