course Mth 158 ܽOڌ
......!!!!!!!!...................................
13:27:54 R.9.6. What steps did you follow to simplify (-8)^(-5/3) and what is your result?
......!!!!!!!!...................................
RESPONSE --> This problem cannot be solved because it is an nonreal answer and -8 does not have a square root. confidence assessment: 0
.................................................
......!!!!!!!!...................................
13:29:36 ** (-8)^(-5/3) = [ (-8)^(1/3) ] ^-5. Since -8^(1/3) is -2 we get [-2]^-5 = 1 / (-2)^5 = -1/32. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:32:24 R.9.12. What steps did you follow to simplify (8/27)^(-2/3) and what is your result?
......!!!!!!!!...................................
RESPONSE --> (8/27)^(-2/3)= -16/81 confidence assessment: 0
.................................................
......!!!!!!!!...................................
13:35:37 ** Starting with (8/27)^(-2/3) we can write as (8^(-2/3)/27^(-2/3)). Writing with positive exponents this becomes (27^(2/3)/8^(2/3)) 27^(2/3) = [ 27^(1/3) ] ^2 = 3^2 = 9 and 8^(2/3) = [ 8^(1/3) ] ^2 = 2^2 = 4 so the result is (27^(2/3)/8^(2/3)) = 9/4. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:43:28 R.9.24. What steps did you follow to simplify 6^(5/4) / 6^(1/4) and what is your result?
......!!!!!!!!...................................
RESPONSE --> 6^(5/4)/ 6^(1/4) = [6^(1/4)] ^5= 1.56508458^5= 9.39050748 6^(1/4)=[6^(1/4)]^1= 1.56508458 SO THE RESULT IS 9.39050748/1.56508458 =6 confidence assessment: 3
.................................................
......!!!!!!!!...................................
13:43:49 ** Use the laws of exponents (mostly x^a / x^b = x^(a-b) as follows: 6^(5/4) / 6^(1/4) = 6^(5/4 - 1/4) = 6^1 = 6. **
......!!!!!!!!...................................
RESPONSE --> alright self critique assessment: 2
.................................................
......!!!!!!!!...................................
13:48:05 R.9.36. What steps did you follow to simplify (x^3)^(1/6) and what is your result, assuming that x is positive and expressing your result with only positive exponents?
......!!!!!!!!...................................
RESPONSE --> (x^3)^(1/6)= x^3*(1/6)= x^(1/2) confidence assessment: 3
.................................................
......!!!!!!!!...................................
13:48:07 R.9.36. What steps did you follow to simplify (x^3)^(1/6) and what is your result, assuming that x is positive and expressing your result with only positive exponents?
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
13:48:22 ** Express radicals as exponents and use the laws of exponents. (x^3)^(1/6) = x^(3 * 1/6) = x^(1/2). **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:59:21 R.9.48. What steps did you follow to simplify (x^(1/2) / y^2) ^ 4 * (y^(1/3) / x^(-2/3) ) ^ 3 and what is your result, assuming that x is positive and expressing your result with only positive exponents?
......!!!!!!!!...................................
RESPONSE --> x^(1/2)/y^2(4)=x^(1/2)/y^8=x/y^(15/2)*y^(1/3)/x^(-2/3)^3=y^(1/3)-(-2)/x^(-2)=y^(7/3)/x= x/y^(15/2)* y^(7/3)/x= xy^(7/3)/ y^(15/2)x confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:01:27 ** (x^(1/2) / y^2) ^ 4 * (y^(1/3) / x^(-2/3) ) ^ 3 = x^(1/2 * 4) / y^(2* 4) * y^(1/3 * 3) / x^(-2/3 * 3)= x^2 / y^8 * y / x^(-2) = x^2 * x^2 / y^7 = x^4 / y^7. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:02:44 Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> This is a very hard section that i need to spend more time on and use more critique with my answers. confidence assessment: 3
.................................................
ƧÅlwGε[|ÆQ assignment #009 009. `query 9 College Algebra 07-04-2007 O㑇[ծ assignment #010 010. `query 10 College Algebra 07-04-2007
......!!!!!!!!...................................
14:12:47 1.1.20 (was 1.1.12). Explain, step by step, how you solved the equation 5y + 6 = -18 - y
......!!!!!!!!...................................
RESPONSE --> First you add y to both sides and you get 6y+6=-18 Next, you subtract 6 from both sides and get 6y=-24 Next, you divide each side by six to find y and get 6y/6=-24/6=-4 y=-4 confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:14:16 ** STUDENT SOLUTION WITH INSTRUCTOR COMMENT: 5y + 6 = 18 - y Subtract 6 from both sides, giving us 5y = 12 - y Add y to both sides, 5y + y = 12 or 6y = 12 divide both sides by 6 y = 2 INSTRUCTOR COMMENT: This is correct for equation 5y + 6 = 18 - y but the equation as I note it is 5y + 6 = -18 - y. If that's the correct equation the solution is found by practically the same steps but you end up with y = -4. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 2
.................................................
......!!!!!!!!...................................
14:17:46 1.1.44 (was 1.1.30). Explain, step by step, how you solved the equation (2x+1) / 3 + 16 = 3x
......!!!!!!!!...................................
RESPONSE --> First, to reduce the fraction, (2x+1)/3 You must multiply each side by 3 which gives us 2x+1+16=9x Then, you subtract 2x from each side and get 17=7x Next, you divide each side by 7 and get 17/7=x confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:17:47 1.1.44 (was 1.1.30). Explain, step by step, how you solved the equation (2x+1) / 3 + 16 = 3x
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
14:19:15 ** STUDENT SOLUTION: (2x + 1) / 3 + 16 = 3x First, multiply both sides of the equation by 3 2x +1 + 48 =9x or 2x + 49 = 9x subtract 2x from both sides. 49 = 7x Divide both sides by 7 7 = x **
......!!!!!!!!...................................
RESPONSE --> oh ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:25:29 was 1.1.36. Explain, step by step, how you solved the equation (x+2)(x-3) = (x+3)^2
......!!!!!!!!...................................
RESPONSE --> x^2-6+2x-3x=x^2-x-6 confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:27:17 ** STUDENT SOLUTION: (x+2)(x+3) = (x+3)^2 First, we use the distributive property to remove the parenthesis and get x^2 - x - 6 = x^2 + 6x + 9 subtract x^2 from both sides, -x - 6 = 6x + 9 Subtract 9 from both sides - x - 6 - 9 = 6x or -x - 15 = 6x add x to both sides -15 = 7x Divide both sides by 7 -15/7 = x **
......!!!!!!!!...................................
RESPONSE --> oh ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:30:24 1.1.52 (was 1.1.48). Explain, step by step, how you solved the equation x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9)/
......!!!!!!!!...................................
RESPONSE --> i am not sure how to do this one confidence assessment: 0
.................................................
......!!!!!!!!...................................
14:33:37 ** Starting with x / (x^2 -9) + 4 / (x+3) = 3 / (x^2 -9), first factor x^2 - 9 to get x / ( (x-3)(x+3) ) + 4 / (x+3) = 3 / ( (x-3)(x+3) ). Multiply both sides by the common denominator ( (x-3)(x+3) ): ( (x-3)(x+3) ) * x / ( (x-3)(x+3) ) + ( (x-3)(x+3) ) * 4 / (x+3) = ( (x-3)(x+3) ) * 3 / ( (x-3)(x+3) ). Simplify: x + 4(x-3) = 3. Simplify x + 4x - 12 = 3 5x = 15 x = 3. If there is a solution to the original equation it is x = 3. However x = 3 results in denominator 0 when substituted into the original equation, and division by 0 is undefined. So there is no solution to the equation. When you multiplied both sides by x-3, if x = 3 you were multiplying by zero, which invalidated your solution **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 2
.................................................
......!!!!!!!!...................................
14:41:36 ** STUDENT SOLUTION: 1) clear the equation of fractions by multiplying both sides by the LCM (10w - 7)(5W + 7) After cancellation the left side reads: (5w+7)(8w + 5) After cancellation the right side reads: (10w - 7)(4w - 3) multiply the factors on each side using the DISTRIBUTIVE LAW Left side becomes: (40w^2) + 81w + 35 Right side becomes: (40w^2) - 58w + 21 3) subtract 40w^2 from both sides add 58w to both sides subtract 35 from both sides Rewrite: 139w = - 14 Now divide both sides by 139 to get w = - (14 / 139) DER**
......!!!!!!!!...................................
RESPONSE --> oh ok . self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:47:28 1.1.70 (was 1.1.78). Explain, step by step, how you solved the equation 1 - a x = b, a <> 0.
......!!!!!!!!...................................
RESPONSE --> 1-x=1 Subtract one from each side and you get -x=0 divide each side by -1 and you get 0 for the answer x=0 confidence assessment: 1
.................................................
......!!!!!!!!...................................
14:50:24 ** Start with 1 -ax = b, a <> 0. Adding -1 to both sides we get 1 - ax - 1 = b - 1, which we simplify to get -ax = b - 1. Divide both sides by -a, which gives you x = (b - 1) / (-a). Multiply the right-hand side by -1 / -1 to get x = (-b + 1) / a or x = (1 - b) / a. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 2
.................................................
......!!!!!!!!...................................
14:51:56 1.1.82 (was 1.1.90). Explain, step by step, how you solved the equation v = g t + v0 for t.
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 0
.................................................
......!!!!!!!!...................................
14:53:35 ** NOTE: v0 stands for v with subscript 0; the whole expression v0 stands for the name of a variable. It doesn't mean v * 0. Starting with v = g t + v0, add -v0 to both sides to get v - v0 = gt. Divide both sides by g to get (v - v0) / g = t }or t = (v - v0) / g. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 2
.................................................
......!!!!!!!!...................................
14:54:09 Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> this chapter was easy in the beginning but towards the end it got confusing and i need to study this chapter a lot confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:11:07 1.1.90 (was 1.2.18). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) scores 86, 80, 84, 90, scores to ave B (80) and A (90).
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 0
.................................................
......!!!!!!!!...................................
15:11:45 ** This can be solved by trial and error but the only acceptable method for this course, in which we are learning to solve problems by means of equations, is by an equation. Let x be the score you make on the exam. The average of the four tests is easy to find: 4-test average = ( 86 + 80 + 84 + 90 ) / 4 = 340 / 4 = 85. The final grade can be thought of as being made up of 3 parts, 1 part being the test average and 2 parts being the exam grade. We would therefore have final average = (1 * test average + 2 * exam grade) / 3. This gives us the equation final ave = (85 + 2 * x) / 3. If the ave score is to be 80 then we solve (85 + 2 * x) / 3 = 80. Multiplying both sides by 3 we get 85 + 2x = 240. Subtracting 85 from both sides we have 2 x = 240 - 85 = 155 so that x = 155 / 2 = 77.5. We can solve (340 + x) / 5 = 90 in a similar manner. We obtain x = 92.5. Alternative solution: If we add 1/3 of the test average to 2/3 of the final exam grade we get the final average. So (using the fact that the test ave is 85%, as calculated above) our equation would be 1/3 * 85 + 2/3 * x = final ave. For final ave = 80 we get 1/3 * 85 + 2/3 * x = 80. Multiplying both sides by 3 we have 85 + 2 * x = 240. The rest of the solution goes as before and we end up with x = 77.5. Solving 1/3 * 85 + 2/3 * x = 90 we get x = 92.5, as before. **
......!!!!!!!!...................................
RESPONSE --> ohh ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:15:39 1.7.20 (was 1.2.30). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) million to lend at 19% or 16%, max lent at 16% to average 18%.
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
15:15:52 ** Good. The details: If x is the amount lent at 16%, then the amount lent at 19% is 1,000,000 - x. Interest on x at 16% is .16 x, and interest on 1,000,000 - x at 19% is .19 (1,000,000 - x). This is to be equivalent to a single rate of 18%. 18% of 1,000,000 is 180,000 so the total interest is 1,000,000. So the total interest is .16 x + .19(1,000,000 - x), and also 180,000. Setting the two equal gives us the equation .16 x + .19(1,000,000 - x) = 180,000. Multiplying both sides by 100 to avoid decimal-place errors we have 16 x + 19 ( 1,000,000 - x) = 18,000,000. Using the distributive law on the right-hand side we get 16 x + 19,000,000 - 19 x = 18,000,000. Combining the x terms and subtracting 19,000,000 from both sides we have -3 x = 18,000,00 - 19,000,000 so that -3 x = -1,000,000 and x = -1,000,000 / (-3) = 333,333 1/3. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
15:17:33 1.7.36 (was 1.2.36). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) 3 mph current, upstream takes 5 hr, downstream 2.5 hr. Speed of boat?
......!!!!!!!!...................................
RESPONSE --> 10 percent confidence assessment: 0
.................................................
......!!!!!!!!...................................
15:17:41 STUDENT SOLUTION: Speed of the boat is 9 mph, I used the equation 5(x - 3) = 2.5(x + 3) Reasoning is that it took 5 hours for the boat to travel against the 3mph current, and then traveled the same distance with the 3mph current in 2.5 hours. INSTRUCTOR COMMENT: Good. The details: If we let x be the water speed of the boat then its actual speed upstream is x - 3, and downstream is x + 3. Traveling for 5 hours upstream, at speed x - 3, we travel distance 5 ( x - 3). Traveling for 2.5 hours downstream, at speed x + 3, we travel distance 2.5 ( x + 3). The two distance must be the same so we get 5 ( x - 3) = 2.5 ( x + 3) or 5 x - 15 = 2.5 x + 7.5. Adding -2.5 x + 15 to both sides we get 2.5 x = 22.5 so that x = 22.5 / 2.5 = 9. So the water speed is 9 mph. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 0
.................................................
......!!!!!!!!...................................
15:19:25 1.7.32 (was 1.2.42). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) pool enclosed by deck 3 ft wide; fence around deck 100 ft. Pond dimensions if pond square, if rectangular 3/1 ratio l/w, circular; which pond has most area?
......!!!!!!!!...................................
RESPONSE --> the rectangular pond confidence assessment: 0
.................................................
......!!!!!!!!...................................
15:19:33 ** If the deck is circular then its circumference is C = 2 pi R and its radius is r = C / (2 pi). C is the 100 ft length of the fence so we have R = 100 ft / ( 2 pi ) = 50 ft / pi. The radius of the circle is 3 ft less, due to the width of the deck. So the pool radius is r = 50 ft / pi - 3 ft. This gives us pool area A = pi r^2 = pi ( 50 / pi - 3)^2 = pi ( 2500 / pi^2 - 300 / pi + 9) = 524, approx.. If the pool is square then the dimensions around the deck are 25 x 25. The dimensions of the pool will be 6 ft less on each edge, since each edge spans two widths of the deck. So the area would be A = 19 * 19 = 361. The perimeter of the rectangular pool spans four deck widths, or 12 ft. The perimeter of a rectangular pool is therefore 12 ft less than that of the fence, or 100 ft - 12 ft = 88 ft. If the pool is rectangular with length 3 times width then we first have for the 2 l + 2 w = 88 or 2 (3 w) + 2 w = 88 or 8 w = 88, giving us w = 11. The width of the pool will be 11 and the length 3 times this, or 33. The area of the pool is therefore 11 * 33 = 363. The circular pool has the greatest area, the rectangular pool the least. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
15:21:52 1.7.44 (was 1.2.54). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) 20 lb bag 25% cement 75% sand; how much cement to produce 40% concentration?
......!!!!!!!!...................................
RESPONSE --> about 12 and half pounds of cement confidence assessment: 0
.................................................
......!!!!!!!!...................................
15:21:56 ** If x stands for the amount of cement added then we have the following: Original amount of cement in bag is 25% of 20 lb, or 5 lb. Original amount of sand in bag is 75% of 20 lb, or 15 lb. The final amount of cement will therefore be 5 lb + x, the final amount of sand will be 15 lb and the final weight of the mixture will be 20 lb + x. The mix has to be 40%, so (amt of cement) / (total amt of mixture) = .40. This gives us the equation (5 + x) / (20 + x) = .40. Multiplying both sides by 20 + x we have 5 + x = .40 ( 20 + x ). After the distributive law we have 5 + x = 80 + .40 x. Multiplying by 100 we get 500 + 100 x = 800 + 40 x. Adding -40 x - 500 to both sides we have 60 x = 300 so that x = 300 / 60 = 5. We should add 5 lbs of cement to the bag. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................
......!!!!!!!!...................................
15:24:35 1.7.52 (was 1.2.60). Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text): without solving what's wrong with prob how many liters 48% soln added to 20 liters of 25% soln to get 58% soln?
......!!!!!!!!...................................
RESPONSE --> 0.25/20=.48/100 We crossmultiply to get 25*9.6= 240liters confidence assessment: 0
.................................................
......!!!!!!!!...................................
15:24:50 ** Solution from Previous Student and Instructor Comment: It's not possible, adding a 25% solution to a 48% solution is only going to dilute it, I don't really know how to prove that algebraically, but logically that's what I think. (This is much like the last problem, that I don't really understand). INSTRUCTOR COMMENT: Right but the 48% solution is being added to the 25% solution. Correct statement, mostly in your words Adding a 48% solution to a 25% solution will never give you a 58% solution. Both concentrations are less than the desired concentration. **
......!!!!!!!!...................................
RESPONSE --> ohhh self critique assessment:
.................................................
......!!!!!!!!...................................
15:25:28 Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> I am awful at word problems. They are my biggest problem when it comes to algebra and i need to spend more time practicing on them. confidence assessment:
.................................................
ۂT{ assignment #011 011. `query 11 College Algebra 07-04-2007 ܽOڌ assignment #009 009. `query 9 College Algebra 07-04-2007
......!!!!!!!!...................................
13:27:54 R.9.6. What steps did you follow to simplify (-8)^(-5/3) and what is your result?
......!!!!!!!!...................................
RESPONSE --> This problem cannot be solved because it is an nonreal answer and -8 does not have a square root. confidence assessment: 0
.................................................
......!!!!!!!!...................................
13:29:36 ** (-8)^(-5/3) = [ (-8)^(1/3) ] ^-5. Since -8^(1/3) is -2 we get [-2]^-5 = 1 / (-2)^5 = -1/32. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:32:24 R.9.12. What steps did you follow to simplify (8/27)^(-2/3) and what is your result?
......!!!!!!!!...................................
RESPONSE --> (8/27)^(-2/3)= -16/81 confidence assessment: 0
.................................................
......!!!!!!!!...................................
13:35:37 ** Starting with (8/27)^(-2/3) we can write as (8^(-2/3)/27^(-2/3)). Writing with positive exponents this becomes (27^(2/3)/8^(2/3)) 27^(2/3) = [ 27^(1/3) ] ^2 = 3^2 = 9 and 8^(2/3) = [ 8^(1/3) ] ^2 = 2^2 = 4 so the result is (27^(2/3)/8^(2/3)) = 9/4. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:43:28 R.9.24. What steps did you follow to simplify 6^(5/4) / 6^(1/4) and what is your result?
......!!!!!!!!...................................
RESPONSE --> 6^(5/4)/ 6^(1/4) = [6^(1/4)] ^5= 1.56508458^5= 9.39050748 6^(1/4)=[6^(1/4)]^1= 1.56508458 SO THE RESULT IS 9.39050748/1.56508458 =6 confidence assessment: 3
.................................................
......!!!!!!!!...................................
13:43:49 ** Use the laws of exponents (mostly x^a / x^b = x^(a-b) as follows: 6^(5/4) / 6^(1/4) = 6^(5/4 - 1/4) = 6^1 = 6. **
......!!!!!!!!...................................
RESPONSE --> alright self critique assessment: 2
.................................................
......!!!!!!!!...................................
13:48:05 R.9.36. What steps did you follow to simplify (x^3)^(1/6) and what is your result, assuming that x is positive and expressing your result with only positive exponents?
......!!!!!!!!...................................
RESPONSE --> (x^3)^(1/6)= x^3*(1/6)= x^(1/2) confidence assessment: 3
.................................................
......!!!!!!!!...................................
13:48:07 R.9.36. What steps did you follow to simplify (x^3)^(1/6) and what is your result, assuming that x is positive and expressing your result with only positive exponents?
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
13:48:22 ** Express radicals as exponents and use the laws of exponents. (x^3)^(1/6) = x^(3 * 1/6) = x^(1/2). **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
13:59:21 R.9.48. What steps did you follow to simplify (x^(1/2) / y^2) ^ 4 * (y^(1/3) / x^(-2/3) ) ^ 3 and what is your result, assuming that x is positive and expressing your result with only positive exponents?
......!!!!!!!!...................................
RESPONSE --> x^(1/2)/y^2(4)=x^(1/2)/y^8=x/y^(15/2)*y^(1/3)/x^(-2/3)^3=y^(1/3)-(-2)/x^(-2)=y^(7/3)/x= x/y^(15/2)* y^(7/3)/x= xy^(7/3)/ y^(15/2)x confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:01:27 ** (x^(1/2) / y^2) ^ 4 * (y^(1/3) / x^(-2/3) ) ^ 3 = x^(1/2 * 4) / y^(2* 4) * y^(1/3 * 3) / x^(-2/3 * 3)= x^2 / y^8 * y / x^(-2) = x^2 * x^2 / y^7 = x^4 / y^7. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:02:44 Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> This is a very hard section that i need to spend more time on and use more critique with my answers. confidence assessment: 3
.................................................
ƧÅlwGε[|ÆQ assignment #009 009. `query 9 College Algebra 07-04-2007 O㑇[ծ assignment #010 010. `query 10 College Algebra 07-04-2007
......!!!!!!!!...................................
14:12:47 1.1.20 (was 1.1.12). Explain, step by step, how you solved the equation 5y + 6 = -18 - y
......!!!!!!!!...................................
RESPONSE --> First you add y to both sides and you get 6y+6=-18 Next, you subtract 6 from both sides and get 6y=-24 Next, you divide each side by six to find y and get 6y/6=-24/6=-4 y=-4 confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:14:16 ** STUDENT SOLUTION WITH INSTRUCTOR COMMENT: 5y + 6 = 18 - y Subtract 6 from both sides, giving us 5y = 12 - y Add y to both sides, 5y + y = 12 or 6y = 12 divide both sides by 6 y = 2 INSTRUCTOR COMMENT: This is correct for equation 5y + 6 = 18 - y but the equation as I note it is 5y + 6 = -18 - y. If that's the correct equation the solution is found by practically the same steps but you end up with y = -4. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 2
.................................................
......!!!!!!!!...................................
14:17:46 1.1.44 (was 1.1.30). Explain, step by step, how you solved the equation (2x+1) / 3 + 16 = 3x
......!!!!!!!!...................................
RESPONSE --> First, to reduce the fraction, (2x+1)/3 You must multiply each side by 3 which gives us 2x+1+16=9x Then, you subtract 2x from each side and get 17=7x Next, you divide each side by 7 and get 17/7=x confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:17:47 1.1.44 (was 1.1.30). Explain, step by step, how you solved the equation (2x+1) / 3 + 16 = 3x
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
14:19:15 ** STUDENT SOLUTION: (2x + 1) / 3 + 16 = 3x First, multiply both sides of the equation by 3 2x +1 + 48 =9x or 2x + 49 = 9x subtract 2x from both sides. 49 = 7x Divide both sides by 7 7 = x **
......!!!!!!!!...................................
RESPONSE --> oh ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:25:29 was 1.1.36. Explain, step by step, how you solved the equation (x+2)(x-3) = (x+3)^2
......!!!!!!!!...................................
RESPONSE --> x^2-6+2x-3x=x^2-x-6 confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:27:17 ** STUDENT SOLUTION: (x+2)(x+3) = (x+3)^2 First, we use the distributive property to remove the parenthesis and get x^2 - x - 6 = x^2 + 6x + 9 subtract x^2 from both sides, -x - 6 = 6x + 9 Subtract 9 from both sides - x - 6 - 9 = 6x or -x - 15 = 6x add x to both sides -15 = 7x Divide both sides by 7 -15/7 = x **
......!!!!!!!!...................................
RESPONSE --> oh ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:30:24 1.1.52 (was 1.1.48). Explain, step by step, how you solved the equation x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9)/
......!!!!!!!!...................................
RESPONSE --> i am not sure how to do this one confidence assessment: 0
.................................................
......!!!!!!!!...................................
14:33:37 ** Starting with x / (x^2 -9) + 4 / (x+3) = 3 / (x^2 -9), first factor x^2 - 9 to get x / ( (x-3)(x+3) ) + 4 / (x+3) = 3 / ( (x-3)(x+3) ). Multiply both sides by the common denominator ( (x-3)(x+3) ): ( (x-3)(x+3) ) * x / ( (x-3)(x+3) ) + ( (x-3)(x+3) ) * 4 / (x+3) = ( (x-3)(x+3) ) * 3 / ( (x-3)(x+3) ). Simplify: x + 4(x-3) = 3. Simplify x + 4x - 12 = 3 5x = 15 x = 3. If there is a solution to the original equation it is x = 3. However x = 3 results in denominator 0 when substituted into the original equation, and division by 0 is undefined. So there is no solution to the equation. When you multiplied both sides by x-3, if x = 3 you were multiplying by zero, which invalidated your solution **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 2
.................................................
......!!!!!!!!...................................
14:41:36 ** STUDENT SOLUTION: 1) clear the equation of fractions by multiplying both sides by the LCM (10w - 7)(5W + 7) After cancellation the left side reads: (5w+7)(8w + 5) After cancellation the right side reads: (10w - 7)(4w - 3) multiply the factors on each side using the DISTRIBUTIVE LAW Left side becomes: (40w^2) + 81w + 35 Right side becomes: (40w^2) - 58w + 21 3) subtract 40w^2 from both sides add 58w to both sides subtract 35 from both sides Rewrite: 139w = - 14 Now divide both sides by 139 to get w = - (14 / 139) DER**
......!!!!!!!!...................................
RESPONSE --> oh ok . self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:47:28 1.1.70 (was 1.1.78). Explain, step by step, how you solved the equation 1 - a x = b, a <> 0.
......!!!!!!!!...................................
RESPONSE --> 1-x=1 Subtract one from each side and you get -x=0 divide each side by -1 and you get 0 for the answer x=0 confidence assessment: 1
.................................................
......!!!!!!!!...................................
14:50:24 ** Start with 1 -ax = b, a <> 0. Adding -1 to both sides we get 1 - ax - 1 = b - 1, which we simplify to get -ax = b - 1. Divide both sides by -a, which gives you x = (b - 1) / (-a). Multiply the right-hand side by -1 / -1 to get x = (-b + 1) / a or x = (1 - b) / a. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 2
.................................................
......!!!!!!!!...................................
14:51:56 1.1.82 (was 1.1.90). Explain, step by step, how you solved the equation v = g t + v0 for t.
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 0
.................................................
......!!!!!!!!...................................
14:53:35 ** NOTE: v0 stands for v with subscript 0; the whole expression v0 stands for the name of a variable. It doesn't mean v * 0. Starting with v = g t + v0, add -v0 to both sides to get v - v0 = gt. Divide both sides by g to get (v - v0) / g = t }or t = (v - v0) / g. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 2
.................................................
......!!!!!!!!...................................
14:54:09 Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> this chapter was easy in the beginning but towards the end it got confusing and i need to study this chapter a lot confidence assessment: 3
.................................................
zk僤x띖 assignment #011 011. `query 11 College Algebra 07-04-2007 匐`ǭ assignment #011 011. `query 11 College Algebra 07-04-2007 z|ɼВƓ assignment #011 011. `query 11 College Algebra 07-04-2007
......!!!!!!!!...................................
15:11:07 1.1.90 (was 1.2.18). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) scores 86, 80, 84, 90, scores to ave B (80) and A (90).
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 0
.................................................
......!!!!!!!!...................................
15:11:45 ** This can be solved by trial and error but the only acceptable method for this course, in which we are learning to solve problems by means of equations, is by an equation. Let x be the score you make on the exam. The average of the four tests is easy to find: 4-test average = ( 86 + 80 + 84 + 90 ) / 4 = 340 / 4 = 85. The final grade can be thought of as being made up of 3 parts, 1 part being the test average and 2 parts being the exam grade. We would therefore have final average = (1 * test average + 2 * exam grade) / 3. This gives us the equation final ave = (85 + 2 * x) / 3. If the ave score is to be 80 then we solve (85 + 2 * x) / 3 = 80. Multiplying both sides by 3 we get 85 + 2x = 240. Subtracting 85 from both sides we have 2 x = 240 - 85 = 155 so that x = 155 / 2 = 77.5. We can solve (340 + x) / 5 = 90 in a similar manner. We obtain x = 92.5. Alternative solution: If we add 1/3 of the test average to 2/3 of the final exam grade we get the final average. So (using the fact that the test ave is 85%, as calculated above) our equation would be 1/3 * 85 + 2/3 * x = final ave. For final ave = 80 we get 1/3 * 85 + 2/3 * x = 80. Multiplying both sides by 3 we have 85 + 2 * x = 240. The rest of the solution goes as before and we end up with x = 77.5. Solving 1/3 * 85 + 2/3 * x = 90 we get x = 92.5, as before. **
......!!!!!!!!...................................
RESPONSE --> ohh ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:15:39 1.7.20 (was 1.2.30). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) million to lend at 19% or 16%, max lent at 16% to average 18%.
......!!!!!!!!...................................
RESPONSE --> confidence assessment:
.................................................
......!!!!!!!!...................................
15:15:52 ** Good. The details: If x is the amount lent at 16%, then the amount lent at 19% is 1,000,000 - x. Interest on x at 16% is .16 x, and interest on 1,000,000 - x at 19% is .19 (1,000,000 - x). This is to be equivalent to a single rate of 18%. 18% of 1,000,000 is 180,000 so the total interest is 1,000,000. So the total interest is .16 x + .19(1,000,000 - x), and also 180,000. Setting the two equal gives us the equation .16 x + .19(1,000,000 - x) = 180,000. Multiplying both sides by 100 to avoid decimal-place errors we have 16 x + 19 ( 1,000,000 - x) = 18,000,000. Using the distributive law on the right-hand side we get 16 x + 19,000,000 - 19 x = 18,000,000. Combining the x terms and subtracting 19,000,000 from both sides we have -3 x = 18,000,00 - 19,000,000 so that -3 x = -1,000,000 and x = -1,000,000 / (-3) = 333,333 1/3. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
15:17:33 1.7.36 (was 1.2.36). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) 3 mph current, upstream takes 5 hr, downstream 2.5 hr. Speed of boat?
......!!!!!!!!...................................
RESPONSE --> 10 percent confidence assessment: 0
.................................................
......!!!!!!!!...................................
15:17:41 STUDENT SOLUTION: Speed of the boat is 9 mph, I used the equation 5(x - 3) = 2.5(x + 3) Reasoning is that it took 5 hours for the boat to travel against the 3mph current, and then traveled the same distance with the 3mph current in 2.5 hours. INSTRUCTOR COMMENT: Good. The details: If we let x be the water speed of the boat then its actual speed upstream is x - 3, and downstream is x + 3. Traveling for 5 hours upstream, at speed x - 3, we travel distance 5 ( x - 3). Traveling for 2.5 hours downstream, at speed x + 3, we travel distance 2.5 ( x + 3). The two distance must be the same so we get 5 ( x - 3) = 2.5 ( x + 3) or 5 x - 15 = 2.5 x + 7.5. Adding -2.5 x + 15 to both sides we get 2.5 x = 22.5 so that x = 22.5 / 2.5 = 9. So the water speed is 9 mph. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 0
.................................................
......!!!!!!!!...................................
15:19:25 1.7.32 (was 1.2.42). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) pool enclosed by deck 3 ft wide; fence around deck 100 ft. Pond dimensions if pond square, if rectangular 3/1 ratio l/w, circular; which pond has most area?
......!!!!!!!!...................................
RESPONSE --> the rectangular pond confidence assessment: 0
.................................................
......!!!!!!!!...................................
15:19:33 ** If the deck is circular then its circumference is C = 2 pi R and its radius is r = C / (2 pi). C is the 100 ft length of the fence so we have R = 100 ft / ( 2 pi ) = 50 ft / pi. The radius of the circle is 3 ft less, due to the width of the deck. So the pool radius is r = 50 ft / pi - 3 ft. This gives us pool area A = pi r^2 = pi ( 50 / pi - 3)^2 = pi ( 2500 / pi^2 - 300 / pi + 9) = 524, approx.. If the pool is square then the dimensions around the deck are 25 x 25. The dimensions of the pool will be 6 ft less on each edge, since each edge spans two widths of the deck. So the area would be A = 19 * 19 = 361. The perimeter of the rectangular pool spans four deck widths, or 12 ft. The perimeter of a rectangular pool is therefore 12 ft less than that of the fence, or 100 ft - 12 ft = 88 ft. If the pool is rectangular with length 3 times width then we first have for the 2 l + 2 w = 88 or 2 (3 w) + 2 w = 88 or 8 w = 88, giving us w = 11. The width of the pool will be 11 and the length 3 times this, or 33. The area of the pool is therefore 11 * 33 = 363. The circular pool has the greatest area, the rectangular pool the least. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment:
.................................................
......!!!!!!!!...................................
15:21:52 1.7.44 (was 1.2.54). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) 20 lb bag 25% cement 75% sand; how much cement to produce 40% concentration?
......!!!!!!!!...................................
RESPONSE --> about 12 and half pounds of cement confidence assessment: 0
.................................................
......!!!!!!!!...................................
15:21:56 ** If x stands for the amount of cement added then we have the following: Original amount of cement in bag is 25% of 20 lb, or 5 lb. Original amount of sand in bag is 75% of 20 lb, or 15 lb. The final amount of cement will therefore be 5 lb + x, the final amount of sand will be 15 lb and the final weight of the mixture will be 20 lb + x. The mix has to be 40%, so (amt of cement) / (total amt of mixture) = .40. This gives us the equation (5 + x) / (20 + x) = .40. Multiplying both sides by 20 + x we have 5 + x = .40 ( 20 + x ). After the distributive law we have 5 + x = 80 + .40 x. Multiplying by 100 we get 500 + 100 x = 800 + 40 x. Adding -40 x - 500 to both sides we have 60 x = 300 so that x = 300 / 60 = 5. We should add 5 lbs of cement to the bag. **
......!!!!!!!!...................................
RESPONSE --> self critique assessment:
.................................................
......!!!!!!!!...................................
15:24:35 1.7.52 (was 1.2.60). Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text): without solving what's wrong with prob how many liters 48% soln added to 20 liters of 25% soln to get 58% soln?
......!!!!!!!!...................................
RESPONSE --> 0.25/20=.48/100 We crossmultiply to get 25*9.6= 240liters confidence assessment: 0
.................................................
......!!!!!!!!...................................
15:24:50 ** Solution from Previous Student and Instructor Comment: It's not possible, adding a 25% solution to a 48% solution is only going to dilute it, I don't really know how to prove that algebraically, but logically that's what I think. (This is much like the last problem, that I don't really understand). INSTRUCTOR COMMENT: Right but the 48% solution is being added to the 25% solution. Correct statement, mostly in your words Adding a 48% solution to a 25% solution will never give you a 58% solution. Both concentrations are less than the desired concentration. **
......!!!!!!!!...................................
RESPONSE --> ohhh self critique assessment:
.................................................
......!!!!!!!!...................................
15:25:28 Add comments on any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> I am awful at word problems. They are my biggest problem when it comes to algebra and i need to spend more time practicing on them. confidence assessment:
.................................................
ۂT{ assignment #011 011. `query 11 College Algebra 07-04-2007"