Assignment 3

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course Mth 158

6/15 10:23am

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=196+2304

C= 2500

C=sqrt(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:

YES.

26^2=10^2+24^2

676=100+576

676=676

When both sides equal the same number 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:

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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:

Volume and surface area equals 113.0973355

Plug in 3 in the following equations

4/3(pi)3^3 (Volume) and 4(pi)3^2 (Surface area)

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 believe I misused the calculator in the equation. I included the pi in the calculations. I understand my mistake but also a little confused as to why pi is not factored in.

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Self-critique Rating:3

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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=pi*13^2 (pool+deck=13 feet)

A= pi*169

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. **

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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:

A=Pi*r^2

A=pi*13^2 (pool+deck=13 feet)

A= pi*169

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 (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:

A=Pi*r^2

A=pi*13^2 (pool+deck=13 feet)

A= pi*169

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. **

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Self-critique (if necessary):

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Self-critique rating:

#*&!#*&!

@&

Any decimal you use to represent pi is at some point inaccurate.

Expressing the result as a multiple of pi is exact.

It's also easier to see what 36 pi has to do with radius 6 than to identify the radius associated with 113.0973355.

*@

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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:

A=Pi*r^2

A=pi*13^2 (pool+deck=13 feet)

A= pi*169

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. **

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Self-critique (if necessary):

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Self-critique rating:

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