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Solutions 1 Intermediate Algebra Activities Using WA

WA Activity # 1 ‐‐ Real Numbers (Solutions)

1. Following the order of operations, WA multiplied 3 times 4 then added 2. The answer is 14.

2. Using parentheses, type “(2 + 3)*4” into the WA search engine. The answer is 20. This is a different answer because you changed the order of operations.

3. The Exact result = 13/5. The Egyptian fraction is 2 + ½ + 1/10 = 20/10 + 5/10 + 1/10 = 26/20 = 13/5.

4. Type “(2 + 3)/5” into the WA search engine. The answer is 1.

5. A real number is any number which can be represented by a point on the number line.

6. Answers will vary.

7. Every positive integer has a unique prime factorization, which is the product of prime numbers which when multiplied together give the original integer.


9. MDCCLXXVI = 1776.


11. Exact result = 19/12 (fraction). Other forms of the answer are (mixed number),

1.583333 (decimal approximation to six places) or 1.58 (using the repeat bar).

12. You can multiply top and bottom of a fraction by 2 or 3 or any other number (as long as the number is the same on the top and bottom) because any number divided by itself is 1; you are allowed to multiply a number by 1 without changing it (1 is the multiplicative identity) .

13. 0.06

14. 3/50 is exactly equal to 0.06.

15. 3/50 means 3 divided by 50. Then change that decimal answer into a percent. 3/50 = 0.06 = 6%.

16. 2.7

17. WA expresses 6% as a fraction (6/100) and multiplies that by 45. The answer is 27/10 = 2.7.

18. The answer in Exact form is a fraction; in Approximate form it’s a decimal. Both answers are exact!

19. 31.25%

20. 5/16 = 0.3125 = 31.25%

21. ‐27

Solutions 2 Intermediate Algebra Activities Using WA 22. 684

23. 2.526316

24. 25

25. 145

26. 192/7

27. Yes. This is still 24 divided by 7/8.

28. No. This is not the same as the previous question. WA follows the order of operations (dividing from left to right) and calculates 24 divided by 7 then takes that answer and divides it by 8.

29. Yes, this is the same as the previous question. WA follows the order of operations (work inside the parentheses first) and calculates 24 divided by 7 then takes that answer and divides it by 8.

30. ‐8 = (‐2)(‐2)(‐2) meaning use ‐2 as a factor three times, that is, multiply ‐2 times itself three times.

31. ‐8 = ‐(2^3) = ‐1(2)(2)(2) meaning ‐1 times the result of 2 being multiplied times itself three times. Same answer, different question from the previous one.

Type "sqrt 8” which is short for “square root of 8” into the WA search engine. Use the results to answer the following questions.

32. =

33. Irrational. WA gives a Continued fraction, meaning it’s a fraction that goes on and on forever and ever and never ends. As you go the answer gets closer and closer to but never gets there!

34. 14/5 = 2.8

35. 59/21 = 2.809524

36. 82/29 = 2.827559

37. = 2.828427 to six places. The further out you go with the continued fraction the closer you get to this number.


WA Activity # 2 ‐‐ Linear Functions (Solutions)

1. 13/7, 1 , 1.8571

2. Answers may vary. Most students would not subtract 13x + 4 from both sides then subtract 13x again from both sides. Most students would place terms containing the variable x on one side of the equation, the numbers on the other side, then divide both sides by the coefficient.

3. The blue line represents the graph of y = 3 – 2(‐3 – 7) or y = 6x + 17. That’s because at x = 1, y = 23 and the blue line goes through that point (1, 23) and has slope 6. The red line represents the graph of y = 13x + 4. That’s because at x = 1, y = 17 and the red line goes through that point (1, 17) and it has slope 13. The red dot represents the point of intersection of the two graphs, that is, the point at which 3 ‐ 2(‐3x ‐ 7) = 4 + 13x. The x‐coordinate of that point is the solution to the equation, 13/7.

4. The result appears with more digits past the decimal. 1.857143.

5. y = x – 3; ‐5x + 4y + 12 = 0; 5x – 4y – 12 = 0.

6.

7. Answers may vary. Some points are (‐4, ‐8), (0, ‐3), (4, 2), etc.

8. Let y = 0 and solve for x, that is, solve the equation x – 3 = 0. You get x = or 2.4.

9. Let x = 0 and solve for y, that is, solve the equation y = (0) – 3. You get y = ‐3.

10. Draw a right triangle using the line segment between the two given points as the hypotenuse. Then use the Pythagorean Theorem to find the distance. Or use the formula for the distance between two points d = .

11. A table of values for the equation from x = ‐10 to x = 10 in simplest fraction form and in decimal form, plus a plot of all the points on the graph of the line.

12. f(x) = 2.01106x + 0.911397

13. 2.01106

14. y = 5; x = 2

15. The four data points; (2, 5.17); (4, 7.9)

16. x = ‐0.453192; let y = 0 and solve for x, that is, solve the equation 2.01106x + 0.911397 = 0.

17. y = 0.911397; let x = 0 and solve for y, that is, solve the equation y = 2.01106(0) + 0.911397.


WA Activity # 3 ‐‐ Systems of Linear Equations and Inequalities (Solutions)

1. In simple terms it means “and.” A more formal definition is the value that is produced if and only if the two statements before and after the symbol are true.

2. (1, 5)

3. Yes; 2(1) + 3 = 5 or 5 = 5 and 1 + 5 = 6 or 6 = 6.

Type “1.3x‐0.2y=‐3, ‐0.1x+0.5y=1.2” into the WA search engine. Notice with the comma WA no longer uses the logical conjunction symbol. Use the results to answer the following questions.

4. Yes

5. Xmin = ‐4, Xmax = 4, Ymin = 1.6, Ymax = 3.2; no

6. (‐2, 2)

7. Yes; 1.3(‐2) – 0.2(2) = ‐3 ‐2.6 – 0.4 = ‐3 ‐3 = ‐3 and ‐0.1(‐2) + 0.5(2) = 1.2 0.2 + 1 = 1.2 1.2 = 1.2

Type "4x‐3y=‐1; 2x‐(3/2)y=3” into the WA search engine. Use the results to answer the following questions.

8. Two lines that appear to be parallel, that is, they do not intersect; Xmin = ‐1, Xmax = 1, Ymin = ‐3.4, Ymax = ‐0.6

9. One way to prove the lines are parallel is to rewrite them in slope‐intercept form and compare the slopes. The first equation would be y = x + . The second equation would be

y = x ‐ 2. The slopes are equal therefore the lines are parallel.

Type "3x+4y=12; 2y=6‐(3/2)x” into the WA search engine. Use the results to answer the following questions.

10. A single line; below that you see alternate forms of the equations. The first set of alternate forms are the same two equations. The second set of alternate forms show the same equation in two different forms.

11. The system has infinitely many solutions.

12. Yes.

13. Answers may vary; some examples are (0, 3), (4, 0) and (8, ‐3).

14. Answers may vary; some examples are (‐4, 6), (‐2, 4.5) and (3, 0.75).

Type “2x‐3<5” into the WA search engine. Use the results to answer the following questions.

15. The blue line is the graph of the equation y = 2x – 3; the red line is the graph of the equation y = 5.

Solutions 5 Intermediate Algebra Activities Using WA 16. The shaded area represents the region for all the (infinitely many) points the x‐coordinates of which, if substituted into the inequality, would satisfy the inequality.

17. The x‐coordinates of all the points in the shaded region are all less than 4.

18. x < 4 or ( , 4)

19. The blue line represents the graph of the equation y = 5x + 8; the red line represents the graph of the equation y = 7(1 + x) or y = 7x + 7.

20. The shaded area represents the region for all the (infinitely many) points the x‐coordinates of which, if substituted into the inequality, would satisfy the inequality.

21. [ ,

22. Yes.

23. Yes.

24. The shaded region.

25. The line for y x – 3 would be solid; the line for y < x would be dashed. WA appears to show that but it’s hard to see.

Type “x<=3 and y>‐1” into the WA search engine. Use the results to answer the following questions.

26. Yes.

27. Yes but it’s just a repeat of the original system.

28. The shaded region.

29. The line for x 3 would be solid; the line for y > ‐1 would be dashed. WA may show that but it’s hard to see.


WA Activity # 4 ‐‐ Polynomial Functions (Solutions)

Type “let x = ‐ 4 in (3x^2 ‐7x – 41)” into the WA search engine. Use the results to answer the following questions.

1. 35

2. 3(‐4)^2 – 7(‐4) – 41 = 3(16) + 28 – 41 = 48 – 13 = 35

Type “(3ab‐7b‐2a+20) ‐ (‐11ab‐4a+b‐13)” into the WA search engine. Use the results to answer the following questions.

3. Yes.

4. Yes.

5. (3ab – 7b – 2a + 20) – (‐11ab – 4a + b – 13) = 3ab – 7b – 2a + 20 + 11ab + 4a – b + 13 = 14ab + 2a ‐8b + 33.

Type “(3x+2)(2x‐5)” into the WA search engine. Use the results to answer the following questions.

6. Yes.

7. The graph of the equation y = (3x + 2)(2x – 5), a parabola.

8. The answer as we would normally express it is given as an Alternate form. The answer is – 11x – 10.


10. Yes.

11. . WA wrote the answer in descending order of

exponents but either form of the answer is okay.

Type “(6x^2‐11x‐7) / (3x‐7)” into the WA search engine. Use the results to answer the following questions.

12. Yes.

13. The blue line represents the graph of the equation y = .

14. 2x + 1

15. The quotient is 2x + 1; the remainder is 0. Yes, WA shows you this information.

16. The root of the equation is x = ½. .

Solutions 7 Intermediate Algebra Activities Using WA Type “(3x^3‐2x^2+5) / (x+3)” into the WA search engine. Use the results to answer the following questions.

17. 3

18. No.

20. The quotient is 3 ; the remainder is 94.

21. The Real root is x = ‐1. .

22. The x‐intercept is (‐1, 0)

Type “gcd 16x^5y^2, 20x^4y^6, 24x^6y^8” into the WA search engine. Use the results to answer the following questions.

23. Greatest common divisor.

24. . No remainders.

25. (2x + 3) (3x – 1)

26. (2x + 3) (3x – 1) = 6 + 7x – 3; yes it checks.

27. 3x(x – 3)(3x + 4)

Type “factor 4x^2 ‐ 25” into the WA search engine. Use the results to answer the following questions.

28. (2x + 5)(2x – 5)

29. Difference of two squares.

30. Two.

Type “factor 4x^2 + 25” into the WA search engine. Use the results to answer the following questions.

31. The Result is 4 + 25, same as the original problem. This expression is not factorable over the set of real numbers.

32. There are no x‐intercepts. The horizontal axis of the plots appear to be the lines y = 20 and y = 40. The x‐axis does not show in either of the plots.

33. x = ‐3

34. The blue line represents the graph of the equation y = x + 3.

Solutions 8 Intermediate Algebra Activities Using WA 35. The red dot is the x‐intercept of the graph of y = x + 3 which is the root or the solution to the equation x + 3 = 0.

36. Xmin = ‐4, Xmax = ‐2, Ymin = ‐1, Ymax = 1. The WA Plot shows the x‐axis but not the y‐axis.

37. The blue line represents the graph of the parabola y = 5 .

38. The red line represents the graph of the line y = ‐16x – 3.

39. The red dots represent the intersection points of the two graphs, which are the solutions to the equation.

40. Two.

41. 5 = ‐16(‐3) – 3 5(9) = 48 – 3 45 = 45 and

; both solutions check.


WA Activity # 5 ‐‐ Quadratic Functions (Solutions)

Type "x^2 = 9” into the WA search engine. Use the results to answer the following questions.

1. The blue line represents the graph of the equation y = .

2. The red line represents the graph of the equation y = 9.

3. The red dots represent the intersection of the two graphs. The x‐coordinates of those points are the solutions to the equation.

4. Square Root Property: . Factoring: (x +3)(x – 3) = 0, x = 3.

Type “x^2 – 4x – 5 = 0” into the WA search engine. Use the results to answer the following questions.

5. There are two solutions, x = ‐1 and x = 5.

6. Quadratic Formula: x = = .

Factoring: (x + 1)(x – 5) = 0 x + 1 = 0 x = ‐1 or x – 5 = 0 x = 5.

Type “(1/4)x^2 = 2x – 9/2” into the WA search engine. Use the results to answer the following questions.

7. Complex. The results are labeled “complex” and of course one can see they are part real, part imaginary.

8. No, the graph does not show the x‐axis. The graph does not intersect the x‐axis. There is no x‐intercept. Of course we knew this because the solutions were complex.

9. One method: Clear the fraction and re‐write in standard form then use the Quadratic Formula: .

.

Type “quadratic fit {(1.3, 13.2), (3.1, 25.8), (3.7, 39.2), (5.2, 63.8)}” into the WA search engine. Use the results to answer the following questions.

10. y = 2.1954 – 1.09219x + 10.5748

11. The scatter plot appears to be in the shape of a parabola. The equation appears to model the data reasonably well.

12. The linear equation of best fit is y = 13.0375x – 7.84957. The linear equation appears to also model the data reasonably well.

Solutions 10 Intermediate Algebra Activities Using WA 13. It would depend on the data or the situation being modeled. If the trend is for the change (the slope) to increase as the values of x increase then a quadratic model would be appropriate; if the trend is for the change (the slope) to stay constant a linear equation would be a good model.

14. x =

15. x = ‐2.69 or x = 0.19

Type the function “3x^2 +2x – 4” into the WA search engine. Use the results to answer the following questions.

16. x =

17. x = ‐1.535 or x = 0.869

18.

19. The parabola opens upward so it has a minimum value of .

20. The x‐intercepts are and or approximately (‐1.535, 0) and (0.869, 0)

to the nearest thousandth.


WA Activity # 6 ‐‐ Exponential Functions (Solutions)

1. The blue graph is y = because it goes through the point (1, 2). The red graph is y = because it goes through the point (1, 3). Answers may vary.

2. Yes, these are both functions because for each input there is one and only one output. The domain for both functions is ( and the range for both functions is (0,

3. For y = : (‐1, ½), (0, 1), (1, 2). For y = : (‐1, 1/3), (0, 1), (1, 3). Answers may vary.

4. Yes, the parentheses make a difference. If the exponent is NOT in parentheses the power of 2 becomes 0.5 not 0.5 times x. For example if x = 3 then with the parentheses 2^(0.5*3) 2.828 and without the parentheses 2^0.5*3 4.243. The function and its graph are completely different.

5. The blue graph is y = because it goes through the point (1, 2). The red graph is y = because it goes through the point ( 1, ) or approximately (1, 1.414). Answers may vary.


7. For y = : (‐1, ½), (0, 1), (1, 2). For y = : (‐1, 0.707), (0, 1), (2, 2). Answers may

vary.

8. No, the parentheses do not make a difference. Because the numerator is 1 both functions are

the same. For example if x = 3 with parentheses and without parentheses

.

9. Yes, the parentheses make a difference. For example when x = 2 with parentheses

and without parentheses . The functions are very different

because of the notation.

10. No, the parentheses do not make a difference in this case because the numerator is 1. If you type the command with OUT parentheses you get identical results.

11. The blue graph is y = because it goes through the point (1, ½ ). The red graph is y = because it goes through the point (1, 1/3) which is underneath the blue graph. Answers may vary.


13. For y = : (‐2, 4), (‐1, 2), (0, 1). For y = : (‐1, 3), (0, 1), (1, 1/3). Answers may vary.

14. Yes, these functions are the same. In both cases the exponent is calculated first then the parentheses and the asterisk symbol * both mean multiply that answer times 2. Type

Solutions 12 Intermediate Algebra Activities Using WA “2(2^x),2*2^x”into the WA search engine. Are these the same functions or are they different? Explain your answer.

15. because of the properties of exponents.

16. The blue graph is y = because it goes through the point (1, 4). The red graph is y = because it goes through the point (1, 6) which is higher than and above the blue graph.

Answers may vary.

17. as in #15 above but and nothing can be further simplified. Because the bases, 2 and 3, are different the properties of exponents do not apply.

18. For : (‐1, 1), (0, 2), (1,4). For : (‐1, 3/2), (0, 3), (1, 6). Answers may vary.

19. The blue graph is y = because it goes through the point (1, 2) which is above the x‐axis. The red graph is y = because it goes through the point (1, ‐2) which is below the x‐axis. Answers may vary.

20. The domain for both functions is . The range for y = is (0, ) and the range for y = is ( , 0).

21. The y‐intercept for y = is (0, 1). The y‐intercept for y = is (0, ‐1). Substitute x = 0 into both functions and solve for y.

22. Yes, the graphs are reflections of each other (symmetric) about the x‐axis which is the line y = 0.

23. For y = : (0, 1), (1, 2), (2, 4). For y = : (0, ‐1), (1, ‐2), (2, ‐4).

24. The blue graph is y = because it goes through the point (1, 2). The red graph is y = because it goes through the point (1, ½ ) which is less than and below the blue graph. Answers may vary.

25. The domain for both functions is ( and the range for both functions is (0,

26. The y‐intercept for both functions is (0, 1). Substitute x = 0 into both functions and solve for y.

27. Yes, the graphs are reflections of each other (symmetric) about the y‐axis which is the line x = 0.

28. For y = : (0, 1), (1, 2), (2, 4). For y = : (0, 1), (1, ½ ), (2, ¼). Answers may vary.


WA Activity # 7 ‐‐ Logarithmic Functions (Solutions)

1. The base of the logarithm is 10. The result is 2 because .

2. The base of the logarithm on the left hand side is 10; the base of the logarithm on the right hand side is e. The two expressions are equivalent because of the Change of Base formula.

3. The base of the logarithm is e.

4. The answer to six places is 4.605170. That is the result because .

5. Using the Properties of Logarithms, 2 log(2) + 2 log(5) = .

6. The base of the logarithm is 10.

7. The blue graph is of the equation y = . The red graph is of the equation y = 2. The red dot represents the intersection of the two graphs or the solution to the original equation in the WA command line.

8. The blue graph appears to be a straight line because of the very small viewing window. Change the window to Xmin=0 and Xmax=50. Use your graphing calculator or WA. The WA command and the result is shown below. This command gives just the graph, not the solution to the equation, but it is a much better visual.

9. Substitute 34 into the equation for x. You get: . The answer checks.

Solutions 14 Intermediate Algebra Activities Using WA 10. Using the properties of logarithms:

.

Type the logarithmic equation “6 log(9, x) + 1 = 4” into the WA search engine. Use the results to answer the following questions.

11. The base of the logarithm is 9.

12. The blue graph is of the equation y = . The red graph is of the equation y = 4. The red dot represents the intersection of the two graphs which is the solution to the original equation in the WA command line.

13. Substitute 3 into the original equation for x: and the

answer checks.

14.

Type the logarithmic equation “2 log(9,x^3) – 3 log(9,2x) = 2” into the WA search engine. Use the results to answer the following questions.

15. The base of the logarithms is 9.

16. Use the Change of Base formula to go from base 9 to base e, then use the properties of

logs: .

17.

. The decimal approximation does match the x‐coordinate of the intersection point of the graph of the two equations in the plot.

Type the logarithmic equation “log(4x^5) ‐ 2 log(x^2) = 5” into the WA search engine. Use the results to answer the following questions.

18. These are natural logarithms. The base is e.

19. The blue graph is of the equation y = and the red graph is of the equation y = 5. The red dot represents the intersection of the two graphs which is the solution to the original equation in the WA command line.

Solutions 15 Intermediate Algebra Activities Using WA 20.

.

21. which is too small, so the number we’re looking for is greater than 2.

22. which is too large, so the number we’re looking for is less than 3. It’s probably closer to 2.

23. The answer is 2.26186 which is a number between 2 and 3, closer to 2.

24. The answer is a natural logarithm; the base of the logarithm is e.

25. Take the log or the natural log of both sides of the equation and proceed using the properties of logarithms and algebra:

.


Type “solve 2e^(3x – 1) = 54 in reals” into the WA search engine. Use the results to answer the following questions.

27. The result is . The base of the logarithm is e.

28. The result as a single fraction is .

29.

.

30. This is a natural logarithm. The base is e.


Solutions 16 Intermediate Algebra Activities Using WA 32.

.


WA Activity # 8 ‐‐ Rational Functions (Solutions)

1. 87/32

2. The domain is all real numbers except x = 5.

3. In the plot there doesn’t appear to be a hole in the graph at x = 5. Of course it wouldn’t show up on your graphing calculator, either. It’s an “infinitely small” hole in the graph.

4. Factoring the numerator and using algebra: .

5. x = 5 is not in the domain. It gives you a zero in the denominator of the original expression. That’s why x = 5 can not be one of the Roots.

6.

7. Factoring the numerator and using algebra: . Either form

of the answer is okay.

8. WA changed the format of the input by multiplying by “1” as follows: .

9. Starting the different format: . Starting with the original

expression: . Either way. That’s the beauty of math. There’s often

more than one way to solve a problem but usually just one right answer. You get to choose the method or the way you like the most or understand the best.

10. The expression is undefined for x = 2 and x = ‐ ½ .

11. Find a common denominator and combine the two expressions:

. Since factoring out the 2 in the numerator doesn’t allow for any further simplification you could leave the answer as either one of the last two expressions.

12. The Roots are x = ‐3 and x = 1. If you substitute those values for x into the original expression you get a zero in the numerator meaning f(x) = 0. If necessary type “plot x/(x ‐ 2) + 3/(2x + 1), x=‐5 to x=5” into the WA search engine to more clearly see the two solutions as x‐intercepts of the graph.

Solutions 18 Intermediate Algebra Activities Using WA 12. WA shows “m” could also be used as units of measurement such as meters (unit of length) or minutes (unit of time).

13. There are two solutions to this equation. The graphs intersect each other in two distinct points. When you solve this equation by clearing the fractions you get a quadratic, second degree equation which can have two solutions and in fact that’s what happens. See the next question.

14. Multiply both sides by the LCD to clear the fractions then solve the quadratic equation by factoring:

.

Type “solve (1/x + 1/y = 1/z) for y” into the WA search engine. Use the results to answer the following questions.

15. The solution is .

16. x – z cannot be equal to zero because then the expression would be undefined, you’d have a zero in the denominator.

17. The reason xz cannot be equal to zero is a little less obvious. If xz = 0 then either x must equal zero or y must equal zero. If so then one or both of the fractions in the original expression would be undefined!

16. WA Show steps may not be too helpful, in this case. As in the problem above, multiply both sides by the LCD to clear the fractions then use algebra solve for y:

. The second answer is preferred because the numerator is positive, the whole expression has fewer negative signs and it “looks nicer” doesn’t it? We like that in mathematics. But the other form of the answer would be okay, too.


WA Activity # 9 ‐‐ Radical Functions (Solutions)

1. 3

2. The blue graph is of the equation and the red graph is of the equation y = 9. The red dots represent the intersection of the two graphs which is the solution to the original equation in the WA command line.

3. There are two solutions.

4. 3 and ‐3.

5. The radical expression means give the “principal” or positive square root of 9 which is 3. The equation means find all the values of x for which the equation is true, and there are two solutions.

6.

7. No, this is not the result we usually get in an Intermediate Algebra class. (P.S. Perhaps by the time you read this WA may give a different result. Let us know.)

8. We would normally give the radical expression for an answer.

9. The two expressions are equivalent: .

10.

11. WA is assuming a > 0 and b > 0 because otherwise the original expression could have a zero in the denominator and be undefined.

12.

Type “simplify (11 ‐ 2sqrt(x)) (3 + sqrt(x))” into the WA search engine. Use the results to answer the following questions.

13.

14. WA multiplied the two expressions using the distributive property or the FOIL method, as many students know it:

15.

16. No, WA did not multiply the two expressions together. They just gave a different form of the input command line. (P.S. Perhaps by the time you read this the problem has been fixed? Let us know.)

Solutions 20 Intermediate Algebra Activities Using WA 17. Yes, one of the Alternate forms looks familiar to us as the answer. It is .

18. WA multiplied the two expressions using the distributive property or the FOIL method, as many students know it:

19. The blue graph is of the equation and the red graph is of the equation y = 4. The red dot represents the intersection of the two graphs which is the solution to the original equation in the WA command line.

20. The blue graph appears to be a straight line, but the graph of the square root function is not a straight line. To see what might be a better plot of the equation type “plot sqrt(x + 3) = 4, x=0 to x=25” into the WA search engine, for example. Your window may vary.

21. 13

22. Square both sides of the equation and solve for x. Check your answer:

Type “sqrt(x + 13) ‐ 4 = x + 3” into the WA search engine. Use the results to answer the following questions.

23. The blue graph is of the equation and the red graph is of the equation y = x + 3. The red dot represents the intersection of the two graphs which is the solution to the original equation in the WA command line.

24.

. Check the answers. Check x = ‐9: . FALSE. Now check x = ‐4:

. TRUE.

25. Yes, there was an extraneous solution. WA did not show it.


WA Activity # 10 ‐‐ Additional Topics: Sequences and Series (Solutions)

1. Yes.

2. This is a geometric sequence because it is an ordered list of numbers with a common ratio of 2. That is, to get from one term to the next multiply the previous term by 2.

3.

4. Yes.

5. This an arithmetic sequence because it is an ordered list of numbers with a common difference of ‐4. That is, to get from one term to the next add ‐4.

6.

7. WA gives two formulas for this sequence: and . The first formula is probably more useful.

8. 1,562,500. Use the formula above or look at the WA continuation list and count to the 9th term.

9. This is a geometric series because it is the sum of an ordered list of numbers with a common ratio of 5.

10. 1,953,124

11. 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55.

12. A sequence of numbers in which each number equals the sum of the two preceding numbers.

13. Yes.

14. 20

15. 733. Answers will vary.

Additional Topics: Complex Numbers (Solutions) 16. ‐1

17. ‐i

18. 1

19. i. The pattern is i, ‐1, ‐i, 1, and the pattern repeats.

Solutions 22 Intermediate Algebra Activities Using WA 20. ‐1

21. 4i

22.

23.

24.

25.

26.

Additional Topics: Conic Sections (Solutions) 27. WA used the formula for the distance between two points:

28. WA used the formula for the midpoint of a line segment:

29. Equation of the circle is . Area of the circle is . WA calls the circumference of the circle the “perimeter.” Same thing, right? The distance around the outside of the figure? The circumference of the circle is 12 .

30. Center of the circle is (‐5, 2). The radius is 4. Rearrange the order of the terms and use the technique of completing the square:

and from there you can see the center and radius of the circle.

31. The equation of the circle is . WA got the result by using the formula for the equation of a circle and substituting to find the radius squared, then you can write the equation of the circle because you are given the center:

. Therefore the equation of the circle must be .

32. This is a parabola that opens sideways, to the right. The vertex of the parabola is (1, ‐2). The focus is (2, ‐2). The directrix is the line x = 0.

33. This is a parabola that opens downward. The vertex of the parabola is (2, 4). The focus is (2, 13/4). The directrix is the line y = 19/4.

Solutions 23 Intermediate Algebra Activities Using WA 34. The graph does not look like an ellipse; it looks more like a circle. That is because the scale on the x‐axis is not the same as on the y‐axis. With the graphing calculator you can choose ZSquare from the Zoom window, as shown below, to better see the elliptical shape of this figure. The center of the ellipse is (0, 0). The foci are (‐9/4, 0) and (9/4, 0).

35. This is a hyperbola that opens sideways. The center of the hyperbola is (0, 0). The foci are and .

wolfram|alpha activity - topic, intermediate algebra activities -- solutions

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