Assignment 3 R3

#$&*

course Mth 158

6/19/13 10:45pm

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:

You must use the Pythagorean Theorem c^2 = a^2 + b^2 where the hypotenuse is c

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:

You must see if the squares of two sides are equal to the square of the last side (this would be the hypotenuse).

So I found the square of each side

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

100 + 576 = 676

This means that it is a right triangle and 26 is the hypotenuse

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): 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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

V = 4/3 * pi * r^3

V = 4/3 * pi * 3^3 I canceled out one set of 3's

V = 4 * pi * 3^2

V = 4 * pi * 9

V = 36 pi

SA = 4 * pi * r^2

SA = 4 * pi * 3^2

SA = 4 * pi * 9

SA = 36 pi

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 left off the unit of measurement. ??? What is it in this case? I don't remember what the m stands for actually.

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Self-critique Rating: 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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

The formula for the area of a circle is A = pi r^2

I must first find the area of the entire pool and deck area. The radius for this is 1/2 the diameter of the pool + the 3 ft of deck (r = 13)

A = pi 13^2

A = 169 pi ft^2

I must then find the area of the pool.

A = pi 10^2 (The radius was 1/2 of the 20 ft diameter)

A = 100 pi ft^2

The pool area must then be subtracted from the total area.

169 pi ft^2 - 100 pi ft^2 = 69 ft^2

This is the area of the deck alone.

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. **

Self-Critique:

??? Why is the pi left in the answer? Would it not be subtracted out in 169 pi - 100 pi?"

@&

169 pi - 100 pi = (169 - 100) pi, by the distributive law.

a * b - c * d is not equal to (a - c) + (b - c), as of course you know.

*@

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:

The formula for the area of a circle is A = pi r^2

I must first find the area of the entire pool and deck area. The radius for this is 1/2 the diameter of the pool + the 3 ft of deck (r = 13)

A = pi 13^2

A = 169 pi ft^2

I must then find the area of the pool.

A = pi 10^2 (The radius was 1/2 of the 20 ft diameter)

A = 100 pi ft^2

The pool area must then be subtracted from the total area.

169 pi ft^2 - 100 pi ft^2 = 69 ft^2

This is the area of the deck alone.

confidence rating #$&*: 3

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

.............................................

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:

??? Why is the pi left in the answer? Would it not be subtracted out in 169 pi - 100 pi?"

@&

169 pi - 100 pi = (169 - 100) pi, by the distributive law.

a * b - c * d is not equal to (a - c) + (b - c), as of course you know.

*@

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:

The formula for the area of a circle is A = pi r^2

I must first find the area of the entire pool and deck area. The radius for this is 1/2 the diameter of the pool + the 3 ft of deck (r = 13)

A = pi 13^2

A = 169 pi ft^2

I must then find the area of the pool.

A = pi 10^2 (The radius was 1/2 of the 20 ft diameter)

A = 100 pi ft^2

The pool area must then be subtracted from the total area.

169 pi ft^2 - 100 pi ft^2 = 69 ft^2

This is the area of the deck alone.

confidence rating #$&*: 3

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

.............................................

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:

??? Why is the pi left in the answer? Would it not be subtracted out in 169 pi - 100 pi?"

@&

169 pi - 100 pi = (169 - 100) pi, by the distributive law.

a * b - c * d is not equal to (a - c) + (b - c), as of course you know.

*@

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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:

The formula for the area of a circle is A = pi r^2

I must first find the area of the entire pool and deck area. The radius for this is 1/2 the diameter of the pool + the 3 ft of deck (r = 13)

A = pi 13^2

A = 169 pi ft^2

I must then find the area of the pool.

A = pi 10^2 (The radius was 1/2 of the 20 ft diameter)

A = 100 pi ft^2

The pool area must then be subtracted from the total area.

169 pi ft^2 - 100 pi ft^2 = 69 ft^2

This is the area of the deck alone.

confidence rating #$&*: 3

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

.............................................

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:

??? Why is the pi left in the answer? Would it not be subtracted out in 169 pi - 100 pi?"

@&

169 pi - 100 pi = (169 - 100) pi, by the distributive law.

a * b - c * d is not equal to (a - c) + (b - c), as of course you know.

*@

#*&!

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