course Mth 158 Z?|???z ??H??????assignment #002002. `query 2
......!!!!!!!!...................................
09:35:08 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.
......!!!!!!!!...................................
RESPONSE --> Substitute the x and y values into the expression so [2(-2)-3] / 3 work in the parenthesis first giving 2*-2 = -4 then subtract 3 so the numerator is -7 divide -7 into 3 giving an answer of -2.33
.................................................
......!!!!!!!!...................................
09:37:14 query R.2. 55 (was R.2.45) Evaluate for x = 3 and y = -2: | |4x| - |5y| | and explan how you got your result.
......!!!!!!!!...................................
RESPONSE --> substitute x and y appropriately into expression so 4(3) - 5(-2) giving 12 - (-10) but because of the absolute value bars -10 becomes 10 so 12-10 giving an answer of 2
.................................................
......!!!!!!!!...................................
09:39:16 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)
......!!!!!!!!...................................
RESPONSE --> the values are present and seen in the expression
.................................................
......!!!!!!!!...................................
09:41:08 ** 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. **
......!!!!!!!!...................................
RESPONSE --> right, a denominator can not be 0 and when you factor out the denominator so x = 0 is the only value
.................................................
......!!!!!!!!...................................
09:44:26 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?
......!!!!!!!!...................................
RESPONSE --> rewrite the expression because of the -2 exponent so - 1 / 4^2 and follow through with the law of exponents and solve -1/16
.................................................
......!!!!!!!!...................................
09:52:03 ** (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. **
......!!!!!!!!...................................
RESPONSE --> group factors: 3^-2/3^2 * 5^3/5 so 3^(-2-2) giving 3^-4 5^(3-1) giving 5^2 so 1/3^4 * 5^2 giving 1/81 * 25 hence .31 or 25/81
.................................................
......!!!!!!!!...................................
09:58:44 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.
......!!!!!!!!...................................
RESPONSE --> work within the parenthesis first so (1/5x^2 / 1/6y^2) ^ -3 so now work with the exponent -3 giving [1/ (1/5x^2 / 1/6y^2 ) ] ^ 3
.................................................
......!!!!!!!!...................................
10:00:34 [ 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.
......!!!!!!!!...................................
RESPONSE --> right, simplifying by grouping works and then re-arranging and simplifying the expression.
.................................................
......!!!!!!!!...................................
10:02:52 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.
......!!!!!!!!...................................
RESPONSE --> keeping -8x^3 together, [1/ (-8x^3) ] ^2 getting rid of the negative exponent
.................................................
......!!!!!!!!...................................
10:05:06 ** 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 ). **
......!!!!!!!!...................................
RESPONSE --> right I am not solving the expression so in the denominator -8^2 giving 64 (x^3)^2 hence x^6 so 1/ (64x^6)
.................................................
......!!!!!!!!...................................
10:09:42 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.
......!!!!!!!!...................................
RESPONSE --> the way I am seeing the problem to make the exponent positive first so 1/x^2y so with the 1/(xy^2) equals 1/ (x^2y)(xy^2)
.................................................
......!!!!!!!!...................................
10:11:08 ** (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).**
......!!!!!!!!...................................
RESPONSE --> Yes this is what I wasn't sure of because there were not the * marks on the problem so in this case (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).
.................................................
......!!!!!!!!...................................
10:14:07 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.
......!!!!!!!!...................................
RESPONSE --> make the exponents positive and simplify through so z^5 / 4x^2 * yz * 25 * x^4 * y^2
.................................................
......!!!!!!!!...................................
10:15:20 ** 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 ) **
......!!!!!!!!...................................
RESPONSE --> grouping properly is the key and grouping like factors.
.................................................
......!!!!!!!!...................................
10:16:46 query R.2.122 (was R.4.72). Express 0.00421 in scientific notation.
......!!!!!!!!...................................
RESPONSE --> 4.21 * 10^-3
.................................................
......!!!!!!!!...................................
10:18:04 query R.2.128 (was R.4.78). Express 9.7 * 10^3 in decimal notation.
......!!!!!!!!...................................
RESPONSE --> 9,700
.................................................
......!!!!!!!!...................................
10:20:30 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?
......!!!!!!!!...................................
RESPONSE --> 97-98.6 = -1.6 but has absolute value bars so 1.6 and 1.6 is greater than 1.5 hence unhealthy however 100-98.6 = 1.4 and 1.4 is not greater than 1.5 so I believe it is not unhealthy
.................................................
M?????~R?????? assignment #002 002. `query 2 College Algebra 05-31-2009