course MTH 151 Note that there was no asst or query 30 and that asst 29 is being submitted here. Asst 28 was previously submitted.
.................................................
......!!!!!!!!...................................
08:31:13 `q001. Note that there are 19 questions in this section. Solve the equation 2 x + 7 = 21.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:31:13 `q001. Note that there are 19 questions in this section. Solve the equation 2 x + 7 = 21.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
08:31:18 To solve a linear equation we can use the basic properties of equivalents and equality. Specifically we can simplify either or both sides of an equation, we can add or subtract the same quantity from both sides of the equation, or we can multiply or divide both sides of the equation by the same quantity. To solve the present equation we need to obtain any equivalent equation with just x on one side or the other. We proceed as follows: Starting with the original equation 2 x + 7 = 21, we first subtract 7 from both sides to obtain 2x + 7 - 7 = 21 - 7. We simplify both sides to obtain 2x = 14, then we divide both sides of the equation by 2 to get 2x / 2 = 14 / 2. Simplifying both sides we have x = 7.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:31:21 `q002. Solve the equation 2 (x + 7) = 30.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:31:25 Starting with 2 (x+7) = 30 we first simplify the left-hand side using the distributive law of multiplication over addition. We obtain 2x + 14 = 30. We subtract 14 from both sides to obtain 2x = 16, then we divide both sides by 2 to obtain{}{}2x / 2 = 16 / 2 or after simplifying x = 8.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:31:34 `q003. Using x for the variable, how do we write the expression 'double a number plus five'?
......!!!!!!!!...................................
RESPONSE --> 0k
.................................................
......!!!!!!!!...................................
08:31:37 If x stands for the number, then double the number is 2x, and double the number plus 5 is therefore 2x + 5.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:31:42 `q004. Using x for the variable, how do we express the statement 'double a number plus five is 19'?
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:31:45 As seen in the preceding question, 'double a number plus five' can be expressed as 2 x + 5, so the statement 'double a number plus five is 19' can be written as the equation 2x + 5 = 19.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:31:49 `q005. Using x for the length of the side of a square, how do we express the perimeter of the square?
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:31:51 The perimeter of a figure is the distance around it. A square consists of 4 equal sides. If each side has length x, then the distance around it is 4 * x, or just 4 x.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:31:55 `q006. Using an appropriate variable for the length of the side of a square, how do we express the area of a square in terms of the length of its side?
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:31:57 The area of a square is the product of the lengths of its sides. If x stands for the length of a side, then the area of the square is x * x or x^2.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:32:00 `q007. Using an appropriate variable how do we express the perimeter of a rectangle whose length is double its width?
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:32:03 The perimeter is the sum of the lengths of the sides of the rectangle. Since we are given the relationship between the lengths of the sides, we first express this relationship in symbols. If we let x stand for the width of the rectangle, then the length of the rectangle is 2x. Since both the length and the width occur twice as we move around the perimeter of the rectangle, it follows that the perimeter is 2 * length + 2 * width = 2 * 2x + 2 * x. This expression simplifies to 4 x + 2 x, which further simplifies to 6x.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:32:07 `q008. Using an appropriate variable, how do we express the area of a rectangle whose length is double its width?
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:32:10 The area of a rectangle is the product of its length then width. Since we are given the relationship between length in width, we begin by expressing the length and width of the rectangle in terms of a symbol. If we let x stand for the width of the rectangle, then 2 x stands for its length. Now since the area is the product of length and width, we see that the area must be x * 2 x, or 2 x^2.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:32:13 `q009. Express in terms of an appropriate variable the statement that the area of a rectangle whose length is double its width is 72. Solve the resulting equation to determine the dimensions of the rectangle.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:32:16 The area of a rectangle whose length is double its width was seen in the preceding problem to be 2 x^2, where x is the width. If the area is 72, we see that 2 x^2 = 72. We solve this equation as follows. Starting with 2 x^2 = 72 we divide both sides by 2 to get x^2 = 36. We then take the square root of both sides of the equation to obtain x = `sqrt(36), or x = 6. Note that the equation x^2 = 36 actually has two solutions, x = 6 and x = -6. However the length of a side cannot be negative so we chose the positive solution.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:32:19 `q010. Suppose that $15,000 is invested, part at five percent annual interest and the rest at 8 percent annual interest. If x represents the amount invested at five percent, then what expression represents the amount invested at 8 percent? What expression represents the amount of the annual interest?
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:32:23 If x represents the amount invested at five percent, then the remaining amount is $15,000 - x, which is invested at eight percent. The annual interest on amount x invested at five percent is .05 x, while the annual interest on amount ($15,000 -x) invested at eight percent is .08 ( $15,000 - x). Thus the total interest on the $15,000 is total interest = .05 x + .08 ( $15,000 - x).
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:32:26 `q011. Continuing the preceding problem, if it is known that the annual interest totals $1,000, then what equation would we solve for x in order to determine the amount invested at five percent? Solve your equation. What is your result?
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:32:29 If the total interest is $1000, then knowing that the expression for the total interest is .05 x + .08 ( $15,000 - x) we obtain the equation .05 x + .08 ( $15,000 - x) = $1000. To solve the equation .05 x + .08 ( $15,000 - x) = $1000 we first apply the distributive law on the left-hand side to obtain .05 x + $1200 - .08 x = $1000. Simplifying the left-hand side we obtain -.03 x + $1200 = $1000. Subtracting $1200 from both sides we get -.03 x = -$200. Dividing both sides by -.03 we obtain{}{}x = -$200 / -.03 = $6667, rounded to the nearest dollar. This means that of the $15,000, the amount invested at five percent is $6667.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:32:32 `q012. Solve the equation x/3 + 9 = 27.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:32:35 A good rule of thumb is that whenever an equation involves denominators, we multiply both sides by a quantity that eliminates the denominators. In this case if we multiply both sides by 3 we will eliminate the denominator of the term x/3. Multiplying both sides of x/3 + 9 = 27 by 3 we obtain 3(x/3+9) = 3 * 27. Applying the distributive law to the left-hand side and multiplying the numbers on the right-hand side this becomes 3 * (x/3) + 3 * 9 = 81, which simplifies to give us x + 27 = 81. We now subtract 27 from both sides to obtain the solution x = 54.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:32:38 `q013. Solve the equation 2 x^2 + 9 x - 68 = 0.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:32:42 This is a quadratic equation, solve using the quadratic formula. Recall that the quadratic formula tells us that the equation a x^2 + b x + c = 0 has solutions x = [ -b + - `sqrt( b^2 - 4 a c) ] / (2 a), where the + - indicates '+ or -', meaning that both + and - symbols in that position will give us correct solutions. In the present case we see that a = 2, b = 9, and c = -68, so that the solutions are x = [ -9 + - `sqrt( 9^2 - 4 * 2 * (-68) ) ] / ( 2 * 2 ) = [ -9 + - `sqrt( 625 ) ] / 4 = [ -9 + - 25 ] / 4. Choosing the + we have x = [-9 + 25 ] / 4 = 16 4 = 4. Choosing the - we have [ -9 - 25 ] / 4 = -34 / 4 = -8.5.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:32:45 `q014. Solve the equation 3/x + x/2 = 31/10.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:32:48 A common denominator for this equation would be 10 x. Thus multiplying both sides by 10 x would eliminate all denominators. We obtain 10 x ( 3/x + x/2) = 10 x (31/10). Applying the distributive law to the left-hand side we obtain 10 x ( 3/x) + 10 x (x/2) = 10 x ( 31/10). Simplifying we have 30 + 5 x^2 = 31 x. This might appear difficult to solve, but if we subtract the 31 x from both sides and rearrange the order of the left-hand side we have 5 x^2 - 31 x + 30 = 0, which we now recognize as a quadratic equation with a = 5, b = -31 and c = 30. Thus are solutions are x = [ -b + - `sqrt( b^2 - 4 a c) ] / (2 a) = [ -(-31) + - `sqrt( (-31)^2 - 4 * 5 * 30 ] / (2 * 5) = [ 31 + - `sqrt(361) ] / 10 = [ 31 + - 19 ] / 10. Choosing the + solution we have x = [ 31 + 19 ] / 10 = 50 / 10 = 5. Choosing the - solution we have x = [ 31 - 19] / 10 = 1.2.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:32:50 `q015. Express in lowest terms the ratio of 2 feet to 72 inches.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:32:54 In order to express the ratio both quantities must be given in the same units. Since 2 ft. is the same as 24 inches, the ratio is 24 inches to 72 inches, or 24 in / 72 in = 1/3. We could alternatively have expressed the 72 inches as 6 ft. to get the ratio of 2 ft./6 ft. = 1/3.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:32:57 `q016. Express in lowest terms the ratio of five gallons to 72 pints.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:33:00 There are 4 quarts in a gallon and 2 pints in a quart, so there are eight pints in a gallon. In five gallons we therefore have 40 pints. The desired ratio is therefore 40 pints / 72 pints = 5/8.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:33:03 `q017. Express in lowest terms the ratio of 450 cm to three meters.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:33:06 There are 100 cm in a meter. So 450 cm is 4.5 meters and the ratio is 4.5 meters/(3 meters) = 1.5.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:33:10 `q018. Express in lowest terms the ratio of the area of a square three meters on a side to the area of a square two meters on a side.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:33:12 A square three meters on a side has area (3 meters)^2 = 9 m^2, while a square 2 meters on a side has area (2 meters)^2 = 4 m^2, so the ratio of areas is 9 m^2 / (4 m^2) = 9/4. The
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:33:20 `q019. Express in lowest terms the ratio of the area of a square 4 cm on a side to the area of a square two meters on a side.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:33:23 A square 4 cm on a side has area (4 cm) * (4 cm) = 16 cm^2. A square 2 meters on a side has area (200 cm) * (200 cm) = 40,000 cm^2. The ratio of the areas is therefore 16 cm^2/(40,000 cm^2) = 1 / 2500.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:33:49 029. Variation
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
08:43:31 `q001. Note that there are five questions in this set. If y is proportional to x, and if y = 9 when x = 12, then what is the value of y when x = 32?
......!!!!!!!!...................................
RESPONSE --> Ok, I think this would be, ok I'm having a little trouble getting started on this one.
.................................................
Ú։zx Student Name: assignment #029
.................................................
......!!!!!!!!...................................
08:59:19 To say that y is proportional to x is to say that there exists some constant number k such that y = k x. Using the given values of y and x we can determine the value of k: Since y = 9 when x = 12, y = k x becomes 9 = k * 12. Dividing both sides by 12 we obtain 9 / 12 = k. Reducing and reversing sides we therefore obtain k =.75. Now our proportionality reads y = .75 x. Thus when x = 32 we have y = .75 * 32 = 24.
......!!!!!!!!...................................
RESPONSE --> Ok, I'm still a litte foggy on this. I think I'm making it harder than it needs to be.
.................................................
......!!!!!!!!...................................
09:01:51 `q002. If y is proportional to the square of x, and y = 8 when x = 12, then what is the value of y when x = 9?
......!!!!!!!!...................................
RESPONSE --> I need to see how to get started.
.................................................
......!!!!!!!!...................................
09:09:06 To say that y is proportional to x is to say that there exists some constant number k such that y = k x^2. Using the given values of y and x we can determine the value of k: Since y = 8 when x = 12, y = k x^2 becomes 8 = k * 12^2, or 8 = 144 k. Dividing both sides by 144 we obtain k = 8 / 144 = 1 / 18. Now our proportionality reads y = 1/18 x^2. Thus when x = 9 we have y = 1/18 * 9^2 = 81 / 18 = 4.5.
......!!!!!!!!...................................
RESPONSE --> I think my problem is that I'm really not sure on how to set up the problem. I think once I figure that out, I can go from there.
.................................................
......!!!!!!!!...................................
09:15:35 `q003. If y is inversely proportional to x and if y = 120 when x = 200, when what is the value of y when x = 500?
......!!!!!!!!...................................
RESPONSE --> Ok, I'm just going to guess: then y = 3000
.................................................
......!!!!!!!!...................................
09:22:54 To say that y is inversely proportional to x is to say that there exists some constant number k such that y = k / x. Using the given values of y and x we can determine the value of k: Since y = 120 when x = 200, y = k / x becomes 120 = k / 200. Multiplying both sides by 200 we obtain k = 120 * 200 = 24,000. Now our proportionality reads y = 24,000 / x. Thus when x = 500 we have y = 24,000 / 500 = 480.
......!!!!!!!!...................................
RESPONSE --> Ok, I think I'm a little closer to getting this but isn't 24,000 / 500 = 48 instead of 480 or where am I going wrong and didn't get 480.
.................................................
......!!!!!!!!...................................
09:26:08 `q004. If y is inversely proportional to the square of x and if y = 8 when x = 12, then what is the value of y when x = 16?
......!!!!!!!!...................................
RESPONSE --> I need to see how this is being worked out to get started.
.................................................
......!!!!!!!!...................................
09:31:10 To say that y is inversely proportional to the square of x is to say that there exists some constant number k such that y = k / x^2. Using the given values of y and x we can determine the value of k: Since y = 8 when x = 12, y = k / x^2 becomes 8 = k / 12^2, or 8 = k / 144. Multiplying both sides by 144 we obtain k = 8 * 144 = 1152. Now our proportionality reads y = 1152 / x^2. Thus when x = 16 we have y = 1152 / (16)^2 = 4.5.
......!!!!!!!!...................................
RESPONSE --> Ok, I think I'm starting to get it. I worked this out on my paper and for the first time, I was able to come up witht he correct answser of 4.5.
.................................................
......!!!!!!!!...................................
09:35:17 `q005. If y is proportional to the square of x and inversely proportional to z, then if y = 40 when x = 10 and z = 4, what is the value of y when x = 20 and z = 12?
......!!!!!!!!...................................
RESPONSE --> Ok, the answer I got was y = 333.33. I'll check my work with the answer.
.................................................
......!!!!!!!!...................................
09:43:31 To say that y is proportional to the square of x and inversely proportional to z is to say that the there exists a constant k such that y = k x^2 / z. Substituting the given values of x, y and z we can evaluate k: y = k x^2 / z becomes 40 = k * 10^2 / 4. Multiplying both sides by 4 / 10^2 we obtain 40 * 4 / 10^2 = k, or k = 1.6. Our proportionality is now y = 1.6 x^2 / z, so that when x = 20 and z = 12 we have y = 1.6 * 20^2 / 12 = 1.6 * 400 / 12 = 53 1/3.
......!!!!!!!!...................................
RESPONSE --> I think I was thrown off by the x, y, and z. I think that ""inversely proportional to z"" means ""over z"" or divided by z. I will definately need to go back and look over chapter 7 agian.
.................................................
"