course mth 158 ƗľՖsٱɦassignment #001
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17:27:55 Query Introduction to General Themes; Examples (no summary needed) What were some of the things in this introduction that you found interesting or surprising?
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RESPONSE --> Determining the value of an expression and knowing what to do without having already memorized a formula. Evaluating an expression and turning it into a function and solving. Princple behind math and how it applies to every day as in our jobs and life. Being able to base life off of math and know how to solve a simple daily problem with its application. confidence assessment: 3
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17:27:59 ** Continue to the next question **
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RESPONSE --> self critique assessment: 3
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dy assignment #001 001. `query 1 College Algebra 09-19-2007
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17:36:38 R.1.14 (was R.1.6) Of the numbers in the set {-sqrt(2), pi + sqrt(2), 1 / 2 + 10.3} which are counting numbers, which are rational numbers, which are irrational numbers and which are real numbers?
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RESPONSE --> From the list: Real, all are real Irrational, all are irrtional confidence assessment: 3
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17:38:38 ** Counting numbers are the numbers 1, 2, 3, .... . None of the given numbers are counting numbers Rational numbers are numbers which can be expressed as the ratio of two integers. 1/2+10.3 are rational numbers. Irrational numbers are numbers which can be expressed as the ratio of two integers. {-sqrt(2)}, pi+sqrt(2) are irrational numbers.. **
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RESPONSE --> I understand that 1/2+10.3 would not be included in the set of irrational but would be expressed as rational numbers because they are the ratio of the two integers 1/2 and 10.3 self critique assessment: 3
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17:39:05 R.1.32 (was R.1.24) Write in symbols: The product of 2 and x is the product of 4 and 6
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RESPONSE --> 2*x=4*6 confidence assessment: 3
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17:39:20 ** The product of 2 and x is 2 * x and the product of 4 and 6 iw 4 * 6. To say that these are identical is to say that 2*x=4*6. **
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RESPONSE --> understood self critique assessment:
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17:44:21 R.1.50 (was R.1.42) Explain how you evaluate the expression 2 - 5 * 4 - [ 6 * ( 3 - 4) ]
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RESPONSE --> This must be solved according to the order of operations, so it is necessary to begin with the outer most brackets as indicated in the expression. [6*(3-4)] it is appointed in this expression by the existance of the paranthesis, that 3-4 must be the first solved. So, 3-4=-1 Leaving the expression to read, 2-5*4-(6*-1) Now we must finish what is left of what originated inside the brackets by solving 6*-1 giving us a -6. This leaves, 2-5*4-(-6)= 2-5*4+6 As we know in order of operations, next of importance of our equation is multiplication, so we multiply 5*4=20, 2-20+6 remains. Since addition and subtraction are held on the same level in terms of order, we solve the rest from left to right. 2-20= -18+6= -12 confidence assessment: 3
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17:44:41 **Starting with 2-5*4-[6*(3-4)]. First you evaluate the innermost group to get 2-5*4-[6*-1] . Then multiply inside brackets to get 2-5*4+6. Then do the multiplication to get 2-20+6. Then add and subtract in order, obtaining -12. **
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RESPONSE --> I wrote outermost, I meant inner!* self critique assessment: 3
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17:48:36 R.1.80 (was R.1.72) Explain how you use the distributive property to remove the parentheses from the express (x-2)(x-4).
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RESPONSE --> you must use the FOIL method here; first, outer, inner, last. By this we begin with (x-2) an apply the x to every aspect of (x-4). So we multiply x*x=x^2 and then x*-4= -4 Next we must do the same with the -2, multiplying -2*x=-2x and then -2*-4=8 When putting it all together we get the expression, x^2-2x-4x+8, simplifying to x^2-6x+8 confidence assessment: 3
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17:49:55 ** COMMON ERROR: Using FOIL. FOIL is not the Distributive Law. FOIL works for binomial expressions. FOIL follows from the distributive law but is of extremely limited usefulness and the instructor does not recommend relying on FOIL. Starting with (x-2)(x-4) ; one application of the Distributive Property gives you x(x-4) - 2(x-4) . Applying the property to both of the other terms we get x^2 - 4x - (2x -8). Simplifying: x^2 - 4x - 2x + 8 or x^2 - 6x + 8. *
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RESPONSE --> I understand that the FOIL method is not the same as the distributive property in that it only works with binomials. self critique assessment: 3
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17:54:10 R.1.86 (was R.1.78) Explain why (4+3) / (2+5) is not equal to 4/2 + 3/5.
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RESPONSE --> order of operations. The expession does not show 4 and 2 as numerators, nor 3 and 5 to be denominators. The entire numerator here is (4+3) and denominator (2+5). The parenthesis indicate that you must solve what lies within them first. So you would end up with (4+3)=7/(2+5)=7= 7/7=1 Even if these were not in parenthesis, we would know that you could not split the equation as shown, that by order of operations, you would divide 3 and 2 first. and add in order from left to right. 4+3/2+5= 4+(3/2)+5 confidence assessment: 3
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17:54:51 ** Good answer but at an even more fundamental level it comes down to order of operations. (4+3)/(2+5) means 7/7 which is equal to 1. By order of operations, in which multiplications and divisions precede additions and subtractions, 4/2+3/5 means (4/2) + (3/5), which gives us 2+3/5 = 2 3/5 **
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RESPONSE --> understood self critique assessment: 3
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17:55:37 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Very necessary to go through this review and use it as a building block for successive lessons since I have not taken a math class in so long. confidence assessment: 3
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~_} assignment #002 002. `query 2 College Algebra 09-19-2007
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18:06:04 R.2.46 (was R.2.36) Evaluate for x = -2, and y = 3 the expression (2x - 3) / y and explan how you got your result.
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RESPONSE --> To solve this, it is necessary to take the information given, x= -2, y=3 and plug them into the appropriate places in the expression. This will read, [(2*-2)-3]/3. So we work out the innermost parenthesis first, 2*-2= -4 leaving, (-4-3)/3. Next we finish the parenthesis and solve -4-3= -7 -7/3 now we divide -7 by 3. This can either be left as a fraction sine it does not evenly divide, or be expressed as a decimal, -2.333 repeating. confidence assessment: 3
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18:06:11 ** Starting with (2x-3)/y we substitute x=-2 and y=3 to get (2*(-2) - 3)/3 = (-4-3)/3= -7/3. **
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RESPONSE --> understood self critique assessment: 3
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XRԃ֛ assignment #002 002. `query 2 College Algebra 09-23-2007
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17:40:39 R.2.46 (was R.2.36) Evaluate for x = -2, and y = 3 the expression (2x - 3) / y and explan how you got your result.
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RESPONSE --> confidence assessment:
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17:40:57 ** Starting with (2x-3)/y we substitute x=-2 and y=3 to get (2*(-2) - 3)/3 = (-4-3)/3= -7/3. **
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RESPONSE --> self critique assessment:
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17:45:27 R.2. 55 (was R.2.45) Evaluate for x = 3 and y = -2: | |4x| - |5y| | and explan how you got your result.
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RESPONSE --> first plug in 3 for x and -2 for y. giving: | |4*3| - |5*-2| | We solve what is in the absolute value symbols first by order of operations. giving us the |4*3|=|12|=12 and |5*-2|=|-10|=10, leaving us with |10-12|=|-2| and take the absolute value of 2, with the final result of the expression to be 2. Absolute value is the total distance the number is from 0 on a number line, which is why negatives do not apply because you cannot have a negative count. confidence assessment: 3
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17:45:37 ** Starting with | | 4x |- | 5y | | we substitute x=3 and y=-2 to get | | 4*3 | - | 5*-2 | | = | | 12 | - | -10 | | = | 12-10 | = | 2 | = 2. **
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RESPONSE --> understood self critique assessment: 3
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17:51:45 R.2.64 (was R.2.54) Explain what values, if any, must not be present in the domain of the expression (-9x^2 - x + 1) / (x^3 + x)
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RESPONSE --> x cannot be equal to 0 in the domain of this expression. If 0 is present in the domain, then when solved, 0^3+0=0. This will leave us with 0 in the denominator which gives us an undefined result. confidence assessment: 2
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17:53:28 ** The denominator of this expression cannot be zero, since division by zero is undefined. Since x^3 + x factors into (x^2 + 1) ( x ) we see that x^3 + x = 0 only if x^2 + 1 = 0 or x = 0. Since x^2 cannot be negative x^2 + 1 cannot be 0, so x = 0 is indeed the only value for which x^3 + x = 0. **
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RESPONSE --> came to the same conclusion. I understand also that you can find this by factoring x^3 +x into (x^2+1)(x) to give the equivalency of 0 as its parent expression. self critique assessment: 3
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18:01:30 R.2.73 (was R.4.6). What is (-4)^-2 and how did you use the laws of exponents to get your result?
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RESPONSE --> a^m-n=1/a^n-m. By using this rule from the law of exponents, we can apply (-4)^-2 and convert this expression to 1/(-4)^2. By moving the negative exponent to the denominator rather than the numerator-- ((-4)^-2)/1 -- it elimantes the negative, making the remainding expression solvable. 1/(-4)^2=1/16 confidence assessment: 3
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18:01:44 **Since a^-b = 1 / (a^b), we have (-4)^-2 = 1 / (-4)^2 = 1 / 16. **
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RESPONSE --> understood self critique assessment: 3
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18:04:22 Extra Problem. What is (3^-2 * 5^3) / (3^2 * 5) and how did you use the laws of exponents to get your result?
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RESPONSE --> By the law of exponents we can solve the numerator first. 3^-2*5^3=15^-2+3=15^1=15 and conclude with the denominator, 3^2=9*5=45, leaving 15/45 which simplifies to 1/3 confidence assessment: 3
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18:10:24 ** (3^(-2)*5^3)/(3^2*5). Grouping factors with like bases we have 3^(-2)/3^2 * 5^3 / 5. Using the fact that a^b / a^c = a^(b-c) we get 3^(-2 -2) * 5^(3-1), which gives us 3^-4 * 5^2. Using a^(-b) = 1 / a^b we get (1/3^4) * 5^2. Simplifying we have (1/81) * 25 = 25/81. **
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RESPONSE --> I see that I solved for an exponential equation by the wrong rule. I solved for it if had been possibly just the numerator and was not being divided by 3^2*5. I understand that you would group the factors with like bases, and that when working with a^b/a^c, you will convert this to a^(b-c). This will give 3^(-2-2)*5^(3-1) [The 1 being understood to be carried by the 5 in the denominator 5^1=5]. This will give 3^-4*5^2=1/3^4*5^2 1/3^4=1/81 5^2=25 25*1/81=25/81 self critique assessment: 3
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18:27:26 R.2.94. Express [ 5 x^-2 / (6 y^-2) ] ^ -3 with only positive exponents and explain how you used the laws of exponents to get your result.
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RESPONSE --> we can break this equation in half in order to make it easier to work with by numerator and denominator. by using the law of exponents, 5x^-3(x-2)-3=5^1*-3(x-2*-3)=5^-3(x6)=1/5^3(x^6)=1/125(x^6) We multiply the outer exponent by the inner exponent to simplify the expression. When multiplying the exponent of 5, being 1, by -3 we get 5^-3, and again to get rid of this remaining negative exponent we convert this into 1/5^3=1/125. The x is already being raised to the -2 power, so -3*-2=6, leaving us with a positive integer for x, x^6. Same applies for the denominator; (6y^-2)^-3=6^(1*-3)*y^(-2*-3)=6^-3*y^6=1/6^3*y^6=1/216*y^6. when combining again the numerator and denominator we are left with (1/125*x^6)/(1/216*y^6) To get rid of the fractions, we need to multiply, leaving 216*x^6/125*y^6. confidence assessment: 3
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18:27:53 [ 5 x^-2 / (6 y^-2) ] ^ -3 = (5 x^-2)^-3 / (6 y^-2)^-3, since (a/b)^c = a^c / b^c. This simplifies to 5^-3 (x^-2)^-3 / [ 6^-3 (y^-2)^-3 ] since (ab)^c = a^c b^c. Then since (a^b)^c = a^(bc) we have 5^-3 x^6 / [ 6^-3 y^6 ] . We rearrange this to get the result 6^3 x^6 / (5^3 y^6), since a^-b = 1 / a^b.
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RESPONSE --> received same result self critique assessment: 3
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18:30:33 Extra Problem. Express (-8 x^3) ^ -2 with only positive exponents and explain how you used the laws of exponents to get your result.
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RESPONSE --> First we distribute to get rid of the outer most exponent. -8^(1*-2)*x^(3*-2)=-8^-2*x^-6. We must now get rid of our negative exponents by moving the corresponding exponents to the denominator under the number 1. 1/(8^2*x^6)=1/64*x^6 confidence assessment: 3
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18:32:22 ** ERRONEOUS STUDENT SOLUTION: (-8x^3)^-2 -1/(-8^2 * x^3+2) 1/64x^5 INSTRUCTOR COMMENT:1/64x^5 means 1 / 64 * x^5 = x^5 / 64. This is not what you meant but it is the only correct interpretation of what you wrote. Also it's not x^3 * x^2, which would be x^5, but (x^3)^2. There are several ways to get the solution. Two ways are shown below. They make more sense if you write them out in standard notation. ONE CORRECT SOLUTION: (-8x^3)^-2 = (-8)^-2*(x^3)^-2 = 1 / (-8)^2 * 1 / (x^3)^2 = 1/64 * 1/x^6 = 1 / (64 x^5). Alternatively (-8 x^3)^-2 = 1 / [ (-8 x^3)^2] = 1 / [ (-8)^2 (x^3)^2 ] = 1 / ( 64 x^6 ). **
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RESPONSE --> understood, I left out my parenthesis in my result, 1/(64*x^6) to illustrate the order of operations and if you were solving for x, the denominator is grouped together. self critique assessment: 3
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18:37:51 R.2.90 (was R.4.36). Express (x^-2 y) / (x y^2) with only positive exponents and explain how you used the laws of exponents to get your result.
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RESPONSE --> (x^-2*y)/(x*y^2)=x^(-2-1)=x^-3*y^(1-2)=y^-1*x^-3 leaving 1/(x^3*y) by the law of exponents you must factor like groups, meaning you need to subtract the denominators exponent from its corresponding base from the numerators exponent. ex, (x^-2)/(x)=x^(-2-1) (1 being the understood exponent for the solitary x, x^1=x) confidence assessment: 3
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18:38:02 ** (1/x^2 * y) / (x * y^2) = (1/x^2 * y) * 1 / (x * y^2) = y * 1 / ( x^2 * x * y^2) = y / (x^3 y^2) = 1 / (x^3 y). Alternatively, or as a check, you could use exponents on term as follows: (x^-2y)/(xy^2) = x^-2 * y * x^-1 * y^-2 = x^(-2 - 1) * y^(1 - 2) = x^-3 y^-1 = 1 / (x^3 y).**
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RESPONSE --> understood self critique assessment: 3
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18:48:22 Extra Problem. . Express 4 x^-2 (y z)^-1 / [ (-5)^2 x^4 y^2 z^-5 ] with only positive exponents and explain how you used the laws of exponents to get your result.
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RESPONSE --> 4x^-2 (y^1*-1 z^1*-1)=4x^-2 (y^-1 z^-1)/(25x^4 y^2 z^-5) 4x^(-2-4)=(4/25)x^-6 y^-1-2=y^-3 z^-1+5=z^4 =(4/25)x^-6 y^-3 z^4=z^4/(4/25)x^6 y^3 confidence assessment: 2
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18:51:05 ** Starting with 4x^-2(yz)^-1/ [ (-5)^2 x^4 y^2 z^-5] Squaring the -5 and using the fact that (yz)^-1 = y^1 * z^-1: 4x^-2 * y^-1 * z^-1/ [25 * x^4 * y^2 * z^-5} Grouping the numbers, and the x, the y and the z expression: (4/25) * (x^-2/x^4) * (y^-1/y^2) * (z^-1/z^-5) Simplifying by the laws of exponents: (4/25) * x^(-2-4) * y^(-1-2) * z^(-1+5) Simplifying further: (4/25) * x^-6 * y^-3 * z^4 Writing with positive exponents: 4z^4/ (25x^6 * y^3 ) **
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RESPONSE --> I understand the solution and how it differs from my own in that the 4 will remain in the numerator of the equation because when factoring the 4x in the original numerators and -5^2 or 25, from the denominator are factored together and away from the x in order to complete the law of exponents properly. self critique assessment: 3
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18:51:31 R.2.122 (was R.4.72). Express 0.00421 in scientific notation.
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RESPONSE --> 4.21*10^-3 confidence assessment: 3
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18:51:42 ** 0.00421 in scientific notation is 4.21*10^-3. This is expressed on many calculators as 4.21 E-4. **
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RESPONSE --> understood self critique assessment: 3
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18:51:55 R.2.128 (was R.4.78). Express 9.7 * 10^3 in decimal notation.
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RESPONSE --> 9700 confidence assessment: 3
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18:52:28 ** 9.7*10^3 in decimal notation is 9.7 * 1000 = 9700 **
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RESPONSE --> understood, 10^3=1000*9.7=9700 self critique assessment: 3
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18:57:19 R.2.150 (was R.2.78) If an unhealthy temperature is one for which | T - 98.6 | > 1.5, then how do you show that T = 97 and T = 100 are unhealthy?
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RESPONSE --> plug in 97 and 100 for T. as long as the expression remainding is not equal to 1.5, which would indicate healthy, we will know is its unhealthy. |97-98.6|>1.5 |-1.6|>1.5 1.6>1.5 is true, so we know that 97 is unhealthhy. |100-98.6|>1.5 |100-98.6|>1.5 1.4>1.5 untrue, so it remains that 100 is an unhealthy temperature as well. confidence assessment: 2
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18:59:09 ** You can show that T=97 is unhealthy by substituting 97 for T to get | -1.6| > 1.5, equivalent to the true statement 1.6>1.5. But you can't show that T=100 is unhealthy, when you sustitute for T then it becomes | 100 - 98.6 | > 1.5, or | 1.4 | > 1.5, giving us 1.4>1.5, which is an untrue statement. **
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RESPONSE --> I understand that it is not that the equivelancy of the equation be equal to 1.5 to show unhealthy, it is if the statement after the expression has been solved is true or not. when using 97, we can prove that this temperature is unhealthy, but when plugging in 100 with the expression 1.4>1.5 we cannot prove it by this method because we know this is not a true statement. self critique assessment: 3
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h~W} assignment #003 003. `query 3 College Algebra 09-23-2007
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19:01:55 R.3.12 (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 --> by using the pythagorean theorem, a^2+b^2=c^2 you can find the hypotenuse of the following right triangle. 14^2+48^2=c^2 196+2304=c^2 'sqrt 2500= 'sqrt c c=50 confidence assessment: 3
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19:03:47 ** 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 --> I understand that when taking the square root, it can be positive or negative by rules of multiplication (+)*(+)=(+) (-)*(-)=(+) but since we are dealing with triangles a side cannot be negative so it can only include the positive form of itself, 50. self critique assessment: 3
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19:07:20 R.3.18 (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 --> you must apply these numbers into the equation a^2+b^2=c^2 We know that c represents the hypotenuse, and that this is the longest side. Which means 10^2+24^2=26^2 100+576=676 676=676 confidence assessment: 3
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19:07:33 ** 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 --> understood self critique assessment: 3
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19:13:54 R.3.30 (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 --> by using the equation v=4/3pi*r^3 to solve for the volume and S=4pi*r^2 for surface area. We know our radius is 3 meters so when we apply that for volume V=4/3pi*3m^3=4/3pi* 3^3*m^3 =4/3pi*27m^3 =pi36m^3 we use a cube here to represent the three variables, length width and heighth. With surface area we just want the surface of one side of one facing of the sphere, so we will only multiply length and height. S=4pi*3m^2=4pi*3^2*m^2 =4pi*9m^2 =pi36m^2 confidence assessment: 3
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19:14:04 ** 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 --> understood self critique assessment: 3
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19:29:14 R.3.42 (was R.3.36). A pool of radius 10 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 --> We know that the radius of the pool is 10 ft, making its diameter or length from one point of the pool to the point directly across from the corresponding point on the opposite side, 20 ft. We also know that around the entire pool is 3 feet of deck from each side, so this would contribute a total of 6 ft to the 20ft diameter of the pool. We need this information to first figure out the area of the pool and the deck combined. so we add the 20 ft + 6 ft to get a total diameter of the entire area= 26ft Now we can use the formula, A=p*r^2, the radius being 13 ft. A=pi*13ft^2 =pi169ft^2 Now that we know the area of the entire fixture, we need to find the area of the pool itself. we know its radius is 10ft, so A=pi*10ft^2 =pi100ft^2 So with the pool being pi 100ft^2 and the whole area deck and pool pi169ft^2, we can subtract the area of the pool from the whole area to get the remaining area of the deck. pi169ft^2-pi100ft^2=pi 69 ft^2=A of deck confidence assessment: 3
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19:29:21 ** 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 --> understood! self critique assessment: 3
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19:29:30 005. `query 5
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RESPONSE --> self critique assessment: 3
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19:42:40 R.4.36 (was R.5.30). What is the single polynomial that is equal to 8 ( 4 x^3 - 3 x^2 - 1 ) - 6 ( 4 x^3 + 8 x - 2 )?
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RESPONSE --> distrubute 8 and -6 to corresponding expressions: 8(4x^3-3x^2-1)=32x^3-24x^2-48x-8 -6(4x^3+8x-2)=-24x^3-48x+12 32x^3-24x^2-48x-8-24x^3-48x+12 (32x^3-24x^3)=8x^3 (-8+12)=4 8x^3-24x^2-48x+4 you can now factor out a 4 4(2x^3-6x^2-12x+1) confidence assessment: 3
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19:43:55 ** ERRONEOUS STUDENT SOLUTION: To make this problem into a single polynomial, you can group like terms together. (8-6)+ (4x^3-4x^3) + (-3x^2) + (8x) + (-1+2). Then solve from what you just grouped...2 (-3x^2+8x+1). INSTRUCTOR CORRECTION: 8 is multiplied by the first polynomial and 6 by the second. You can't isolate them like that. Starting with 8 ( 4 x^3 - 3 x^2 - 1 ) - 6 ( 4 x^3 + 8 x - 2 ) use the Distributive Law to get 32 x^3 - 24 x^2 - 8 - 24 x^3 - 48 x + 12. Then add like terms to get 8x^3 - 24x^2 - 48x + 4 **
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RESPONSE --> I got the same answer only I simplified it too far and should have left the expression without factoring out the 4 again. self critique assessment: 3
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19:49:31 R.4.60 (was R.5.54). What is the product (-2x - 3) ( 3 - x)?
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RESPONSE --> first you must use the distributive property to solve (-2x-3)(3-x) -2x*3=-6x -2x*-x=2x^2 2x^2-6x -3*3=9 =-3*-x=3x 3x+9 and now you can combine like terms, -6x+3x=-3x 2x^2-3x-9 confidence assessment: 3
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19:49:45 ** Many students like to use FOIL but it's much better to use the Distributive Law, which will later be applied to longer and more complicated expressions where FOIL does not help a bit. Starting with (-2x - 3) ( 3 - x) apply the Distributive Law to get -2x ( 3 - x) - 3 ( 3 - x). Then apply the Distributive Law again to get -2x(3) - 2x(-x) - 3 * 3 - 3 ( -x) and simiplify to get -6x + 2 x^2 - 9 + 3x. Add like terms to get 2 x^2 - 3 x - 9. **
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RESPONSE --> understood self critique assessment: 3
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19:52:09 R.4.66 (was R.5.60). What is the product (x - 1) ( x + 1) and how did you obtain your result using a special product formula?
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RESPONSE --> this represents the difference of squares. when we see a square being multiplied by its negative and positive we know its the difference of squares. this will give us x^2-1^2 or x^2-1. Using the distributive law to check we get the same thing (x-1)(x+1) x^2-x+x-1 x^2-1 confidence assessment: 3
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19:52:19 ** Starting with (x-1)(x+1) use the Distributive Law once to get x ( x + 1) - 1 ( x+1) then use the Distributive Law again to get x*x + x * 1 - 1 * x - 1 * 1. Simplify to get x^2 +- x - x + - 1. Add like terms to get x^2 - 1. **
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RESPONSE --> understood self critique assessment: 3
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19:57:52 R.4.84 (was R.5.78). What is (2x + 3y)^2 and how did you obtain your result using a special product formula?
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RESPONSE --> This is the squaring of binomials, or perfect squares. We know that when this occurs, (x+a)^2=x^2+2ax+a^2 This will give us 2x^2 =2^2*x^2=4x^2 2(2)(3)(x)(y)=12xy 3y^2=3^2y^2=9y^2 all together this leaves, 4x^2+12xy+9y^2 This can be checked with the distributive property 4x^2+6xy+6xy+9y^2= 4x^2+12xy+9y^2 confidence assessment: 3
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19:58:04 ** The Special Product is (a + b)^2 = a^2 + 2 a b + b^2. Letting a = 2x and b = 3y we get (2x)^2 + 2 * (2x) * (3y) + (3y)^2, which we expand to get 4 x^2 + 12 x y + 9 y^2. **
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RESPONSE --> understood self critique assessment: 3
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20:03:09 R.4.90 (was R.5.102). Explain why the degree of the product of two polynomials equals the sum of their degrees.
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RESPONSE --> I don't understand how this is worded confidence assessment: 1
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20:05:22 ** STUDENT ANSWER AND INSTRUCTOR COMMENTS: The degree of the product of two polynomials equals the sum of their degrees because you use the law of exponenents and the ditributive property. INSTRUCOTR COMMENTS: Not bad. A more detailed explanation: The Distributive Law ensures that you will be multiplying the highest-power term in the first polynomial by the highest-power term in the second. Since the degree of each polynomial is the highest power present, and since the product of two powers gives you an exponent equal to the sum of those powers, the highest power in the product will be the sum of the degrees of the two polynomials. Since the highest power present in the product is the degree of the product, the degree of the product is the sum of the degrees of the polynomials. **
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RESPONSE --> i understand that the degree of a polynomial is the highest power present in the polynomial and that this occurs in occordance to the distributive law. because the product of two powers equals the highest power, the product will then be the same as the sum of the degrees of those polynomials. self critique assessment: 2
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20:05:44 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> interesting how easy it was to pick back up confidence assessment: 3
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20:05:49 006. `query 6
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RESPONSE --> confidence assessment:
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\֑P꺄؟ˌי` assignment #005 005. `query 5 College Algebra 09-23-2007
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21:37:29 R.5.22 (was R.6.18). What do you get when you factor 36 x^2 - 9 and how did you get your result?
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RESPONSE --> This is the difference of squares, we we know automatically that our answer will be the square root of x and its constant and plus and minus the square root of the second number. sqrt 36=6 sqrt x^2=x sqrt -9 + or - 9 =(6x-3)(6x+3) this is evident in the distributive property as well this can be checked with the distributive property, (6x-3)(6x+3) 36x^2+18x-18x-9=36x^2-9 confidence assessment: 3
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21:37:38 ** 36x^2-9 is the difference of two squares. We write this as (6x)^2-3^2 then get (6x-3)(6x+3), using the special formula difference of two squars. **
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RESPONSE --> understood self critique assessment: 3
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21:40:39 R.5.28 (was R.6.24 What do you get when you factor x^2 + 10 x + 1 and how did you get your result?
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RESPONSE --> This is a perfect square. We see this because it fits x^2+2ax+a^2=(x+a)^2 x^2+2(5)x+5^2=x^2+10x+25 So we know that this will equal, (x+5)^2 confidence assessment: 3
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21:46:09 ** STUDENT SOLUTION: x^2+10x+1 is prime because there are no integers whose product is 10 and sum is 1 INSTRUCTOR COMMENTS: The sum should be 10 and the product 1. I agree that there are no two integers with this property. Furthermore there are no two rational numbers with this property. So you would never find the factors by inspection. However that doesn't mean that there aren't two irrational numbers with the property. For example 10 and 1/10 come close, with product 1 and sum 10.1. The quadratic formula tells you in fact that the two numbers are ( -10 + sqrt( 10^4 - 4 * 1 * 1) ) / (2 * 1) and ( -10 - sqrt( 10^4 - 4 * 1 * 1) ) / (2 * 1) . Since 10^2 - 4 = 96 is positive, these are real numbers, both irrational. So the polynomial isn't prime. **
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RESPONSE --> In the book the problem #28 is x^2+10x+25, not +1. This is why I solved it the way I did, I was looking at my work on paper. By this, I understand x^2+10x+1 does not have an integer solution, nor a rational solution. By means of the quadratic equation and using 1/10 and 10 with a near sum of 10.1 giving it an irrational solution. By means of following the sections, which quadratic formula is not introduced, I see how most students would consider this prime. self critique assessment: 3
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21:48:54 R.5.34 (was R.6.30). What do you get when you factor x^3 + 125 and how did you get your result?
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RESPONSE --> This is the sum of two cubes. (x+a)(x^2-ax+a^2)=x^3+a^3 so we are given, the latter of the equation, x^3+125 (x^3+5^3), so we can plug in the equivalency. (x+5)(x^2-5x+25) confidence assessment: 3
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21:49:01 ** x^3+125 is the sum of two cubes, with 125 = 5^3. We know that a^3 + b^3 = (a+b) ( a^2 - 2 a b + b^2). So we write x^3+5^3 = (x+5)(x^2-5x+25). **
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RESPONSE --> understood self critique assessment: 3
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21:53:35 R.5.46 (was R.6.42). What do you get when you factor x^2 - 17 x + 16 and how did you get your result?
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RESPONSE --> Factor by grouping: we need the multiples of 16 that will add to give us -17. These will be -1 and -16, -1*-16=16 so we will plug these into our equation replacing -17x, x^2-x-16x+16 Now we can factor by grouping, (x^2-x)(-16x+16) we can factor out like terms from each (x^2-x)=x(x-1) (-16x+16)=-16(x-1) we get (x-1) from each of these so we can combine what we factored from each equation giving us (x-16)(x-1) confidence assessment: 3
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21:54:02 ** x^2-17x+16 is of the form (x + a) ( x + b) = x^2 + (a + b) x + ab, with a+b = -17 and ab = 16. If ab = 16 then we might have a = 1, b = 16, or a = 2, b = 8, or a = -2, b = -8, or a = 4, b = 4, or a = -1, b = -16, or a = -4, b = -4. These are the only possible integer factors of 16. In order to get a + b = -17 we must have at least one negative factor. The only possibility that gives us a + b = -17 is a = -1, b = -16. So we conclude that x^2 - 17 x + 16 = (x-16)(x-1). **
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RESPONSE --> understood self critique assessment: 3
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21:57:00 R.5.52 (was R.6.48). What do you get when you factor 3 x^2 - 3 x + 2 x - 2 and how did you get your result?
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RESPONSE --> 3x^2-3x+2x-2 we can factor by grouping here by combining variables with like terms (3x^2-3x)(2x-2) now we factor each (3x^2-3x)=3x(x-1) (2x-2)=2(x-1) since both give us (x-1) we can combine what we factored from each (3x+2) =(3x+2)(x-1) confidence assessment: 3
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21:57:10 ** This expression can be factored by grouping: 3x^2-3x+2x-2 = (3x^2-3x)+(2x-2) = 3x(x-1)+2(x-1) = (3x+2)(x-1). **
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RESPONSE --> understood self critique assessment: 3
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22:04:57 R.5.64 (was R.6.60). What do you get when you factor 3 x^2 - 10 x + 8 and how did you get your result?
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RESPONSE --> 3x^2-10x+8. For this trinomial we can faccctor by grouping. We need a product of 3 and 8 who's multiples will add up to a -10. We know that by multiplying two negatives we get a positive number, and by that, we can use the multiples -4 and -6 of 24 that add to give us -10 3x^2-6x-4x+8 Now we can factor like terms (3x^2-6x)(-4x+8) (3x^2-6x)=3x(x-2) (-4x+8)=-4(x-2) now we can combine our binomial, (3x-4)(x-2) confidence assessment: 3
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22:05:22 ** Possibilities are (3x - 8) ( x - 1), (3x - 1) ( x - 8), (3x - 2) ( x - 4), (3x - 4) ( x - 2). The possibility that gives us 3 x^2 - 10 x + 8 is (3x - 4) ( x - 2). **
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RESPONSE --> understood self critique assessment: 3
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22:12:15 R.5.82 (was R.6.78). What do you get when you factor 14 + 6 x - x^2 and how did you get your result?
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RESPONSE --> This equation we would try factoring by grouping. But here we need multiples of -14 to add up to 6. The only multiples of 14 are 1 and 14, which will not give us positive 6. Or 2 and 7 which will not give us 6 either. By this method this equation would be considered prime. But by means of the quadratic eqaution we know that it is not prime, it just does not have any real or integer solutions. Its solutions will be irrational numbers. confidence assessment: 3
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22:12:36 ** This expression factors, but not into binomtials with integer coefficients. We could list all the possibilities: (x + 7) ( -x + 2), (x + 2) ( -x + 7), (x + 14) ( -x + 1), (x + 1)(-x + 14), but none of these will give us the desired result. For future reference: You won't find the factors in the usual manner. The quadratic formula tells us that there are factors ( -6 + sqrt(6^2 - 4 * 14 * (-1) )) / (2 * -1) and ( -6 - sqrt(6^2 - 4 * 14 * (-1) ) ) / (2 * -1) . Since sqrt(6^2 - 4 * 14 * (-1) ) = sqrt(36 + 56) = sqrt(92) is a real number these solutions are real numbers but again, as in a previous example, they aren't rational numbers and nobody could ever find them by inspection. This is not something you're expected to do at this point. **
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RESPONSE --> understood self critique assessment: 3
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22:12:40 007. `query 7
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RESPONSE --> self critique assessment: 3
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