#$&* course mth151 12/17/14 3:40 If your solution to stated problem does not match the given solution, you should self-critique per instructions
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Given Solution: To solve a linear equation we can use the basic properties of equivalents and equality. Specifically we can simplify either or both sides of an equation, we can add or subtract the same quantity from both sides of the equation, or we can multiply or divide both sides of the equation by the same quantity. To solve the present equation we need to obtain any equivalent equation with just x on one side or the other. We proceed as follows: Starting with the original equation 2 x + 7 = 21, we first subtract 7 from both sides to obtain 2x + 7 - 7 = 21 - 7. We simplify both sides to obtain 2x = 14, then we divide both sides of the equation by 2 to get 2x / 2 = 14 / 2. Simplifying both sides we have x = 7. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q002. Solve the equation 2 (x + 7) = 30. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2(x+7)=30 x+7=30/2 x+7=15 x=15-7 x=8 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q003. Using x for the variable, how do we write the expression 'double a number plus five'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2x+5 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If x stands for the number, then double the number is 2x, and double the number plus 5 is therefore 2x + 5. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q004. Using x for the variable, how do we express the statement 'double a number plus five is 19'? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2x+5=19 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q005. Using x for the length of the side of a square, how do we express the perimeter of the square? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: perimeter is all the sides combined together which would mean we have to multiply x by 4 which can be written as... 4x confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The perimeter of a figure is the distance around it. A square consists of 4 equal sides. If each side has length x, then the distance around it is 4 * x, or just 4 x. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q006. Using an appropriate variable for the length of the side of a square, how do we express the area of a square in terms of the length of its side? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: area of a square if length times width. for a square, the length and width are the side x*x which can also be written as x^2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q007. Using an appropriate variable how do we express the perimeter of a rectangle whose length is double its width? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: perimeter is all sides combined. width will be x. since length is double width then length=2x 2x+2x+x+x 4x+2x 6x confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The perimeter is the sum of the lengths of the sides of the rectangle. Since we are given the relationship between the lengths of the sides, we first express this relationship in symbols. If we let x stand for the width of the rectangle, then the length of the rectangle is 2x. Since both the length and the width occur twice as we move around the perimeter of the rectangle, it follows that the perimeter is 2 * length + 2 * width = 2 * 2x + 2 * x. This expression simplifies to 4 x + 2 x, which further simplifies to 6x. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q008. Using an appropriate variable, how do we express the area of a rectangle whose length is double its width? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: area is length times width width is x length is 2x 2x*x=2x^2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2x^2=72 x^2=36 x=sqrt36 x=6 width= 6 length is 2x=12 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: $15000-x is how much would be invested at eight percent say that you invested $3000 at 5% that $3000 would be x but when you calculate how much annual interest you have you need to say $3000 at 5% which is or .05 we can use .08 to represent the amount invested at eight percent we know that the amount that is invested at eight percent is ($15000-x) so we can write the equation as such.. annual interest = .05x +.08($15,000-x) confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: $1,000 = .05x + .08($15,000-x) $1,000= .05x + $1200 -.08x $1,000= -.03x +$1200 -$200=-.03x -200/-.03=x $6,666.67=x confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 $6667. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q012. Solve the equation x/3 + 9 = 27. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x/3+9=27 x/3=27-9 x/3=18 x=18*3 here to get rid of the x/3 we multiply 3 on both sides x=54 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q013. Solve the equation 2 x^2 + 9 x - 68 = 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2x^2+9x-68=0 quadratic equation for x^2+bx+c=0 is x=[-b +- sqrt(b^2-4ac)]/2a x=[-9 +- sqrt(9^2-4(2*-68)]/(2*2) x=[-9 +- sqrt(81-(-544))]/4 x=[-9 +- sqrt(625)]/4 x=(-9 +- 25)/4 x=(-9 + 25)/4 x=16/4 x=4 or x=(-9-25)/4 x=-34/4 x=-8.5 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: This is a quadratic equation, solve using the quadratic formula. Recall that the quadratic formula tells us that the equation a x^2 + b x + c = 0 has solutions x = [ -b + - `sqrt( b^2 - 4 a c) ] / (2 a), where the + - indicates '+ or -', meaning that both + and - symbols in that position will give us correct solutions. In the present case we see that a = 2, b = 9, and c = -68, so that the solutions are x = [ -9 + - `sqrt( 9^2 - 4 * 2 * (-68) ) ] / ( 2 * 2 ) = [ -9 + - `sqrt( 625 ) ] / 4 = [ -9 + - 25 ] / 4. Choosing the + we have x = [-9 + 25 ] / 4 = 16 4 = 4. Choosing the - we have [ -9 - 25 ] / 4 = -34 / 4 = -8.5. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q014. Solve the equation 3/x + x/2 = 31/10. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3/x + x/2=31/10 common denominator is 10x so we divide by 10x on both sides 10x(3/x+x/2)=10x(31/10) 30+5x^2=31x now we need to get it all on one side so we can work the quadratic eqation 5x^2-31x+30=0 x=[-b +- sqrt(b^2-4ac)]/2a x=[-(-31) +- sqrt(-31^2-(4*5*30)]/2*5 x=[31 +- sqrt(961-600)]/10 x=[31 +- srt(361)]/10 x=[31 +- 19]/10 x=(31+19)/10 x=50/10 x=5 or x=(31-19)/10 x=12/10 x=1.2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q015. Express in lowest terms the ratio of 2 feet to 72 inches. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: we need to convert to the same unit of measurement first 72in is 6ft so 2ft/6ft=1/3 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: In order to express the ratio both quantities must be given in the same units. Since 2 ft. is the same as 24 inches, the ratio is 24 inches to 72 inches, or 24 in / 72 in = 1/3. We could alternatively have expressed the 72 inches as 6 ft. to get the ratio of 2 ft./6 ft. = 1/3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q016. Express in lowest terms the ratio of five gallons to 72 pints. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 8 pints in a gallon 72 pints=9 gallons 5gallons/8gallons=5/8 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q017. Express in lowest terms the ratio of 450 cm to three meters. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 100m in a meter 3 meters is equal to 300cm 450cm/300cm=1.5 Confindence Assessment:3
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: first need to find the area of each square area is length *width 3m*3m=9m^2 2m*2m=4m^2 ratio is 9m^2/4m^2= 9/4 or 2.25 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A square three meters on a side has area (3 meters)^2 = 9 m^2, while a square 2 meters on a side has area (2 meters)^2 = 4 m^2, so the ratio of areas is 9 m^2 / (4 m^2) = 9/4. The &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q019. Express in lowest terms the ratio of the area of a square 4 cm on a side to the area of a square two meters on a side. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 4cm*4cm=16cm^2 100 cm in a meter so 200cm in 2m 200cm*200cm=40,000cm^2 16cm^2/40,000cm^2=16/40,000=1/2500 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q020. A rectangle has length triple its width. Its area is 75. Set up and solve an equation to find the dimensions of the rectangle. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: width=x length=3x 3x*x=75 3x^2=75 x^2=75/3 x^2=25 x=sqrt25 x=5 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q021. If $100 000 is invested, with x dollars invested at 5% and the rest at 7%, what is the expression for the resulting interest? What value of x would result in interest $6500? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x is the amount that is invested at 5% .05x + .07($100,000-x) .05x + .07($100,000-x)=$6500 .05x + $7000 - .07x=$6500 -.02x +$7000=$6500 -.02x=-$500 x=-$500/(-.02) x=$25,000 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q022. Solve the equation x^2 - 5 x + 4 = 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: use quadratic equation x=[-b +- sqrt(b^2-(4ac)/2a x=[-(-5) +- sqrt(-5^2-(4*1*4)/2(1) x=[5 +- sqrt(25-16)]/2 x=[5 +- sqrt(9)]/2 x=(5 +- 3)/2 x=(5+3)/2 x=8/2 x=4 or x=(5-3)/2 x=2/2 x=1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q023. One square has edges 5 times as long as those of another. What is the ratio of the areas of the two squares? One cube has edges 5 times as long as those of another. What is the ratio of the volumes of the two cubes? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x*x=x^2 is the area of the second square 5x*5x=25x^2 is the ratio of the first square the ratio is 25x^2/x^2=25/1 or 25 volume is x^3 x*x*x=x^3 is the volume of one cube 5x*5x*5x=125x^3 is the volume of the other cube the ratio is 125x^3/x^3=125/1 or 125 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating:ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!