#$&* course mth 151 7:12pm12/18/13
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Given Solution: `a** COMMON ERROR: 556. INSTRUCTOR COMMENT: The digit 6 does not exist in base six for the same reason as there is no single digit for 10 in base 10 (digits stop at 9; in general there is no digit corresponding to the base). CORRECT SOLUTION: 555{base 6} = 5 * 6^2 + 5 * 6^1 + 5 * 6^0. If you add 1 you get 5 * 6^2 + 5 * 6^1 + 5 * 6^0 + 1, which is equal to 5 * 6^2 + 5 * 6^1 + 6 * 6^0. But 6 * 6^0 is 6^1, so now you have 5 * 6^2 + 6 * 6^1 + 0 * 6^0. But 6 * 6^1 is 6^2 so you have 6 * 6^2 + 0 * 6^1 + 0 * 6^0. But 6 * 6^2 is 6^3 so the number is 6 * 6^3 + 0 * 6^2 + 0 * 6^1 + 0 * 6^0. So the number following 555{base 6} is 1000{base 6}. The idea is that we can use a number greater than 5 in base 6, so after 555 we go to 1000, just like in base 10 we go from 999 to 1000. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery 4.3.20 34432 base five YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3 * 5^4 + 4 * 5^3 + 4 * 5^2 + 3 * 5^1 + 2 * 5^0= 3 * 625 + 4 * 125 + 4 * 25 + 3 * 5 + 2 * 1 = 1875 + 500 + 100 + 15 + 2 = 2492 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a**34432 base five means 3 * 5^4 + 4 * 5^3 + 4 * 5^2 + 3 * 5^1 + 2 * 5^0. 5^2 = 625, 5^3 = 125, 5^2 = 25, 5^1 = 5 and 5^0 = 1 so 3 * 5^4 + 4 * 5^3 + 4 * 5^2 + 3 * 5^1 + 2 * 5^0= 3 * 625 + 4 * 125 + 4 * 25 + 3 * 5 + 2 * 1 = 1875 + 500 + 100 + 15 + 2 = 2492. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qExplain how you use the calculator shortcut to get the given number. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I do 3 + 5, then 4 + the answer, then 5 * the answer, then 4 + the answer, then 5 * the answer, then 3 + the answer, then 5 * the answer, then 2 + the answer confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 2 + 5 (3 + 5(4 + 5(4 + 3 * 5))) multiplies out to the given number because, for example, the 3 * 5 at the end will be multiplied by the three 5’s to its left to give 3 * 5 * 5 * 5 * 5 = 3 * 5^4, which matches the 3 * 5^4 in the sum 3 * 5^4 + 4 * 5^3 + 4 * 5^2 + 3 * 5^1 + 2 * 5^0. So we do 3 + 5, then 4 + the answer, then 5 * the answer, then 4 + the answer, then 5 * the answer, then 3 + the answer, then 5 * the answer, then 2 + the answer. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery 4.3.40 11028 decimal to base 4 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2230110 which is base 4 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 4^0 = 1 4^1 = 4 4^2 = 16 4^3 = 64 4^4 = 256 4^5 = 1024 4^6 = 4096 (4*7 = 16386, which is larger than the given 11028) So to ‘build up’ 11028 we need 2 * 4^6 = 8192, leaving 2836. 2 * 4^5 = 2048, leaving 788. 3 * 4^4 = 768, leaving 20. 0 * 4^3, because we need only 20, which is less than 64. 1 * 4^2 = 16, leaving 4. 1 * 4^1 = 4, leaving 0.0 * 4^1. Thus our number is 2230110 base 4. STUDENT QUESTION I don’t understand how to get this answer, I looked at the answer and either way it’s not the same number. I always get above or below the 11028. Your answer says 2230110 but I thought you were trying to get 11028. INSTRUCTOR RESPONSE 2230110 in base four is equivalent to the decimal number 2 * 4^6 + 2 * 4^5 + 3 * 4^4 + 0 * 4^3 + 1 * 4^2 + 1 * 4^1 + 0 * 4^0, which if multiplied out and added is 11028 in decimal form. The given solution shows how to reason this out, starting with the decimal number 11028. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qquery 4.3.51 DC in base 16 to binary YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 11011100 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** C stands for decimal 12, which in binary is 1100. D stands for decimal 13, which in binary is 1101. Since our base 16 is the fourth power of 2, our number can be written 11011100, where the 1101 in the first four places stands for D and the 1100 in the last four places stands for C. Note that this method works only when one base is a power of the other.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qIs a base-9 number always even, always odd, or sometimes even and sometimes odd if the number ends in an even number? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 770 in base 9 and even confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** You can investigate this question by trying a variety of examples. For example, 82nine = 8 * 9^1 + 2 * 9^0 = 72 + 2 = 74 in decimal. This is even because you are adding two even multiples of the odd numbers 9^1 = 9 and 9^0 = 1. You have to use an even multiple of 9^0 = 1 because the number ends with an even number. But you don't have to use an even multiple of 9^1 = 9. So we try something like 72nine = 7 * 9^1 + 2 * 9^0 = 63 + 2 = 65 in decimal and we see that the result is odd. The key is that in base nine, the powers of nine are always odd numbers. So when converting you get from a base-9 number with two even digits a number which is even * odd + even = even, while from a base-9 number with an odd digit followed by an even digit you get odd * odd + even = odd; many other possible combinations show the same thing but this is one of the simpler examples that shows why we get odd for some base-9 numbers, and even for others. For a couple of specifics, 70 in base 9 is 63 in base ten, and is odd. However 770 in base 9 is even. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course mth 151 7:17pm12/18/13
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Given Solution: 3 * 9 (mod 4) is the remainder when 3 * 9 is divided by 4. Since 3 * 9 = 27 and 27 / 4 leaves remainder 3, we see that 3 * 9 (mod 4) = 3. 7 * 12 (mod 4) is the remainder when 7 * 12 is divided by 4. Since 7 * 12 = 84 and 84 / 4 leaves remainder 0, we see that 7 * 12 (mod 4) = 0. 11 * 13 (mod 4) is the remainder when 11 * 13 is divided by 4. Since 11 * 13 = 143 and 143 / 4 leaves remainder 3,we see that 11 * 13 (mod 4) = 3. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q002. Make a table for the x * y mod 4 operation, which we will call '* mod 4', operating on the set {0, 1, 2, 3}. Determine which of the properties, including commutativity, associative, identity, inverse and closure properties, are properties of this operation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: {0, 1, 2, 3}. 4 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Whatever x is, 0 * x = x * 0 = 0, which when divided by 4 leaves remainder 0. Whatever x is, 1 * x = x * 1 = x, and if x is in the set {0, 1, 2, 3} we have get remainder x when dividing by 4 (e.g., 4 divides into 0, 1, 2 or 3 zero times, leaving that number as the remainder) and x mod 4 = x. From this we can see that 1 is the identity for this operation. Multiplying 0, 1, 2, and 3 by 2 we get 0, 2, 4, and 6, which when divided by 4 leave remainders 0, 2, 0 and 2, respectively. Multiplying 0, 1, 2, and 3 by 2 we get 0, 3, 6, and 9, which when divided by 4 leave remainders 0, 3, 2 and 1, respectively. The table for this operation is therefore * mod 4 0 1 2 3 0 0 0 0 0 1 0 1 2 3 2 0 2 0 2 3 0 3 2 1 We note that this operation does contain identity 1, but since neither 0 nor 2 can be combined with any of the elements of the set to give us the identity, the operation on this set does not have the inverse property. We do see from the symmetry of the table about the main diagonal that it has the commutative property, which we could in any event have concluded from the fact that multiplication is commutative so that the product we get before calculating the remainder is independent of the order of the two numbers. In a similar matter we can reason that the operation is associative. The operation is also closed, since the remainder upon dividing by 4 must always be 0, 1, 2 or 3 and hence in the set {0, 1, 2, 3}. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. Repeat the preceding exercise for the operation x * y mod 5, defined to give the remainder when x * y is divided by 5, on the set {1, 2, 3, 4}. Determine which of the properties are exhibited by this operation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution:x = 4 : 3x + 7 = 9 (mod 5). 5 1 2 3 4 1 1 2 3 4 2 2 4 1 3 3 3 1 4 2 4 4 3 2 1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: First we might wish to do a couple of example calculations to get familiar with the operation. For example: 2 * 3 mod 5 = 6, which when divided by 5 gives us remainder 1. 3 * 4 mod 5 = 12 which when divided by 5 gives us remainder 2. 2 * 4 mod 5 = 8 which when divided by 5 gives us remainder 3. The table is * mod 5 1 2 3 4 1 1 2 3 4 2 2 4 1 3 3 3 1 4 2 4 4 3 2 1 We immediately see that all the results are in the set {1, 2, 3, 4}, so that the operation is closed. This operation has identity 1, as we can see from the row and the column across from and beneath 1. We easily see from the table that the identity appears exactly once in each row and in each column, which assures us that the operation has the inverse property. Specifically we see that 1 * 1 mod 5 = 1 so that 1 is its own inverse, that 2 * 3 mod 5 = 1 so that 2 and 3 are inverses, and that 4 * 4 mod 5 = 1, so that 4 is its own inverse. The associativity and commutativity of the operation follow from the associative and commutative properties of multiplication on real numbers, as discussed in the preceding problem. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. The equation 3x + 7 = 9 (mod 5) has an integer solution for x = 0, 1, 2, 3 or 4. Which value of x is a solution to this equation? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x = 4; 3x + 7 = 9 (mod 5). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 3x + 7 = 9 (mod 4) means that 3x + 7 - 9 = 0 (mod 5) so 3x - 2 = 0 (mod 5). If 3x - 2 = 0 (mod 5) then when we divide 3x - 2 by 5 we should get remainder 0. So we substitute the different possible values for x into the expression 3x - 2 until we get a number of which when divided by 5 gives us remainder 0. If x = 0 then 3x - 2 = -2, and -2 (mod 5) = 3 (if you don't understand why -2 mod 5 = 3, think of the 5-hour clock in the text; but for now it should be obvious that -2 is not a multiple of 5 so that you cannot get remainder 0 when dividing -2 by 5). If x = 1 then 3x - 2= 1, and 1 (mod 5) = 1. If x = 2 then 3x - 2= 4, and 4 (mod 5) = 4. If x = 3 then 3x - 2= 7, and 7 (mod 5) = 2. If x = 4 then 3x - 2= 10, and 10 (mod 5) = 0. Thus x = 4 is a solution to the equation 3x + 7 = 9 (mod 5). STUDENT QUESTION What is that 5-hour clock? INSTRUCTOR RESPONSE A 5-hour clock is depicted below: We label the usual 12:00 position 0. Moving 1 hour at a time in the clockwise direction we encounter hours 1, 2, 3 and 4 before finally returning after 5 hours to the 0 position. To calculate 3 + 4 on this clock, we start at 0 and move 3 hours, which puts us at position 0. Then we move 4 more hours, which takes us through positions 4, 0 and 1 before ending up at 2. Thus, on this clock, 3 + 4 = 2. It's easy to see how this gives us a picture for the calculation (3 + 4) mod 5, and why the result is the remainder when we divide 3 + 4 by 5. To calculate 3 * 4 on this clock, we would start at 0 and move 4 hours to position 4, then 4 more hours to position 3, then 4 more hours to position 2. Thus on this clock 3 * 4 = 2, and this gives us a picture of (3 * 4) mod 5. It's easy to see that we have moved 3 * 4 = 12 hours, and that where we end up is the remainder when 12 divided by 5. To divide 3 by 4 on this clock, we observe that for ordinary base-10 division, (3 / 4) * 4 = 3. So we ask what number multiplied by 4 on this clock will give us 3. If we start at 0 and move 4 hours around the clock we end up at 4; if we move 4 more hours we end up at 3. Thus 4 * 2 = 3, and we can therefore say that 2 = 3/4. That is, 3 divided by 4 is 2. To solve the equation 3x - 2 = 0: Next we try different numbers for x: If x = 1 we get 3 * 1 - 2 = 1. If x = 2 we get 3 * 2 - 2 = 4. If x = 3 we get 3 * 3 - 2 = 7. Since 7 hours on this clock takes us from our starting position at 0 to position 2, we conclude that when x = 3 our result is 3 * 3 - 2 = 2. If x = 4 we get 3 * 4 - 2 = 10. Since 10 hours on this clock takes us from our starting position at 0, twice around the clock and back to position 0, we conclude that when x = 4 our result is 3 * 4 - 2 = 0. So x = 4 is a solution to our equation 3 * x - 2 = 0 (mod 5). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. You see that x = 4 is a solution to the equation 3x + 7 = 9 (mod 5). One of the numbers x = 5, 6, 7, 8, 9 is also a solution. Which one is it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x = 9 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We recall that 3x + 7 = 9 (mod 5) is equivalent to 3x - 2 = 0 (mod 5). We evaluate 3x - 2 (mod 5) for x = 5, 6, 7, 8 and 9 and we find that the results are 3, 1, 4, 2, and 0. So x = 9 is our next solution. We might also note that the series of results 3, 1, 4, 2, 0 is the same as the series we got for x = 0, 1, 2, 3, 4. Our results therefore seem to indicate a repeating pattern in which the remainder 0 occurs every fifth number starting with 4. This is in fact what happens, and you might wish to think about why this happens. However, you should in a case remember that this is what happens. In general when we have an equation of the form A x + B = C (mod n), integer solutions happen at intervals of n. for some values of A, B and C integer solutions can also occur at shorter intervals, but they always do occur at intervals of n. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q006. What are the first five positive values of x which solve the equation 3x + 7 = 9 (mod 5) of the preceding problem? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x = 4, 9, 14, 19 and 24 goes by 5 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We just saw that x = 4 and x = 9 are solutions, and we saw that because we are solving an equation mod 5, the solutions have to occur at intervals of 5. Thus the first five solutions are x = 4, 9, 14, 19 and 24. ********************************************* Question: `q007. Can you find a solution to the equation 3x + 7 = 9 (mod 6)? Show how you reason out your answer. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating:
#$&* course mth 151 12:03pm12/19/13
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Given Solution: `a** Divide both sides by 4 to get x - 9 = 2(x+3). Then apply the Distributive Law to get x - 9 = 2 x + 6. Then add -x to both sides: x - 9 - x = 2 x + 6 - x. Simplify -9 = y + 6. Add -6 to both sides -9 - 6 = y + 6 - 6. Simplify -15 = x. The solution is x = -15.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q7.1.39 (2r - 3 ) / 7 + 3/7 = -r/3. Show how you solved the equation for r. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 3 (2r - 3) + 9 = -7 r. add 7 r to both sides: 6 r + 7 r = 0 so 13 r = 0 and r = 0 / 13 = 0. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** You don't want to deal with fractions so you do best to multiply both sides by the common denominator 21 to get 21 (2r - 3) / 7 + 21 * 3 / 7 = 21 * (-r/3), which simplifies to 3 (2r - 3) + 9 = -7 r. The expand on the left using the Distributive Law: 6 r - 9 + 9 = - 7 r. -9 + 9 = 0; add 7 r to both sides: 6 r + 7 r = 0 so 13 r = 0 and r = 0 / 13 = 0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q7.1.45 .20 ( 14000) + .14 t = .18 (14000 + t ). Show how you solved the equation for t. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 20 ( 14000) + 14 t = 18 ( 14000 + t). 10 ( 14000) + 7 t = 9 ( 14000 + t). 140,000 + 7 t = 126,000 + 9 t. 140,000 + 7 t - 9 t = 126,000 + 9 t - 9 t. 140,000 - 2 t = 126,000. Add -140,000 to both sides: 140,000 - 2 t = 126,000 - 140,000. -2 t = -14,000. Divide by -2: t = 7000 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aCORRECT STUDENT SOLUTION WITH INSTRUCTOR COMMENT: .20(14,000)+.14t=.18(14,000+t) 2800+.14t=2520+.18t -.14t+2800+.14t=2520+.18t-.14t 2800=2520+.04t -2520+2800=2520+.04t-2520 280=.04t 280/.04=.04t/.04 7000=t INSTRUCTOR COMMENT: Correct solution but you're better off if you can avoid decimals and the attendant risk of misplacing one. One possible sequence of steps: First multiply both sides by 100 to avoid having to deal with decimals: 20 ( 14000) + 14 t = 18 ( 14000 + t). Divide both sides by the common factor 2: 10 ( 14000) + 7 t = 9 ( 14000 + t). Simplify: 140,000 + 7 t = 126,000 + 9 t. Add -9 to both sides: 140,000 + 7 t - 9 t = 126,000 + 9 t - 9 t. Simplify 140,000 - 2 t = 126,000. Add -140,000 to both sides: 140,000 - 2 t = 126,000 - 140,000. Simplify: -2 t = -14,000. Divide both sides by -2: t = 7000. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q7.1.70 A = 1/2 ( B + b ) * h. Show how you solved the equation for B YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: B = 2A / h - b confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** You need to isolate B: First multiply both sides by 2 and get 2A= (B+b) *h . Next divide both sides by h to get 2A / h = B + b. Then subtract b: 2A / h - b = B. Thus B = 2A / h - b. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q7.2.18 20 million more viewers on Thursday than on Saturday; the total number is 102 million. How many watch each night? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2x + 20 = 120. 2x + 20 - 20 = 120 - 20 o 2x = 100. (2x) / 2 = 100 / 2 x = 50. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** If we have 20 million more on Thursday then if x = number of Saturday viewers in millions, the number of Thursday viewers is x + 20, again in millions. The total number is 120 million so we have # of Thursday viewers + # of Saturday viewers = 120; using our expressions we have (x + 20) + x = 120 so that 2x + 20 = 120. Adding -20 to both sides we get 2x + 20 - 20 = 120 - 20 or 2x = 100. Dividing both sides by 2 we get (2x) / 2 = 100 / 2 so that x = 50. Thus the number of Saturday viewers is 50, representing 50 million, and the number of Thursday viewers is x + 20 = 70, representing 70 million.GE and GM total 13.5 billion; GE .3 billion less than GM. Show how you set up and solved an equation to obtain the profits of both companies. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 6.9 billion and GE profits are 6.6 billion confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** If GE makes .3 billion less than GM then if x = GM profits in billions we can say that the GE profits are x - .3. Thus we have x + (x-.3) = 13.5. This gives us 2x - .3 = 13.5 so that 2x = 13.5 + .3 = 13.8 and x = 13.8 / 2 = 6.9. It follows that x - .3 = 6.9 so that x = 6.9-.3 = 6.6. So GM profits are 6.9 billion and GE profits are 6.6 billion. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q7.2.40 Water added to 3 gal of 4% soln to get 3% soln. Show how you solved the problem by setting up and solving an equation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1 exactly confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** You need to first specify what x is. If x is the number of gallons of water added, then : After adding 3 gallons we have 3 + x gallons of 3% solution. Originally we had .04 * 3 gallons of the solute, and since we add only water that's what we still have after adding the water. But now the solute is in 3+x gallons of solution and its concentration is 3%. We have the same amount of solute after as before, since all we added was water, so .04 * 3 = .03 * ( 3 + x ). We solve this equation to get x, the number of gallons of water added. We obtain x = 1. The number of gallons added will be exactly 1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q7.2.50 Equal #'s of nickels and quarters, twice as many half-dollars; total $2.60. Show how you solved the problem by setting up and solving an equation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We have 2 nickles, 2 quarters and 4 half-dollars confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a**If x is the number of nickles then we have x quarters and 2x half-dollars. So the amount of money we have is .05 x + .25 x + .5 (2x). This amount is 2.60 so .05 x + .25 x + .5 (2x) = 2.60. Simplifying the left-hand side: .05 x + .25 x + x = 2.60 1.30 x = 2.60 so that x = 2.60 / 1.30 = 2. We have 2 nickles, 2 quarters and 4 half-dollars.** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q7.2.50 Equal #'s of nickels and quarters, twice as many half-dollars; total $2.60. Show how you solved the problem by setting up and solving an equation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We have 2 nickles, 2 quarters and 4 half-dollars confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a**If x is the number of nickles then we have x quarters and 2x half-dollars. So the amount of money we have is .05 x + .25 x + .5 (2x). This amount is 2.60 so .05 x + .25 x + .5 (2x) = 2.60. Simplifying the left-hand side: .05 x + .25 x + x = 2.60 1.30 x = 2.60 so that x = 2.60 / 1.30 = 2. We have 2 nickles, 2 quarters and 4 half-dollars.** " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!