course Mth 158 Will we be responsible for knowing all of the formulas for volume, surface area, etc. for the tests? Also, did you get my assmt 1 I sent on 9/1?
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08:18:12 query 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 --> First I substituted for x and y, then solved first for the numcerator. 2(-2) - 3/3 = -4 -3/ 3 = -7/3.
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08:18:25 ** 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 --> ok
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08:21:42 query 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 I substituted for x and y. Then I took the absolute value of 4 *3 and 5*-2 to equal 12-10. The answer is |2| or 2.
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08:21:51 ** 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 --> ok
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08:24:06 query 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 --> The values (c) x=0 and (d) x=-1. This is true because the denominator becomes 0 with these values. You can't divide by 0.
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08:26:37 ** 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 --> I see what I did wrong.
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08:27:57 query R.2.76 \ 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 --> The answer is 16. Since -4 is in parentheses, it would be (-4)(-4).
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08:30:02 ** order of operations implies exponentiation before multiplication; the - in front of the 4 is not part of the 4 but is an implicit multiplication by -1. Thus only 4 is raised to the -2 power. -4^(-2) Since a^-b = 1 / (a^b), we have 4^-2 = 1 / (4)^2 = 1 / 16. The - in front then gives us -4^(-2) = - ( 1/ 16) = -1/16. If the intent was to take -4 to the -2 power the expression would have been written (-4)^(-2).**
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RESPONSE --> I'm sorry. I gave the answer for 74. This is exactly what I had for 76. I know that you use the reciprocal and the number should be negative because the -4 is not in parentheses.
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08:34:48 query 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 --> This is how I would do it. 3^(2-2) * 5^(3-1). This would equal 3^0 * 5^2 = 1 * 25 = 25
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08:36:10 ** (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 must have missed the negative on the numerator. I have the right idea.
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08:42:34 query 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 --> You use the reciprocal of negative exponents. This would make it (5y^2/6x^2)^-3 =(6x^2/ 5y^2)^3 216y^6/125x^6.
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08:48:56 [ 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 --> So I needed to multiply exponents first. ok
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08:54:13 query 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 --> (-8)^-2 * (x^3)^-2= 1/64 * x^-6 = 1/64 *1/x^6= 1/64x^6 This follows the (ab)^n = a^nb^n rule.
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08:56:49 ** 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 --> I thought that was the answer that I gave.
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09:01:43 query 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 --> I used a^m/a^n = a^m-n. x^-2-1 * y^1-2= x^-3 * y^-1 = 1/x^3y
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09:03:01 ** (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 --> ok
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09:11:47 query 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 --> I would first use a^m/a^n =a^m-n. 4(x^-2-4)(y^-1-2)(z^-1+5)/-5^2= 4(x^-6)(y^-3)(z^4)/25= 4z^4/25x^6(y^3)
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09:12:23 ** 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 --> ok
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09:12:58 query R.2.122 (was R.4.72). Express 0.00421 in scientific notation.
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RESPONSE --> 4.21 *10^-3
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09:13:15 ** 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 --> ok
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09:13:35 query R.2.128 (was R.4.78). Express 9.7 * 10^3 in decimal notation.
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RESPONSE --> 9,700
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09:13:41 ** 9.7*10^3 in decimal notation is 9.7 * 1000 = 9700 **
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RESPONSE --> ok
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09:18:07 query R.2.152 \ 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 --> I would use the formula |x - 98.6| >1.5 and substitue 97 and 100 for x. Substituting 97, the answer is |-1.6| or 1.6>1.5. This is unhealthy. Substituting 100. the answer is 1.4 which is not > 1.5, so 100 is healthy
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09:18:15 ** 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 --> ok
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???}???????? assignment #003 003. `query 3 College Algebra 09-03-2008
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09:22:23 query R.3.16 \ 12 (was R.3.6) What is the hypotenuse of a right triangle with legs 14 and 48 and how did you get your result?
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RESPONSE --> I used the pythagorean theorem. a^2 +b^2 = c^2 14^2 + 48^2 = c^2 196+2304=sqrt2500 =50
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09:22:35 ** The Pythagorean Theorem tells us that c^2 = a^2 + b^2, where a and b are the legs and c the hypotenuse. Substituting 14 and 48 for a and b we get c^2 = 14^2 + 48^2, so that c^2 = 196 + 2304 or c^2 = 2500. This tells us that c = + sqrt(2500) or -sqrt(2500). Since the length of a side can't be negative we conclude that c = +sqrt(2500) = 50. **
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RESPONSE --> ok
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09:24:02 query R.3.22 \ 18 (was R.3.12). Is a triangle with legs of 10, 24 and 26 a right triangle, and how did you arrive at your answer?
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RESPONSE --> Using the Pythagorean Theorem 10^2 + 24^2 = 26^2
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09:24:25 ** Using the Pythagorean Theorem we have c^2 = a^2 + b^2, if and only if the triangle is a right triangle. Substituting we get 26^2 = 10^2 + 24^2, or 676 = 100 + 576 so that 676 = 676 This confirms that the Pythagorean Theorem applies and we have a right triangle. **
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RESPONSE --> ok
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09:47:38 query R.3.34 \ 30 (was R.3.24). What are the volume and surface area of a sphere with radius 3 meters, and how did you obtain your result?
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RESPONSE --> I obtained my results using the fomulas V=4/3pir^3 and SA=4pir^2 V=4/3 pi3^3 = 4/3 pi 27 V=36pi ft^3 SA = 4pi 3^2 = 36pi ft^2
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09:48:08 ** 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 --> ok
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09:52:49 query R.3.50 \ 42 (was R.3.36). A pool of diameter 20 ft is enclosed by a deck of width 3 feet. What is the area of the deck and how did you obtain this result?
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RESPONSE --> First I figured the area of the pool, then the entire area. I then subtracted the area of the pool. pool area = pir^2= pi 10^2 = 100pi ft^2. Entire area = pi 13^2=169pi ft^2. 169-100 = 69pi ft^2.
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09:53:05 ** 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 --> ok
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09:53:48 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I had a terrible time ( and finally gave up) on problem 58.
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