course mth 151 more difficult then I thought ½¹‹™üw‹…E„–¦¸¨Å‡þ®Z¹¼¡assignment #029
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08:48:31 7.3.18 (1/3) / 6 = 1/18. Is this ratio equation valid or not and how did you determine your answer?
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RESPONSE --> valid, if you work the problem out; (1/3=(1/3) confidence assessment: 1
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08:49:01 **If we multiply both sides by 6 * 18 we get 6 * 18 * (1/3 ) / 6 = 6 * 18 * (1 / 18) or 18 * 1/3 = 6. Note that the effect here is the same as that of 'cross-multiplying', but it's a good idea to remember that 'cross-multiplying' is really a shortcut way to think of multiplying both sides by the common denominator. Since 18 * 1/3 = 18 / 3 = 6, the equation 18 * 1/3 = 6 is true, which verifies the original equality. **
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RESPONSE --> ok self critique assessment: 2
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08:51:06 7.3.20 z/8 = 49/56. Solve this proportionality for z.
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RESPONSE --> z/8=49/56 cross multiply; 56z/56=392/56 z=7 confidence assessment: 2
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08:51:15 **Multiply both sides by 8 * 56 to get 8 * 56 * z / 8 = 8 * 56 * 49 / 56. Simplify to get 56 * z = 8 * 49. Divide both sides by 56 to get z = 8 * 49 / 56. Simplify to get z = 7. **
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RESPONSE --> ok self critique assessment: 2
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09:00:21 7.3.42 8 oz .45; 16 oz. .49; 50 oz. 1.59`sb Which is the best value per unit for green beans and how did you obtain your result?
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RESPONSE --> since i can't find this problem in the book to see what 1.59'sb stands for, .45/8=.06 per unit .49/16=.03 per unit 1.59/50=.031 per unit .49 cents for 16oz. of green beans is best value confidence assessment: 1
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09:00:43 ** 45 cents / 8 oz = 5.63 cents / oz. 49 cents / 16 oz = 3.06 cents / oz. 159 cents / 50 oz = 3.18 cents / oz. 16 oz for .49 is the best value at 3.06 cents / oz. **
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. ok Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems. self critique assessment: 2
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09:03:18 7.3.45 triangles 4/3, 2, x; 4, 6, 3. What is the value of x and how did you use an equation to find it?
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RESPONSE --> not sure how to set this up. this isn't 7.3.45 in book confidence assessment: 0
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09:04:23 ** the 4/3 corresponds to 4, 2 corresponds to 6, and x corresponds to 3. The ratios of corresponding sides are all equal. So 4/3 / 4 = 2 / 6 = x / 3. Just using x / 3 = 2 / 6 we solve to get x = 1. We would have obtained the same thing if we had used x / 3 = 4/3 / 4. **
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RESPONSE --> ok self critique assessment:
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09:11:41 If z = 9 when x = 2/3 and z varies inversely as x, find z when x = 5/4. Show how you set up and used an equation of variation to solve this problem.
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RESPONSE --> 9/x=(2/3) / (5/4) 2/3x=9(5/4) cross multiply x=4.8 after changing to decimals confidence assessment: 0
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09:14:12 ** If z varies inversely as x then z = k / x. Then we have 9 = k / ( 2/3). Multiplying both sides by 2/3 we get 2/3 * 9 = k so k = 6. Thus z = 6 / x. So when x = 5/4 we have z = 6 / (5 /4 ) = 24 / 5 = 4.8. Note that the translations of other types of proportionality encountered in this chapter include: z = k x^2: z varies as square of x. z = k / x^2: z varies inversely as square of x. z = k x: z is proportional to x. **
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RESPONSE --> ok self critique assessment: 2
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09:17:39 7.3.72. Illumination is inversely proportional to the square of the distance from the source. Illumination at 4 ft is 75 foot-candles. What is illumination at 9 feet?
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RESPONSE --> 4/9=75/x 4x/4=675/4 x=168.75 confidence assessment: 1
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09:20:20 **Set up the variation equation I = k / r^2, where I stands for illumination and r for distance (you might have used different letters). This represents the inverse proportionality of illumination with the square of distance. Use I = 75 when r = 4 to get 75 = k / 4^2, which gives you k = 75 * 4^2 = 75 * 16 = 1200. Now rewrite the proportionality with this value of k: I = 1200 / r^2. To get the illumination at distance 9 substitute 9 for r to get I = 1200 / 9^2 = 1200 / 81 = 14.8 approx.. The illumination at distance 9 is about 14.8.
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RESPONSE --> ok, set up wrong self critique assessment: 0
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09:27:13 7.3.66 length inv prop width; L=27 if w=10; w = 18. L = ? Explain how you set up and used a variation equation to obtain the length as a function of width, giving your value of k. Then explain how you used your equation to find the length for width 18
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RESPONSE --> i guessed at 529, but not sure how to set up. confidence assessment: 0
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09:31:29 **Set up the variation equation L = k / w, which is the inverse proportion. Use L = 27 when w = 10 to get 27 = k / 10, which gives you k = 27 * 10 = 270. Now we know that L = 270 / w. So if w = 18 you get L = 270 / 18 = 15. **
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RESPONSE --> ok self critique assessment: 0
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