R3 Assignment

course MTH 158

06/29 10:17pm

If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution:

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

003. `* 3

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Question: * R.3.16 \ 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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Your solution:

C^2=a^2+b^2

C^2=14^2+48^2

C^2=196+2304

C^2=2500

C=50

confidence rating #$&* 3

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Given Solution:

* * ** 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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Question:

* R.3.22 \ 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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Your solution:

10,24,26

10^2 24^2 26^2

100 + 576 = 676

It is a right triangle.

confidence rating #$&* 3

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Given Solution:

* * ** 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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Self-critique (if necessary):

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Self-critique rating #$&* OK

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Question:

* R.3.34 \ 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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Your solution:

v=4/3pi*r^3

V=4/3pi*3^3

V=4/3pi*27m^3

V=108/3pim^3

A=4*pi* r^2

A=4*pi*3^2

A=4*pi*9

A=36pi m^2

confidence rating #$&* 3

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Given Solution:

* * ** 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 m)^3

V = 4/3 * pi * 27 m^3

V = 36pi m^3

Surface Area = 4 * pi * r^2

S = 4 * pi * (3 m)^2

S = 4 * pi * 9 m^2

S = 36pi m^2. **

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Self-critique (if necessary):

I think the answer for the volume is incorrect. Shouldn’t it be 108/3pi m^3?

That would be OK, but 108/3 = 36.

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Self-critique rating #$&* OK

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Question:

* R.3.50 \ 42 (was R.3.36). A pool of diameter 20 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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Your solution:

A=pi*r^2

A=3.141*10^2

A=3.141*100

A=314.1

A=pi*r^2

A=3.141*13^2

A=3.141*169

A=530.829

530.829-314.1=216.729 as the area of the deck around the pool.

confidence rating #$&* 3

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Given Solution:

Think of a circle of radius 10 ft and a circle of radius 13 ft, both with the same center. If you 'cut out' the 10 ft circle you are left with a 'ring' which is 3 ft wide. It is this 'ring' that's covered by the deck. The 10 ft. circle in the middle is the pool.

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

• area = pi r^2 = 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. **

I took it all the way to the numbers using 3.141 for pi.

&#This looks good. See my notes. Let me know if you have any questions. &#

#$&*