course Mth 173 Hi :)
......!!!!!!!!...................................
13:09:32 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 --> (2(-2) - 3)/3 First we do inside the parentheses after we substitute the values of x and y into the expression (-4 - 3)/3 Then we do the outer parentheses to have -7/3
.................................................
......!!!!!!!!...................................
13:09:52 ** Starting with (2x-3)/y we substitute x=-2 and y=3 to get (2*(-2) - 3)/3 = (-4-3)/3= -7/3. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
13:15:48 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 --> First we substitute x for 3 and y for -2 | |4(3)| - |5(-2)| | |12| - |-10| | The negative ten will be 10 (absolute value) | 2 | = 2
.................................................
......!!!!!!!!...................................
13:16:10 ** Starting with | | 4x |- | 5y | | we substitute x=3 and y=-2 to get | | 4*3 | - | 5*-2 | | = | | 12 | - | -10 | | = | 12-10 | = | 2 | = 2. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
13:18:11 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 --> x cannot be equal to 0
.................................................
......!!!!!!!!...................................
13:19:57 ** 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 -->
.................................................
......!!!!!!!!...................................
13:29:44 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 --> -4^-2 First we use the reciprocal to make the exponent positive to have (-1/4)^2 = -1/16 or -0.0625
.................................................
......!!!!!!!!...................................
13:33:15 ** 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).**
......!!!!!!!!...................................
RESPONSE --> I see where I put my parentheses around the whole numbers and I should have only put them around the denominator only.
.................................................
......!!!!!!!!...................................
13:43:58 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?
......!!!!!!!!...................................
RESPONSE --> (125/9)/45 I made 3^-2 =1/3^2 (1/3^2 * 5^3) / (3^2 * 5) (1/9 * 5^3) / (9 * 5) (1/9 * 125) / (45) (125/9) / (45) = .3086419753
.................................................
......!!!!!!!!...................................
13:49:53 ** (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 --> I remember doing this problem this way in the book now that I see it written out again.
.................................................
......!!!!!!!!...................................
14:16:28 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 --> ((5x^-2)^-3)/(6y6-2)^-3 (5^-3(x^6))/(6^-3(y^6) (216x^6)/(125y^6)
.................................................
......!!!!!!!!...................................
14:17:25 [ 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 -->
.................................................
......!!!!!!!!...................................
14:24:48 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 --> (-8x^3)^-2 Distribute the exponent throughout the problem -8^-2 *x^-6 Reciprocal of negative exponents -1/8^2 * 1/x^6 -1/8^2 *x^6 -1/64*x^6
.................................................
......!!!!!!!!...................................
14:27:25 ** 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 --> I see where I made the mistake of not keeping the negative sign with the 8. Because it was in parentheses it did belong just to the 8.
.................................................
......!!!!!!!!...................................
14:33:34 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 --> = (1/(x)^2)(y)/(xy^2) =x^-2-1*y1-2 =x^-3*y^-1 =1/x^3y
.................................................
......!!!!!!!!...................................
14:33:51 ** (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 -->
.................................................
......!!!!!!!!...................................
14:41:00 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 --> = 4/25* x^-2-4*y^-1-2*z^-1-5 =4/25*x^-6*y^-3*z^-6 =4/25(1/x^6y^3z^6) =4/(25x^6y^3z^6)
.................................................
......!!!!!!!!...................................
14:41:56 ** 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 -->
.................................................
......!!!!!!!!...................................
14:42:21 query R.2.122 (was R.4.72). Express 0.00421 in scientific notation.
......!!!!!!!!...................................
RESPONSE --> 4.21 X 10^-3
.................................................
......!!!!!!!!...................................
14:42:33 ** 0.00421 in scientific notation is 4.21*10^-3. This is expressed on many calculators as 4.21 E-4. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
14:42:57 query R.2.128 (was R.4.78). Express 9.7 * 10^3 in decimal notation.
......!!!!!!!!...................................
RESPONSE --> 9700
.................................................
......!!!!!!!!...................................
14:43:02 ** 9.7*10^3 in decimal notation is 9.7 * 1000 = 9700 **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
14:46:29 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| = 1.6> 1.5 |100-98.6|=|1.4|=1.4<1.5
.................................................
......!!!!!!!!...................................
14:46:50 ** 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. **
......!!!!!!!!...................................
RESPONSE -->
.................................................