course MTH 158 There were several questions regarding arithmetic, geometric, and harmonic mean. The geometric was not assigned, but I attempted them regardless trying to get an understanding of the others. After consulting the solution manual I ALMOST understood the arithmetic, but it wasn't a great understanding and as for the others I'm lost. Do we need to know these for a test or anything in the classes future? AND Is there anyway you can explain them IN ENGLISH(!!!) that may help me understand them better (regardless of whether we will be tested on them) for my own sanity?
......!!!!!!!!...................................
02:31:57 **** query 1.5.34 (was 1.5.24). How did you write the interval [0, 1) using an inequality with x? Describe your illustration using the number line.
......!!!!!!!!...................................
RESPONSE --> 0 0 (greater than/equal to) x (greater than 1) The graph of the solution set on the real number line is a closed interval [ on the point 0 extending towards the positive end of the real number line until it reaches the open interval of the point 1 and all points in between. The graph includes the point 0 and includes all points up to, but excluding the point 1.
.................................................
......!!!!!!!!...................................
02:32:21 10-28-2008 02:32:21 ** My notes here show the half-closed interval [0, 1). When sketching the graph you would use a filled dot at x = 0 and an unfilled dot at x = 1, and you would fill in the line from x = 0 to x = 1. **
......!!!!!!!!...................................
NOTES -------> ok
.......................................................!!!!!!!!...................................
02:33:01 **** query 1.5.40 (was 1.5.30). How did you fill in the blank for 'if x < -4 then x + 4 ____ 0'?
......!!!!!!!!...................................
RESPONSE --> If x < -4 then x + 4 < -4 + 4 or x + 4 < 0
.................................................
......!!!!!!!!...................................
02:33:09 10-28-2008 02:33:09 ** if x<-4 then x cannot be -4 and x+4 < 0. Algebraically, adding 4 to both sides of x < -4 gives us x + 4 < 0. **
......!!!!!!!!...................................
NOTES -------> ok
.......................................................!!!!!!!!...................................
02:33:36 **** query 1.5.46 (was 1.5.36). How did you fill in the blank for 'if x > -2 then -4x ____ 8'?
......!!!!!!!!...................................
RESPONSE --> If x > -2, then -4x < (-4)(-2) or -4x < 8
.................................................
......!!!!!!!!...................................
02:33:44 10-28-2008 02:33:44 **if x> -2 then if we multiply both sides by -4 we get -4x <8. Recall that the inequality sign has to reverse if you multiply or divide by a negative quantity. **
......!!!!!!!!...................................
NOTES -------> ok
.......................................................!!!!!!!!...................................
02:35:06 **** query 1.5.58 (was 1.5.48). Explain how you solved the inquality 2x + 5 >= 1.
......!!!!!!!!...................................
RESPONSE --> 2x + 5 > 1 2x + 5 - 5 > 1 - 5 2x > -4 x > -4/2 x > -2 The solution set is {x | x > -2} or using interval notational all numbers in the interval (-2, infinity).
.................................................
......!!!!!!!!...................................
02:35:16 10-28-2008 02:35:16 ** Starting with 2x+5>= 1 we add -5 to both sides to get 2x>= -4, the divide both sides by 2 to get the solution x >= -2. **
......!!!!!!!!...................................
NOTES -------> ok
.......................................................!!!!!!!!...................................
02:41:49 **** query 1.5.64 (was 1.5.54). Explain how you solved the inquality 8 - 4(2-x) <= 2x.
......!!!!!!!!...................................
RESPONSE --> 8 - 4(2 - x) <= -2x 8 - 8 + 4x <= -2x 4x <= -2x 4x + 2x <= 0 6x <= 0 6x/6 <= 0/6 x <= 0 I'm not sure if it was intentional but you have 8 - 4(2 - x) <= 2x, so that solution would be: 8 - 4(2 - x) <= 2x 8 - 8 + 4x <= 2x 4x <= 2x 4x - 2x <= 0 2x <= 0 x <= 0
.................................................
......!!!!!!!!...................................
02:42:01 10-28-2008 02:42:01 ** 8- 4(2-x)<= 2x. Using the distributive law: 8-8+4x<= 2x. Simplifying: 4x<=2x. Subtracting 2x from both sides: 2x<= 0 x<=0 **
......!!!!!!!!...................................
NOTES -------> ok
.......................................................!!!!!!!!...................................
02:45:51 **** query 1.5.76 (was 1.5.66). Explain how you solved the inquality 0 < 1 - 1/3 x < 1.
......!!!!!!!!...................................
RESPONSE --> 0 < 1 - (1/3)x < 1 0 - 1 < 1 - 1 - (1/3)x < 1 - 1 -1 < -(1/3)x < 0 -1 * -3 > -3 * -(1/3)x > 0 * -3 3 > x > 0 0 < x < 3 The solution set is {x | 0 < x < 3} or using interval notation all numbers in the interval (0, 3).
.................................................
......!!!!!!!!...................................
02:46:01 10-28-2008 02:46:01 ** Starting with 0<1- 1/3x<1 we can separate this into two inequalities, both of which must hold: 0< 1- 1/3x and 1- 1/3x < 1. Subtracting 1 from both sides we get -1< -1/3x and -1/3x < 0. We solve these inequalitites separately: -1 < -1/3 x can be multiplied by -3 to get 3 > x (multiplication by the negative reverses the direction of the inequality) -1/3 x < 0 can be multiplied by -3 to get x > 0. So our inequality can be written 3 > x > 0. This is not incorrect but we usually write such inequalities from left to right, as they would be seen on a number line. The same inequality is expressed as 0 < x < 3. **
......!!!!!!!!...................................
NOTES -------> ok
.......................................................!!!!!!!!...................................
02:50:06 **** query 1.5.94 (was 1.5.84). Explain how you found a and b for the conditions 'if -3 < x < 3 then a < 1 - 2x < b.
......!!!!!!!!...................................
RESPONSE --> If -3 < x < 3, then a < 1 - 2x < b. -3 < x < 3 -3(-2) > -2x > 3(-2) 6 > -2x > -6 6 + 1 > 1 - 2x > -6 + 1 7 > 1 - 2x > -5 -5 < 1 - 2x < 7 a = -5 and b = 7
.................................................
......!!!!!!!!...................................
02:51:12 10-28-2008 02:51:12 ** Adding 1 to each expression gives us 1 + 6 > 1 - 2x > 1 - 6, which we simplify to get 7 > 1 - 2x > -5. Writing in the more traditional 'left-toright' order: -5 < 1 - 2x < 7. **
......!!!!!!!!...................................
NOTES -------> I know it's the order of operations, but with inequalities is it necessary to add first or is it okay to multiply first?
.......................................................!!!!!!!!...................................
02:58:36 **** query 1.5.106 (was 1.5.96). Explain how you set up and solved an inequality for the problem. Include your inequality and the reasoning you used to develop the inequality. Problem (note that this statement is for instructor reference; the full statement was in your text) commision $25 + 40% of excess over owner cost; range is $70 to $300 over owner cost. What is range of commission on a sale?
......!!!!!!!!...................................
RESPONSE --> Let a = excess of owner's cost when excess is $200 Let b = excess of owner's cost when excess is $3000 Let x = commission a <= x/0.4 - 25 <= b 0.4a <= x - 25 <= 0.4b 0.4(200) <= x - 25 <= 0.4(3000) 80 <= x - 25 <= 1200 80 + 25 <= x <= 1200 + 25 105 <= x <=1225 The salesperson can expect his commission to vary between $105 and $1225 for every car sold.
.................................................
......!!!!!!!!...................................
03:06:59 10-28-2008 03:06:59 ** If x = owner cost then 70 < x < 300. .40 * owner cost is then in the range .40 * 70 < .40 x < .40 * 300 and $25 + 40% of owner cost is in the range 25 + .40 * 70 < 25 + .40 x < 25 + .40 * 300 or 25 + 28 < 25 + .40 x < 25 + 120 or 53 < 25 + .40 x < 145. **
......!!!!!!!!...................................
NOTES -------> I'm not sure where the 70 or the 300 came from? The text has the excess range as $200 to $3000 over owner cost. I did however rework the problem with your range and achieved the same results as you, so I assume that the equation I used was correct.
.......................................................!!!!!!!!...................................
03:14:02 **** query 1.5.123 \ 112. Why does the inequality x^2 + 1 < -5 have no solution?
......!!!!!!!!...................................
RESPONSE --> I'm glad this question was asked because my answer differed from the one in the solution manual. x^2 + 1 < -5 x^2 < -5 - 1 x^2 < -6 My answer: When you subtract 1 from both sides of the inequality you will have x^2 < -6. This inequality has no real solution because the principle square root of a real number is nonnegative. Also, x^2 is always greater than or equal to 0, because -6 can never be greater than or equal to 0, there is no real solution. Manual answer: Since x^2 >= 0, we have x^2 + 1 >= 0 + 1 x^2 + 1 >= 1 Therefore, the expression x^2 + 1 can never be less than -5. What I am unsure of in this answer is why they added 1 rather than subtracting? I'm aware that subtracting 1 would result in an unreal solution, but that still doesn't explain why they added (at least not to me).
.................................................
......!!!!!!!!...................................
03:14:22 10-28-2008 03:14:22 STUDENT SOLUTION: x^2 +1 < -5 x^2 < -4 x < sqrt -4 can't take the sqrt of a negative number INSTRUCTOR COMMENT: Good. Alternative: As soon as you got to the step x^2 < -4 you could have stated that there is no such x, since a square can't be negative. **
......!!!!!!!!...................................
NOTES -------> ok
.................................................