Assignment 003

course MTH 158

I finished the assignment three for section R3, but instead of ending, it went straight to section R4? I didn't do these because it wasn't part of this assignment, so I just closed it out. I hope I done it right.....

???????T??|d???assignment #003

That's fine. You did the right thing.

003. `query 3

College Algebra

09-09-2007

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18:42:15

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 -->

By using the pythagorean theory, a^2 + b^2=c^2

14^2 + 48^2 = c^2

196 + 2304 = c^2

2500=c^2

50=c

The hypotenuse is 50

confidence assessment: 3

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18:42:24

** 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 -->

ok

self critique assessment: 3

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18:45:08

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 -->

Square the lenths:

10^2=100

24^2=576

26^2=676

Lengths 10 and 24 squared equal 26 squared, so the hypotenuse is 26 and the lenths are 10 and 24. It is a right triangle.

confidence assessment: 3

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18:45:17

** 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 -->

ok

self critique assessment: 3

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18:51:18

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 -->

To find the find the volume of a sphere you use the formula V=4/3*pi*r^2

V=4/3*3.14*3^3

V=113.04 meters squared

To find the surface area you use the formula S=4*pi*r^2

S=4*3.14*3^2

S=113.04 meters squared.

confidence assessment: 3

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18:51:24

09-09-2007 18:51:24

** 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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NOTES ------->

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18:58:17

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 -->

To find the area of the deck you first need the area of the entire space.

A=pi*r^2

A=3.14*13^2

A=530.66 ft^2

Then you find area of the pool.

A=pi*r^2

A=3.14 * 10^2

A=314 feet^2

Subtract 530.66 from 314 to get 216.66 sq feet of deck.

confidence assessment: 2

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18:58:24

09-09-2007 18:58:24

** 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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NOTES ------->

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18:58:36

005. `query 5

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RESPONSE -->

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self critique assessment: 3

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Your work looks very good. Let me know if you have any questions. &#