course Mth 158 X?{?????L???????assignment #013013. `query 13
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13:03:51 query 1.4.12 (was 1.4.6). Explain how you found the real solutions of the equation (1-2x)^(1/3) - 1 = 0
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RESPONSE --> (1 - 2x)^(1/3)-1=0 [(1- 2x)^(1/3)]^(3) - 1=0 1 - 2x - 1=0 1 - -2x - 1 +1=0+1 1 - 2x=1 1 - 2x - 1=1-1 -2x=0 x=0 {0} confidence assessment: 2
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13:04:15 ** Starting with (1-2x)^(1/3)-1=0 add 1 to both sides to get (1-2x)^(1/3)=1 then raise both sides to the power 3 to get [(1-2x)^(1/3)]^3 = 1^3. Since [(1-2x)^(1/3)]^3 = (1 - 2x) ^( 1/3 * 3) = (1-2x)^1 = 1 - 2x we have 1-2x=1. Adding -1 to both sides we get -2x=0 so that x=0. **
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RESPONSE --> yes! a good start! self critique assessment: 2
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13:15:36 1.4.28 (was 1.4.18). Explain how you found the real solutions of the equation sqrt(3x+7) + sqrt(x+2) = 1.
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RESPONSE --> sqrt(3x+7) + sqrt(x+2)=1 sqrt(3x+7)=1-sqrt(x+2) (sqrt(3x+7))^2=(1-sqrt(x+2))^2
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13:21:24 ** Starting with sqrt(3x+7)+sqrt(x+2)=1 we could just square both sides, recalling that (a+b)^2 = a^2 + 2 a b + b^2. This would be valid but instead we will add -sqrt(x+2) to both sides to get a form with a square root on both sides. This choice is arbitrary; it could be done either way. We get sqrt(3x+7)= -sqrt(x+2) + 1 . Now we square both sides to get sqrt(3x+7)^2 =[ -sqrt(x+2) +1]^2. Expanding the right-hand side using (a+b)^2 = a^2 + 2 a b + b^2 with a = -sqrt(x+2) and b = 1: 3x+7= x+2 - 2sqrt(x+2) +1. Note that whatever we do we can't avoid that term -2 sqrt(x+2). Simplifying 3x+7= x+ 3 - 2sqrt(x+2) then adding -(x+3) we have 3x+7-x-3 = -2sqrt(x+2). Squaring both sides we get (2x+4)^2 = (-2sqrt(x+2))^2. Note that when you do this step you square away the - sign, which can result in extraneous solutions. We get 4x^2+16x+16= 4(x+2). Applying the distributive law we have 4x^2+16x+16=4x+8. Adding -4x - 8 to both sides we obtain 4x^2+12x+8=0. Factoring 4 we get 4*((x+1)(x+2)=0 and dividing both sides by 4 we have (x+1)(x+2)=0 Applying the zero principle we end up with (x+1)(x+2)=0 so that our potential solution set is x= {-1, -2}. Both of these solutions need to be checked in the original equation sqrt(3x+7)+sqrt(x+2)=1 It turns out that the -1 gives us sqrt(4) + sqrt(1) = 1 or 2 + 1 = 1, which isn't true, while -2 gives us sqrr(1) + sqrt(0) = 1 or 1 + 0 = 1, which is true. x = -1 is the extraneous solution that was introduced in our squaring step. Thus our only solution is x = -2. **
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RESPONSE --> hmmmmmmm... I got one of the solution set by luck I'm sure. self critique assessment: 2
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13:33:00 1.4.40 (was 1.4.30). Explain how you found the real solutions of the equation x^(3/4) - 9 x^(1/4) = 0.
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RESPONSE --> x^(3/4) - 9x^(1/4)=0 x^(3/4)=9x^(1/4) (x^(3/4))^4=(9x^(1/4))^4 x^3=6561x x^3-6561x=0 x(x-81)(x+81)=0 x+0 x=-81 x=81 {-81,0,81} confidence assessment: 2
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13:34:45 ** Here we can factor x^(1/4) from both sides: Starting with x^(3/4) - 9 x^(1/4) = 0 we factor as indicated to get x^(1/4) ( x^(1/2) - 9) = 0. Applying the zero principle we get x^(1/4) = 0 or x^(1/2) - 9 = 0 which gives us x = 0 or x^(1/2) = 9. Squaring both sides of x^(1/2) = 9 we get x = 81. So our solution set is {0, 81). **
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RESPONSE --> ok so I have somewhat of the same answer...hmmmmmmmmm.. self critique assessment: 2
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13:44:48 1.4.46 (was 1.4.36). Explain how you found the real solutions of the equation x^6 - 7 x^3 - 8 =0
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RESPONSE --> x^6 - 7x^3 - 8=0 (x^(3)-8)(x^(3)+1)=0 x^3=8 or x^3= - 1 x=2 or x=-1/3 {-1/3,2} confidence assessment: 2
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13:45:16 ** Let a = x^3. Then a^2 = x^6 and the equation x^6 - 7x^3 - 8=0 becomes a^2 - 7 a - 8 = 0. This factors into (a-8)(a+1) = 0, with solutions a = 8, a = -1. Since a = x^3 the solutions are x^3 = 8 and x^3 = -1. We solve these equations to get x = 8^(1/3) = 2 and x = (-1)^(1/3) = -1. **
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RESPONSE --> ok sooo...-1 instead of -1/3 self critique assessment: 2
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13:49:48 1.4.64 (was 1.4.54). Explain how you found the real solutions of the equation x^2 - 3 x - sqrt(x^2 - 3x) = 2.
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RESPONSE --> let p=sqrt(x^2-3x) and p=x^2-3x p^2-p=2 p^2-p-2=0 (p-2)(p+1)=0 p=2 or p=-1 {-1,2} confidence assessment: 2
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13:53:22 ** Let u = sqrt(x^2 - 3x). Then u^2 = x^2 - 3x, and the equation is u^2 - u = 2. Rearrange to get u^2 - u - 2 = 0. Factor to get (u-2)(u+1) = 0. Solutions are u = 2, u = -1. Substituting x^2 - 3x back in for u we get sqrt(x^2 - 3 x) = 2 and sqrt(x^2 - 3 x) = -1. The second is impossible since sqrt can't be negative. The first gives us sqrt(x^2 - 3x) = 2 so x^2 - 3x = 4. Rearranging we have x^2 - 3x - 4 = 0 so that (x-4)(x+1) = 0 and x = -4 or x = 1. DER **
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RESPONSE --> ok so i needed to finish the equation...ok self critique assessment: 2
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18:46:19 1.4.90 (was 1.4.66). Explain how you found the real solutions of the equation x^4 + sqrt(2) x^2 - 2 = 0.
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RESPONSE --> ok...I'm stuck with this problem confidence assessment: 0
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18:52:08 ** Starting with x^4+ sqrt(2)x^2-2=0 we let u=x^2 so that u^2 = x^4: u^2 + sqrt(2)u-2=0 using quadratic formula u=(-sqrt2 +- sqrt(2-(-8))/2 so u=(-sqrt2+-sqrt10)/2 Note that u = (-sqrt(2) - sqrt(10) ) / 2 is negative, and u = ( -sqrt(2) + sqrt(10) ) / 2 is positive. u = x^2, so u can only be positive. Thus the only solutions are the solutions to x^2 = ( -sqrt(2) + sqrt(10) ) / 2. The solutions are x = sqrt( ( -sqrt(2) + sqrt(10) ) / 2 ) and x = -sqrt( ( -sqrt(2) + sqrt(10) ) / 2 ). Approximations are x = .935 and x = -.935. **
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RESPONSE --> hmmmmmmm...I will work on these kind of problems... self critique assessment: 2
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18:53:59 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> that last one was a a tough one! confidence assessment: 2
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?????W?{????? assignment #011 011. `query 11 College Algebra 02-25-2007
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20:27:54 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).
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RESPONSE --> x= score on exam A ave=90 B ave=80 3 parts to make final score final ave=[1(test ave) + 2(exam ave)]/3 if final 90 (1(85)+2x)/3=90 multiply both sides by 3 85+2x=270 subtract 85 from both sides 2x=185 divide both sides by 2 x=92.5 if final 80 (1(85)+2x)/3=80 multiply both sides by 3 85+2x=240 subtract 85 from both sides 2x=155 divide both sides by 2 x=77.5 confidence assessment: 2
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20:28:10 ** 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. **
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RESPONSE --> ok self critique assessment: 2
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20:42:02 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%.
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RESPONSE --> x=amt lent at 16%=0.16 ave of 180,000 per yr 1,000,000 - x at 19%=0.19(1,000,00 - x) total interest= 0.16x+0.19(1,000,000 - x) 0.16x+0.19(1,000,000 - x)=180,000 get ride of 16x+19(1,000,000 - x)=1,800,000 decimals 16x+19,000,000 - 19x=1,800,000 -3x+19,000,000=1,800,000 subtract 19,000,000 from -3x= - 1,000,000 both sides, then divide x=333,333 or 1/3 both sides by - 3 confidence assessment: 2
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20:55:51 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?
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RESPONSE --> T = D(=V*T) / V up 5hr = (V-3)(5hr) / V - 3 down 2.5hr = (V+3)(2.5hr) / V + 3 5(x-3)=2.5(x+3) 5x-15=2.5x+7.5 factor 5x-15-2.5x+15=2.5x+7.5-2.5x+15 subtract by2.5x, 2.5x=22.5 and add by 15 x=9 both sides speed of the boat is 9mph confidence assessment: 2
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20:56:15 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. **
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RESPONSE --> ok self critique assessment: 2
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21:03:55 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?
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RESPONSE --> ok this one has me! confidence assessment: 0
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21:24:16 ** 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. **
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RESPONSE --> ok self critique assessment: 1
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21:34:55 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?
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RESPONSE --> ok confidence assessment: 0
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21:40:36 ** 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. **
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RESPONSE --> ok self critique assessment: 2
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21:44:52 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?
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RESPONSE --> ok so I need help with problem solving. When would be a good time to maybe meet with you? I am at the college M,W, and F, 9a.m. til 12p.m. confidence assessment: 1
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21:48:16 ** 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. **
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RESPONSE --> ok self critique assessment: 2
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?|???y??????assignment #016 016. `query 16 College Algebra 02-25-2007
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21:57:31 2.1.28 (was 2.1.18). Dist (a, a) to (0, 0).
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RESPONSE --> d=(p1,p2)=sqrt[(x2-x1)^2 + (y2 - y1)^2] d=sqrt[(0-a)^2+(0-a)^2 d=sqrt(a^2+a^2) confidence assessment: 1
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21:59:22 ** Using the distance formula we get distance = sqrt( (x2 - x1)^2 + (y2 - y1)^2 ) = sqrt((a-0)^2+(a-0)^2) = sqrt(a^2+a^2) = sqrt(2 a^2) = sqrt(2) * sqrt(x^2) = sqrt(2) * a. COMMON ERROR: sqrt(a^2 + a^2) = a + a = 2 a INSTRUCTOR'S CORRECTION: sqrt( x^2 + y^2 ) is not the same thing as x + y. For example sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5 but 3 + 4 = 7. So you can't say that sqrt(a^2 + a^2) = a + a. **
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RESPONSE --> ok so I should have simplified self critique assessment: 2
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22:06:36 2.1.22 (was 2.1.12). Dist (2,-3) to (4,2).
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RESPONSE --> d=sqrt[(4-2)^2 + (2-(-3))^2] d=sqrt[(2^2) + (2-(-3))^2] d=sqrt(4+25) d=sqrt(29) confidence assessment: 1
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22:14:51 ** using the distance formula we get distance = sqrt( (x2 - x1)^2 + (y2 - y1)^2 ) = sqrt((2-4)^2+(-3-2)^2) = sqrt((-4)^2+(-6)^2) = sqrt(16+36) = sqrt(52) = sqrt(4) * sqrt(13) = 2 sqrt(13) **
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RESPONSE --> well I had the concept right. geeez! self critique assessment: 2
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22:45:07 2.1.30 (was 2.1.20). (-2, 5), (12,3), (10, -11) A , B, C.
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RESPONSE --> not a right triangle d(A,B)=sqrt(200) d(B,C)=sqrt(200) d(A,C)=sqrt(400) or 20 [d(A,C)]^2+[d(B,C)]^2=[d(A,B)]^2 (20)^2+[sqrt(200)]^2=(sqrt(200))^2 400+200=200 600=200 not equal confidence assessment: 1
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22:46:06 STUDENT SOLUTION: The triangle is a right triangle if the Pythagorean Theorem holds. d(A,B)= sqrt((-2-12)^2+(5-3)^2) sqrt(196+4)sqrt(200) 10 sqrt2 d(B,C)= sqrt((12-10)^2+(3+11)^2) sqrt(4+196) sqrt200 10 sqrt2 d(A,C)= sqrt((-2-10)^2 + (5+11)^2) sqrt(144+256) sqrt(400) 20 The legs of the triangle are therefore both 10 sqrt(2) while the hypotenuse is 20. The Pythagorean Theorem therefore says that (10sqrt2)^2+(10sqrt2)^2=(20)^2 which simplifies to 10^2 (sqrt(2))^2 + 10^2 (sqrt(2))^2 = 20^2 or 100 * 2 + 100 * 2 = 400 or 200+200=400 and finally 400=400. This verifies the Pythagorean Theorem and we conclude that the triangle is a right triangle. **
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RESPONSE --> ok....I will finish tomorrow! self critique assessment: 2
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