asst 3

course

I am under the impression there is no asst 4 if this is incorrect please let me know.

That is correct.

L錟O[eassignment #003

003. `query 3

College Algebra

09-05-2008

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20:21:19

query 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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RESPONSE -->

c^2=a^2+b^2

c^2=14^2+48^2

c^=196+2304

c^2=2500

c=50

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20:21:36

** 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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RESPONSE -->

ok

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20:24:29

query 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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RESPONSE -->

yes,

c^2=a^2+b^2

26^2=10^2+24^2

676=100+576

676=676

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20:24:38

** 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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RESPONSE -->

ok

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20:28:07

query 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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RESPONSE -->

v=4/3pi r^3

v= 4/3pi 3^3

v=113.10^3

s=4pi r^2

s=4pi 3^2

s=113.10^2

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20:37:44

** To find the volume and surface are a sphere we use the given formulas:

Volume = 4/3 * pi * r^3

V = 4/3 * pi * 3^3

V = 4/3 * pi * 27

V = 36pi m^3

Surface Area = 4 * pi * r^2

S = 4 * pi * 3^2

S = 4 * pi * 9

S = 36pi m^2. **

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RESPONSE -->

I dont understand at all how the answer is 36.

It isn't. The answer is 36 pi m^2.

113.10 is a reasonable approximation of 36 pi, but is not exactly equal to this correct value.

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20:40:34

query 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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RESPONSE -->

a=pi r ^2

pool area=

a=pi 10^2

a=314.16

total pool and deck area=

a=pi 13^2

a=530.93

total area - pool area = deck area

530.93 - 314.16= 216.77

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20:43:08

** 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 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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RESPONSE -->

am I breaking the solution down to far?

You are approximating the value of pi. In this case the object is an actual deck, as opposed to an ideal object (like the sphere of radius three in the preceding problem), and the measurements would not be completely accurate so that an approximation is not inappropriate.

However for an ideal object of these dimensions, 69 pi ft^2 is the correct result. I believe your approximation is accurate to five significant figures, which for a real-world application such as this would be more than sufficient.

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20:43:25

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

All ok

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20:43:37

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