Assign3 R3

course Mth 158

???????|???????assignment #003

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

003. `query 3

College Algebra

01-19-2007

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21:00:13

R.3.12 (was R.3.6) What is the hypotenuse of a right triangle with legs 14 and 48 and how did you get your result?

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RESPONSE -->

a^2+b^2=c^2

14^2 + 48^2=c^2

196 + 2304=c^2

2500=c^2

50=c

confidence assessment: 3

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21:00:42

** The Pythagorean Theorem tells us that

c^2 = a^2 + b^2, where a and b are the legs and c the hypotenuse. Substituting 14 and 48 for a and b we get

c^2 = 14^2 + 48^2, so that

c^2 = 196 + 2304 or

c^2 = 2500.

This tells us that c = + sqrt(2500) or -sqrt(2500). Since the length of a side can't be negative we conclude that c = +sqrt(2500) = 50. **

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RESPONSE -->

self critique assessment: 2

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21:02:58

R.3.18 (was R.3.12). Is a triangle with legs of 10, 24 and 26 a right triangle, and how did you arrive at your answer?

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RESPONSE -->

10^2=100, 24^2=576, 26^2=676

100+576=676

676=676

YES it is a right triangle

confidence assessment: 3

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21:03:13

** Using the Pythagorean Theorem we have

c^2 = a^2 + b^2, if and only if the triangle is a right triangle. Substituting we get

26^2 = 10^2 + 24^2, or

676 = 100 + 576 so that

676 = 676

This confirms that the Pythagorean Theorem applies and we have a right triangle. **

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RESPONSE -->

self critique assessment: 3

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21:06:32

R.3.30 (was R.3.24). What are the volume and surface area of a sphere with radius 3 meters, and how did you obtain your result?

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RESPONSE -->

volume for a sphere of radius r.

V=4/3(pie)r^2

V=4/3(pie)3^2

V=12(pie)m

confidence assessment: 2

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21:08:25

** To find the volume and surface are a sphere we use the given formulas:

Volume = 4/3 * pi * r^3

V = 4/3 * pi * 3^3

V = 4/3 * pi * 27

V = 36pi m^3

Surface Area = 4 * pi * r^2

S = 4 * pi * 3^2

S = 4 * pi * 9

S = 36pi m^2. **

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RESPONSE -->

ooops forgot the S=4*pi*r^2

S=4*pi*3^2

S=4*pi*9

S=36pi m^2

self critique assessment: 1

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21:13:59

R.3.42 (was R.3.36). A pool of radius 10 ft is enclosed by a deck of width 3 feet. What is the area of the deck and how did you obtain this result?

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RESPONSE -->

r=10' w=3' d=2r=2(10)=20+3=23

C=2*pi*10=pi*23

C=20pi=23pi

confidence assessment: 1

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21:16:45

** The deck plus the pool gives you a circle of radius 10 ft + 3 ft = 13 ft.

The area of the deck plus the pool is therefore pi * (13 ft)^2 = 169 pi ft^2.

So the area of the deck must be

deck area = area of deck and pool - area of pool = 169 pi ft^2 - 100 pi ft^2 = 69 pi ft^2. **

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RESPONSE -->

Ok so I completely work that problem wrong. I will go back over that.

self critique assessment: 1

Your response did not agree with the given solution in all details, and you should therefore have addressed the discrepancy with a full self-critique, detailing the discrepancy and demonstrating exactly what you do and do not understand about the given solution, and if necessary asking specific questions.

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21:17:02

005. `query 5

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RESPONSE -->

self critique assessment: 3

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Assign3 R3

Good work overall, but be sure to see my note on self-critique. Let me know if you have questions.