Section R3

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course Mth 158

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:

To solve this we use the formula a^2+b^2=c^2. So we take the length of the two legs and we square them and then add them. 14^2+48^2= 196+2304=2500. Then to find the hypotenuse we take the square root of the final result from our formula which would be 50. Therefore the hypotenuse is 50.

confidence rating #$&*:ok

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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:

Yes this triangle is a right triangle. to find If a triangle is a right triangle or not we must see if it will fit the formula a^2+b^2=c^2. So for this one 10^2+24^2=26^2which equals 100+576=676 and the square root of 676 is in fact 26 so yes this is a right triangle.

confidence rating #$&*:ok

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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):ok

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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:

To solve for the volume of a sphere we are given the formula v=4/3*pi*r^3 so so the volume would be 4/3*pi*27=36pi m^3 and for the surface area we are given the formula 4*pi*r^2 which would be 4*pi*9= 36pi m^2.

confidence rating #$&*:ok

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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 really don’t have a self-critique but I was curious concerning how the answers came out. Will the volume and surface area of a sphere always be equal or is this an “every now and then” scenario????

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Self-critique Rating:ok

@&
This is unique to radius 3.

Note that the radius 3 matches the denominator in the expression 4/3 pi r^3. The reason it works out here comes down to that, and the fact that the area is 4 pi r^2.
*@

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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:

to do this we need to find the area of the pool and the area of the pool plus the deck. So the pool would be a=pi*20^2=400pi ft^2 and the overall area would be a=pi*23^2=529pi ft^2. Not to decipher the area of just the deck we take the overall area and subtract the area of the pool which would be 529pi ft^2-400pi ft^2= 129pi ft^2.

confidence rating #$&*:ok

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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. **

Self-critique (if necessary):

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Self-critique rating:

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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:

confidence rating #$&*:ok

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Given Solution:

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. **

Self-critique (if necessary):

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Self-critique rating:

#*&!

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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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

confidence rating #$&*:ok

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

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. **

Self-critique (if necessary):

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Self-critique rating:

#*&!#*&!

Section R3

@& This is unique to radius 3. Note that the radius 3 matches the denominator in the expression 4/3 pi r^3. The reason it works out here comes down to that, and the fact that the area is 4 pi r^2. *@

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

@&
This is unique to radius 3.

Note that the radius 3 matches the denominator in the expression 4/3 pi r^3. The reason it works out here comes down to that, and the fact that the area is 4 pi r^2.
*@