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course Mth 158

Submitted Sept. 6 at 12:03 pm

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.

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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:

The formula a^2 + b^2 = c^2 can be used to find the length of an unknown side. So using the two known lengths in the formula you can find the unknown length.

14^2 + 48^2 = C^2

Sqrt196 + sqrt2304 = sqrt2500 = c = 50

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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 is a right triangle because the Pythagorean Theorem tell us that if two of lengths of the legs squared equals the third length squared.

The three lengths given are 10, 24 and 26 and if you use these numbers in the formula for the Pythagorean Theorem a^2 +b^2 = c^ it will show that it’s a right triangle

10^2 + 24^2 = 26^2

100 + 576 = 676

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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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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:

The formula to find the area of a sphere is A = 4 * pi * r^2 so since we know the radius is 3 meters we put that in the formula.

A = 4 pi * 3meters^2 = A = 4 pi *9meters

A=36 pi m^2

The formula to find the volume of a sphere is V = 4/3 * pi *r^3 and since we know that the radius is 3 meters we can put that in he formula.

V = 4/3 pi 3meters^3

V = 4/3 pi 27 meters ^3

V = 36 pi m^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):

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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:

The pool is 20 ft in diameter with a 3 ft. wide deck. To find the area of a circle you use the formula A = pi r^2. We first must find the radius and you do that by taking half of the diameter. The pool is 20ft in diameter so the radius would be 20/2 = 10ft. Next find the area of the pool: a = pi (10^2) = 100pi ft^2. Now, find the total area with the deck: A=pi(13^2) = 169pi ft^2. Now subtract the pool area from the deck & pool area to find the area of the deck: 169pi – 100pi = 69 pi ft^2.

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Gven 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. **

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