Friday, May 31, 2019

Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.
After a brief hiatus, the Algebra 2 Problems of the Day are back. Hopefully, daily.

More Algebra 2 problems.

January 2019, Part III

All Questions in Part I are worth 4 credits. Partial credit can be earned.


33.Solve the following system of equations algebraically for all values of a, b, and c.

a + 4b + 6c = 23 a + 2b + c = 2 6b + 2c = a + 14

Answer:
Substitution:
Rewrite the last equation as a = 6b + 2c - 14
Rewrite the first two equations:


6b + 2c - 14 + 4b + 6c = 23
6b + 2c - 14 + 2b + c = 2


10b + 8c - 14 = 23
8b + 3c - 14 = 2


(-3)(10b + 8c = 37)
(8)(8b + 3c = 16)


-30b - 24c = -111
64b + 24c = 128
34b = 17
b = .5


8(.5) + 3c - 14 = 2
4 + 3c - 14 = 2
3c = 12
c = 4


a + 2(.5) + (4) = 2
a + 1 + 4 = 2
a = -3

a = -3, b = 0.5, c = 4

Elimination:
Rewrite the last equation as -a + 6b + 2c = 14


a + 4b + 6c = 23
a + 2b + c = 2
2b + 5c = 21


a + 4b + 6c = 23
-a + 6b + 2c = 14
10b + 8c = 37


2b + 5c = 21
10b + 8c = 37


(5)(2b + 5c = 21)
10b + 8c = 37


10b + 25c = 105
10b + 8c = 37
17c = 68
c = 4


2b + 5(4) = 21
2b + 20 = 21
2b = 1
b = 0.5


a + 2(.5) + (4) = 2
a + 1 + 4 = 2
a = -3

a = -3, b = 0.5, c = 4

Checking the work:
a + 4b + 6c = -3 + 4(.5) + 6(4) = -3 + 2 + 24 = 23 (check)
a + 2b + c = 2 = -3 + 2(.5) + 4 = -3 + 1 + 4 = 2 (check)
6b + 2c = a + 14
6(.5) + 2(4) = (-3) + 14
3 + 8 = 11 (check)





34. Given a(x) = x4 + 2x3 + 4x - 10 and b(x) = x + 2, determine a(x)/b(x) in the form q(x) + r(x)/b(x).
Is b(x) a factor of a(x)? Explain

Answer:
Divide the polynomial and leave the remainder as a fraction over (x + 2).
I used the Reverse Area Model, which is something I've only recently started doing.
Normally, I would only draw one table, but I expanded it here for clarity.
Some students who understood preferred it to long division. Others prefer to stick with long division.

Start by filling in x4 and -10. Label the rows x and +2
To get x4, you have to multiply x by x3, so put that on top of that column.
Multiply +2 by x3 and get 2x3.
The next term in a(x) is 2x3, and 2x3 - 2x3 = 0, so right 0 in the top row, second column. 0 divided by x is 0, so 0 gets written on top. And 0 times +2 is 0, so 0 goes on the bottom.
The next term is 4x, and 4x - 0 = 4x, so write 4x in the next column. 4x / x = 4, so write 4 on top. Then 4 times 2 = 8, so put 8 on the bottom.
We didn't want 8. We needed -10. That means that there is a remainder. Subtract -10 - 8 = -18. That remainder is put in a fraction over (x + 2).

b(x) is NOT a factor of a(x) because there is a remainder.
If something is a factor then there can be no remainder.

If people want to see the long division version of this, I can write it on scratch paper and scan it in.



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Soak

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(C)Copyright 2019, C. Burke.

It's been raining a lot around here lately.




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Thursday, May 30, 2019

Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.
After a brief hiatus, the Algebra 2 Problems of the Day are back. Hopefully, daily.

More Algebra 2 problems.

January 2019, Part II

All Questions in Part I are worth 2 credits. Partial credit can be earned.


31. Point M (t, 4/7) is located in the second quadrant on the unit circle. Determine the exact value of t.

Answer:
Because point M is in Quadrant II, t must be negative.
Because point M is on the unit circle, t2 + (4/7)2 = 12

t2 + (4/7)2 = 1
t2 + 16/49 = 1
t2 = 33/49
t = + SQRT(33/49) = + SQRT(33)/7

You need to specify the exact value of t, so do NOT estimate or round the radical.
Also, it has to be negative so t = - SQRT(31)/7





32. On the grid below, graph the function y = log2(x - 3) + 1

Answer:
Put the equation in the calculator and look at the table of values.
The asymptote is x = 3.
Plot the values (4, 1), (5, 2), (7, 3), and (11, 4). And then draw the curve.





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Wednesday, May 29, 2019

(x, why?) Mini: Ribose

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(C)Copyright 2019, C. Burke.

I'm glad Terbium exists, because there isn't a T to match with Boron.

Not a chemist, so I don't care if the elements won't work that way, or what else the Tb is attached to!
You are welcome to comment, but I probably will not respond.




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Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.
After a brief hiatus, the Algebra 2 Problems of the Day are back. Hopefully, daily.

More Algebra 2 problems.

January 2019, Part II

All Questions in Part I are worth 2 credits. Partial credit can be earned.


29. Rowan is training to run in a race. He runs 15 miles in the first week, and each week following, he runs 3% more than the week before. Using a geometric series formula, find the total number of miles Rowan runs over the first ten weeks of training, rounded to the nearest thousandth.

Answer:
In the back of the booklet, you are given the formula for Geometric Series:

Sn = (a1 - a1rn) / (1 - r)

Substitute n = 10 and r = 1.03 (103%)
Sn = (15 - 15(1.03)10) / (1 - 1.03)
= 171.958





30. The average monthly high temperature in Buffalo, in degrees Fahrenheit, can be modeled by the function

B(t) = 25.29 sin(0.4895t - 1.9752) + 55.2877,

where t is the month number (January = 1). State, to the nearest tenth, the average monthly rate of temperature change between August and November.

Explain its meaning in the given context.

Answer:
If January = 1, then August = 8 and November = 11.
To find the average rate of change, find B(11) - B(8) and divide it by (11 - 8).

B(11) = 25.29*sin(0.4895(11) - 1.9752) + 55.2877 = 48.59796...
B(8) = 25.29*sin(0.4895(8) - 1.9752) + 55.2877 = 78.86622...
B(11) - B(8) = -30.268...
-30.268 / 3 = -10.089 = -10.1

This means that from August to November that the temperature in Buffalo drops an average of 10.1 degrees Fahrenheit per month.



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Tuesday, May 28, 2019

Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.
After a brief hiatus, the Algebra 2 Problems of the Day are back. Hopefully, daily.

More Algebra 2 problems.

January 2019, Part II

All Questions in Part I are worth 2 credits. Partial credit can be earned.


27. Erin and Christa were working on cubing binomials for math homework. Erin believed they could save time with a shortcut. She wrote down the rule below for Christa to follow.

(a + b)3 = a3 + b3

Does Erin’s shortcut always work? Justify your result algebraically.

Answer:
Erin's shortcut does not work.
Substituting non-zero numbers will show this. However, that method will not receive full credit because the question stated "algebraically".

(a + b)3
= (a + b)(a + b)(a + b)
= (a + b)(a2 + ab + ab + b2)
= (a + b)(a2 + 2ab + b2)
= (a3 + 2a2b + ab2 + ba2b + 2ab2 + b3)
= (a3 + 3a2b + ab2 + 3ab2 + b3)
=/= a3 + b3





28. The probability that a resident of a housing community opposes spending money for community improvement on plumbing issues is 0.8. The probability that a resident favors spending money on improving walkways given that the resident opposes spending money on plumbing issues is 0.85. Determine the probability that a randomly selected resident opposes spending money on plumbing issues and favors spending money on walkways.

Answer:
Multiply the probabilities: 0.80 * 0.85 = 0.68





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Monday, May 27, 2019

School Life #9: Memorial Day

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(C)Copyright 2019, C. Burke.

Happy Memorial Day!

Look at Shaun! Talking to girls now.
I was going to have one of the girls comment, "Isn't that an English teacher over there?" as a call back to Judy and Chuck on the beach back in this old comic, #321, which fell on the Fourth of July.




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Friday, May 24, 2019

Weird

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(C)Copyright 2019, C. Burke.

So, yes, you're weird!

Weird numbers are a subset of Abundant numbers.
In brief:
A perfect number is one where the sum of the number's factors, excluding the number itself, equal the number. Ex: 1+2+3 = 6.
An abundant number is one where the sum of the number's factors, excluding the number itself, is greater than the number. Ex: 1+2+3+4+6=16 > 12.
A semiperfect number is one where a subset of the number's factors have a sum equal to the number. Ex: 1+2+3+6 = 12.
A perfect number is also considered to be semiperfect, unlike my wife who is perfect and I would never consider to be semiperfect.
A weird number is abundant but not semiperfect: there is no subset of factors that add up to the number.
Ex: no combination of 1, 2, 5, 7, 10, 14, and 35 add up to 70, but the sum of the factors is 74.

I was familiar with semiperfect, but not the "weird" term until I was looking up what the prefixes for "abundant" numbers were.




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Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.
After a brief hiatus, the Algebra 2 Problems of the Day are back. Hopefully, daily.

More Algebra 2 problems.

January 2019, Part II

All Questions in Part I are worth 2 credits. Partial credit can be earned.


25. Justify why (x2y5)(1/3) / (x3y4)(1/4) is equivalent to x(-1/12)y(2/3) using properties of rational exponents, where x =/= 0 and y =/= 0.


Answer:
As noted above in the way I had to type out the question, the cube root is the same as (1/3) power, and the fourth root is the same as (1/4) power.
Multiply the exponents:

(x(2/3)y(5/3)) / (x(3/4)y(4/4))
Next subtract the exponents of the two x terms and the two y terms
x(2/3) - (3/4) y(5/3) - 1)
x(8/12) - (9/12) y(5/3) - (3/3)
x(-1/12) y(2/3)





26. The zeros of a quartic polynomial function are 2, -2, 4, and -4. Use the zeros to construct a possible sketch of the function, on the set of axes below.

Answer:
You needed to sketch something like what's below, with the end both pointing up or down. There should be three turning points (in this case 2 minimums and 1 maximum). Label the x-axis so that it's obvious that the zeroes are -4, -2, 2 and 4. The y-intercept isn't important but it should be a turning point.





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Thursday, May 23, 2019

Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.
After a brief hiatus, the Algebra 2 Problems of the Day are back. Hopefully, daily.

More Algebra 2 problems.

January 2019, Part I

All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.


22. Consider f(x) = 4x2 + 6x - 3, and p(x) defined by the graph below

The difference between the values of the maximum of p and minimum of f is

(1) 0.25
(2) 1.25
(3) 3.25
(4) 10.25

Answer: (4) 10.25
The minimum point for f(x) occurs on the axis of symmetry, which is x = -b/(2a)
x = -(6)/((2)(4)) = -6/8 = -.75
The maximum of f(x) is f(-.75) = 4(-.75)^2 + 6(-.75) - 3 = -5.25
The maximum of p(x) is 5.
The difference is 5 - 5.25 = 10.25

If you had graph f(x), the minimum point wouldn't be in the table of values, but you could use the functions to find it. However, one you see that the minimum is below zero, there is only one possible answer because the other three are too small.





23. The scores on a mathematics college-entry exam are normally distributed with a mean of 68 and standard deviation 7.2. Students scoring higher than one standard deviation above the mean will not be enrolled in the mathematics tutoring program. How many of the 750 incoming students can be expected to be enrolled in the tutoring program?

(1) 631
(2) 512
(3) 238
(4) 119

Answer: (1) 631
68.27% percent of the data is within one standard deviation of the mean. That means 34.135% score within one standard deviation above as well as below. Since 50 + 34 = 84, 84 percent of the incoming students will be enrolled in the tutoring program.
.84135 * 750 = 631.
Depending upon the number of decimals you used, you may have gotten 630, which makes (1) the best choice.





24. How many solutions exist for 1 / (1 - x2) = -|3x - 2| + 5?

(1) 1
(2) 2
(3) 3
(4) 4

Answer: (4) 4
Fastest solution is to graph both the left and right side of the equation and check for the number of intersections.

End of Part I.



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Wednesday, May 22, 2019

Abundant

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(C)Copyright 2019, C. Burke.

Good answer! Good answer! Incorrect, but good answer!

Unfortunately, 12 is not only abundant but superabundant. You could even go as far as to say colossally abundant, but I wouldn't go and say that, if I were you.




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Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.
After a brief hiatus, the Algebra 2 Problems of the Day are back. Hopefully, daily.

More Algebra 2 problems.

January 2019, Part I

All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.


19. Which graph represents a polynomial function that contains x2 + 2x + 1 as a factor?


Answer: (1) see graph
Factor x2 + 2x + 1 into (x + 1)2.
This means that -1 is a both a zero on the graph and a turning point, which is shown in Choice (1).





20. Sodium iodide-131, used to treat certain medical conditions, has a half-life of 1.8 hours. The data table below shows the amount of sodium iodide-131, rounded to the nearest thousandth, as the dose fades over time.

Number of Half Lives12345
Amount of Sodium Iodide-131139.0069.50034.75017.3758.688

What approximate amount of sodium iodide-131 will remain in the body after 18 hours?

(1) 0.001
(2) 0.136
(3) 0.271
(4) 0.543

Answer: (3) 0.271
Note that the problem says that a half-life is 1.8 hours and that the top row of the table is the number of half-lives, not the number of hours.
There are 10 half-lives in 18 hours, because 18 / 1.8 = 10.
If you continue the table by entering 8.688 into your calculator and dividing by 2 five more times, you will get 4.344, 2.172, 1.086, 0.543, 0.271.

Also, you could put y = 139(1/2)x into your graphing calculator and check the Table of Values. Note: if you do this, you want to look at x = 9. Otherwise, you have to play with your original equation -- using 139 * 3 = 278 as the initial amount, or using (x - 1) as the exponent, with parentheses.





21. Which expression(s) are equivalent to (x2 - 4x) / 2x, where x =/= 0?

I. x / 2 - 2
II. (x - 4) / 2
III. (x - 1) / 2 - 3 / 2


(1) II, only
(2) I and II
(3) II and III
(4) I, II, and III

Answer: (4) I, II, and III
Test-taking tip: All four choices include choice II, so there is no need to check it. It's correct.
That being said, you can check to see if I or III are equivalent to II without using the original expression.

If you split the fraction (x - 4) / 2 = x / 2 - 4 / 2 = x / 2 - 2, which is I, so we can eliminate choices (1) and (3).
If you look at choice III, you can combine the fractions because they have a common denominator of 2:
(x - 1) / 2 - 3 / 2 = (x - 1 - 3) / 2 = (x - 4) / 2, which is II. II and III are the same.
So the answer is choice (4).



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Tuesday, May 21, 2019

Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.
After a brief hiatus, the Algebra 2 Problems of the Day are back. Hopefully, daily.

More Algebra 2 problems.

January 2019, Part I

All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.


16. Savannah just got contact lenses. Her doctor said she can wear them 2 hours the first day, and can then increase the length of time by 30 minutes each day. If this pattern continues, which formula would not be appropriate to determine the length of time, in either minutes or hours, she could wear her contact lenses on the nth day?
(1) a1 = 120; an = an - 1 + 30
(2) an = 90 + 30n
(3) a1 = 120; an = an - 1 + 0.5
(4) an = 2.5 + 0.5n

Answer: (4) an = 2.5 + 0.5n
On the first day, a1 must be 2 hours or 120 minutes.
In choice (4), 2.5 + 0.5(1) = 3 hours. The initial amount was already too big before anything was added to it.
Choice (1) starts with 120 minutes. Then adds 30 minutes to the previous day's total.
Choice (2) starts with 90 minutes plus an increment of 30 for a total of 120 on the first day
Choice (3) is the same as (1) except converted to hours.





17. If f(x) = ax, where a > 1, then the inverse of the function is

(1) f-1(x) = logx a
(2) f-1(x) = a log x
(3) f-1(x) = loga x
(4) f-1(x) = x log a

Answer: (3) f-1(x) = loga x
In ax, a is the base. When you take the log, a is the base.
Ex: If a=2, the f(3) = 23 = 8
f-1 ( f(3) ) = f-1(8) = log 2 8 = 3





18. Kelly-Ann has $20,000 to invest. She puts half of the money into an account that grows at an annual rate of 0.9% compounded monthly. At the same time, she puts the other half of the money into an account that grows continuously at an annual rate of 0.8%.
Which function represents the value of Kelly-Ann’s investments after t years?

(1) f(t) = 10,000(1.9)t + 10,000e0.8t
(2) f(t) = 10,000(1.009)t + 10,000e0.008t
(3) f(t) = 10,000(1.075)12t + 10,000e0.8t
(4) f(t) = 10,000(1.00075)12t + 10,000e0.008t

Answer: (4) f(t) = 10,000(1.00075)12t + 10,000e0.008t
This is almost purely a notation problem.
The choices with .9 and .8 should be eliminated immediately.
Similarly, 0.75 is .9 / 12, so that's eliminated.
Because the first account is compounded monthly and there are 12 months in a year, the 0.009 is divided by 12, giving .00075, and the exponent gets multiplied by 12.



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Monday, May 20, 2019

(x, why?) Mini: Directrix

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(C)Copyright 2019, C. Burke.

Something ... something ... follow directions ... something.

Sometimes the jokes write themselves. Sometimes they don't.




Come back often for more funny math and geeky comics.




Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.
After a brief hiatus, the Algebra 2 Problems of the Day are back. Hopefully, daily.

More Algebra 2 problems.

January 2019, Part I

All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.


13. The function f(x) = a cos bx + c is plotted on the graph shown below.


What are the values of a, b, and c?
(1) a = 2, b = 6, c = 3
(2) a = 2, b = 3, c = 1
(3) a = 4, b = 6, c = 5
(4) a = 4, b = π / 3, c = 3

Answer: (1) a = 2, b = 6, c = 3
Because c is not in parentheses, it is a vertical shift, not horizontal, and it will be equal to the midline of the function, which is 3 (halfway between 1 and 5). This eliminates choices (2) and (3).
Next, a is the amplitude it is equal to the distance from the midline to the maximum point (or one-half the distance from the maximum to the minimum). So a = 2, not 4. Eliminate choice (4).
As for b, it tells us that there will be 6 cycles in a 2π interval, and 2π / 6 = π / 3, which is the period for one cycle.





14. Which equation represents the equation of the parabola with focus (-3,3) and directrix y = 7?

(1) y = 1/8 (x + 3)2 - 5
(2) y = 1/8 (x - 3)2 + 5
(3) y = -1/8 (x + 3)2 + 5
(4) y = -1/8 (x - 3)2 + 5

Answer: (3) y = -1/8 (x + 3)2 + 5
The standard form of a parabola is y = a(x − h)2 + k, with the vertex at (h, k) and the focus at (h, k + 1 / (4a)).
Since the vertex is equidistant between the focus and the directrix, then vertex must be at (-3, 5) because 5 is halfway between 3 and 7. (3 + 7) / 2 = 5.
Flipping the sign, we know that (x + 3) must be in the equation, so eliminate (2) and (4).
Also, the equation ends with + 5, so we can eliminate choice (1) as well.

At this point, we know the answer is (3). We have one more piece of information: because the directrix is above the vertex, which is above the focus, we know that the parabola opens downward, so a must be negative:

1 / (4a) = -2
1 / ((4)(-2)) = a
1 / (-8) = a





15. What is the solution set of the equation

2 / (3x + 1) = 1 / x - 6x / (3x + 1)


(1) { -1/3, 1/2 }
(2) { -1/3 }
(3) { 1/2}
(4) { 1/3, -2 }

Answer: (3) { 1/2}
Strategy: find a common denominator by multiplying each fraction by either (x / x) or (3x + 1)/(3x + 1).
Then eliminate all the denominators by multiplying both sides of the equation by (x)(3x + 1).
However, we will have to check for extraneous solutions. Because of the denominators in the original equations, we know that x =/= 0 and 3x + 1 =/= 0.

2 / (3x + 1) = 1 / x - 6x / (3x + 1)
2x / ((3x + 1)(x)) = 1 / x - 6x(x) / ((3x + 1)(x))
2x / ((3x + 1)(x)) = (3x + 1) / ((3x + 1)(x)) - 6x(x) / ((3x + 1)(x))
2x = 3x + 1 - 6x2
6x2 - x - 1 = 0
(3x + 1)(2x - 1) = 0
3x + 1 = 0 or 2x - 1 = 0
3x = -1 or 2x = 1
x = -1/3 or x = 1/2

However, 3x + 1 =/= 0, and 3(-1/3) + 1 = 0, so we eliminate that answer. That leaves only 1/2, which is Choice (3).



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Friday, May 17, 2019

Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.
After a brief hiatus, the Algebra 2 Problems of the Day are back. Hopefully, daily.

More Algebra 2 problems.

January 2019, Part I

All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.


10. A random sample of 100 people that would best estimate the computations. proportion of all registered voters in a district who support improvements to the high school football field should be drawn from registered voters in the district at a

(1) football game
(2) supermarket
(3) school fund-raiser
(4) high school band concert

Answer: (2) supermarket
At the supermarket, you will encounter a more random set of people.
If you want to improve a high school football field, then football games and school functions will give you biased results.





11. Which expression is equivalent to (2x - i)2 - (2x - i)(2x + 3i), where i is the imaginary unit and x is a real number?

(1) -4 - 8xi
(2) -4 - 4xi
(3) 2
(4) 8x - 4i

Answer: (1) -4 - 8xi
First method: multiply and combine like terms:

(2x - i)2 - (2x - i)(2x + 3i)
4x2 - 4xi + i2 - ( 4x2 + 6xi - 2xi - 3i2)
4x2 - 4xi + i2 - 4x2 - 4xi + 3i2)
4x2 - 4xi - 1 - 4x2 - 4xi - 3
-8xi - 4
-4 - 8xi

Second method: Factor out (2x - i) from each term first.

(2x - i)2 - (2x - i)(2x + 3i)
(2x - i) ( ( 2x - i) - (2x + 3i) )
(2x - i) ( 2x - i - 2x - 3i )
(2x - i) (-4i)
-8xi + 4i2
-8xi - 4
-4 - 8xi





12. Suppose events A and B are independent and P(A and B) is 0.2. Which statement could be true?

(1) P(A) = 0.4, P(B) = 0.3, P(A or B) = 0.5
(2) P(A) = 0.8, P(B) = 0.25
(3) P(A|B) = 0.2, P(B) = 0.2
(4) P(A) = 0.15, P(B) = 0.05

Answer: (2) P(A) = 0.8, P(B) = 0.25
If A and B are independent then P(A and B) = P(A) * P(B)
Choice (1): P(A) * P(B) = 0.4 * 0.3 = 0.12 =/= 0.2.
Choice (2): P(A) * P(B) = 0.8 * 0.25 = 0.2.
Choice (4): P(A) * P(B) = 0.15 * 0.05 = 0.0075 =/= 0.2
In Choice (3), if A and B are independent, then P(A|B) = P(A). So 0.2 * 0.2 = 0.04 =/= 0.2.



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Surface Area of a Cylinder

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(C)Copyright 2019, C. Burke.

Basically, to turn 4 circles into a cylinder, you have to square the circle. Or maybe rectangle two circles.




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Thursday, May 16, 2019

Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.
After a brief hiatus, the Algebra 2 Problems of the Day are back. Hopefully, daily.

More Algebra 2 problems.

January 2019, Part I

All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.


7. Tides are a periodic rise and fall of ocean water. On a typical day at a seaport, to predict the time of the next high tide, the most important value to have would be the

(1) time between consecutive low tides
(2) time when the tide height is 20 feet
(3) average depth of water over a 24-hour period
(4) difference between the water heights at low and high tide

Answer: (1) time between consecutive low tides
If the tides are periodic, then knowing the consecutive low tides will tell you when the next high tide is because it will be at the midpoint of those two times.
The water height by itself will not let you predict the next high tide.





8. An estimate of the number of milligrams of a medication in the bloodstream t hours after 400 mg has been taken can be modeled by the function below.

I(t) = 0.5t4 + 3.45t3 - 96.65t2 + 347.7t, where 0 ≤ t ≤ 6
Over what time interval does the amount of medication in the bloodstream strictly increase?

(1) 0 to 2 hours
(2) 0 to 3 hours
(3) 2 to 6 hours
(4) 3 to 6 hours

Answer: (1) 0 to 2 hours
If you graph the function, you will see that it it's rising from 0 to 2. It reaches a local maximum point at approximately x = 2.15. The graph decreases between x = 2.15 and x = 6.





9. Which representation of a quadratic has imaginary roots?


Answer: (4) 2x2 + 32 = 0
If the quadratic intersects or crosses the x-axis then it does not have imaginary roots.
In other words, if there is some value of x which makes y equal to 0, it has a real root.
Choice (1) has (-2.0, 0) and Choice (3) has (3, 0), so they both can be eliminated.

The equations in (2) and (4) have imaginary roots, if there are no real solutions that make the equation true.
If 2(x + 3)2 = 64
then (x + 3)2 = 32
and x + 3 = SQRT(32)
So x = -3 + SQRT(32), which is a real, irrational value.

If 2x2 + 32 = 0
then 2x2 = -32
and x2 = -16.
This means x = SQRT(-16), which is not a real number.



Comments and questions welcome.

More Algebra 2 problems.

Wednesday, May 15, 2019

Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.
After a brief hiatus, the Algebra 2 Problems of the Day are back. Hopefully, daily.

More Algebra 2 problems.

January 2019, Part I

All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.


4. When a ball bounces, the heights of consecutive bounces form a geometric sequence. The height of the first bounce is 121 centimeters and the height of the third bounce is 64 centimeters. To the nearest centimeter, what is the height of the fifth bounce?
(1) 25
(2) 34
(3) 36
(4) 42

Answer: (2) 34
We know a1 = 121 and a3 = 64.
The common ratio, r = 64 / a2 or a2 / 121
Then r2 = (64 / a2) (a2 / 121) = 64 / 121
And r = SQRT(64/121) = 8/11

To get from a3 to a5, you need to multiply by the common ratio two more times (or multiply by r2).
64 * (8/11)*(8/11) = 33.851... = 34





5. The solutions to the equation 5x2 - 2x + 13 = 9 are
(1) 1/5 + SQRT(21)/5
(2) 1/5 + SQRT(19)/5 i
(3) 1/5 + SQRT(66)/5 i
(4) 1/5 + SQRT(66)/5


Answer: (2) 1/5 + SQRT(19)/5 i
If 5x2 - 2x + 13 = 9
then 5x2 - 2x + 4 = 0
If you graph this, you will see that there are no real roots, and you can eliminate (1) and (4).
Calculate the discriminate, b2 - 4ac = (-2)2 - 4(5)(4) = 4 - 80 = -76
SQRT(-76) = SQRT(-1 * 4 * 19) = 2i * SQRT(19), which elminates choice (3).

x = (-b + SQRT (b2 - 4ac) ) / (2a)
x = ( -(-2) + SQRT (-76) ) / (2*5)
x = ( 2 + 2i SQRT (19)) / (10)
Split the fraction
x = 2/10 + 2i SQRT (19) / 10
x = 1/5 + SQRT(19)/5 i





6. Julia deposits $2000 into a savings account that earns 4% interest per year. The exponential function that models this savings account is y = 2000(1.04)t, where t is the time in years. Which equation correctly represents the amount of money in her savings account in terms of the monthly growth rate?
(1) y = 166.67(1.04)0.12t
(2) y = 2000(1.01)t
(3) y = 2000(1.0032737)12t
(4) y = 166.67(1.0032737)t

Answer: (3) y = 2000((1.0032737)12t
If you take the 12th root (1/12 power) of 1.04, you get 1.00327373978...
Conversely, if you raise 1.0032737 to the 12th power, you will get 1.039999... or 1.04.



Comments and questions welcome.

More Algebra 2 problems.

Tuesday, May 14, 2019

Algebra 2 Problems of the Day

Daily Algebra 2 questions and answers.
After a brief hiatus, the Algebra 2 Problems of the Day are back. Hopefully, daily.

More Algebra 2 problems.

January 2019, Part I

All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.


1. Suppose two sets of test scores have the same mean, but different standard deviations, σ1 and σ2, with σ2 > σ1. Which statement best describes the variability of these data sets?
(1) Data set one has the greater variability.
(2) Data set two has the greater variability.
(3) The variability will be the same for each data set.
(4) No conclusion can be made regarding the variability of either set.

Answer: (2) Data set two has the greater variability.
A lower standard deviation means that the data is closer to the mean, and a higher standard means that the data is more spread out.





2. If f(x) = log3 x and g(x) is the image of f(x) after a translation five units to the left, which equation represents g(x)?
(1) g(x) = log3(x + 5)
(2) g(x) = log3x + 5
(3) g(x) = log3(x - 5)
(4) g(x) = log3x - 5

Answer: (1) g(x) = log3(x + 5)
Adding 5 inside the parentheses shifts the function 5 units to the left. Subtracting 5 moves it to the right.
Adding 5 outside the parentheses shifts the function 5 units up. Subtracting 5 outside (or without) the parentheses moves it down.





3. When factoring to reveal the roots of the equation x3 + 2x2 - 9x - 18 = 0, which equations can be used?

I. x2(x + 2) - 9(x + 2) = 0
II. x(x2 - 9) + 2(x2 - 9) = 0
III. (x - 2)(x2 - 9) = 0

(1) I and II, only
(2) I and III, only
(3) II and III, only
(4) I, II, and III

Answer: (1) I and II, only
If you distribute I, you will get the original four terms.
Likewise, if you distribute II, you will get the four terms, which can be rearranged back into standard form.
If distribute, FOIL, box, or whatever III, you will get -2x2 and +18, which are incorrect.



Comments and questions welcome.

More Algebra 2 problems.

Sunday, May 12, 2019

Happy Mother's Day 2019

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(C)Copyright 2019, C. Burke.

Happy Mother's Day!




Come back often for more funny math and geeky comics.




Thursday, May 09, 2019

August 2018 Common Core Geometry Regents, Part I (multiple choice)

The following are some of the multiple questions from the recent August 2018 New York State Common Core Geometry Regents exam.
The answers to Part II can be found here
The answers to Parts III and IV can be found here

August 2018 Geometry, Part I

Each correct answer is worth up to 2 credits. No partial credit. Work need not be shown.


1. In the diagram below, AEFB || CGD, and GE and GF drawn.

If m∠EFG = 32° and m∠AEG 137°, what is m∠EGF?

Answer: (4) 105°
If ∠AEG = 137, then ∠FEG = 180 – 137 = 43.
The sum of the angles of triangle EFG is 180.
So 180 – (43 + 32) = 105. Angle EGF is 105°.

Alternatively, Angle AEG is an exterior angle to triangle EFG. The Remote Angle Theorem says that it must be the sum of the two remote angles EFG and and EGF.
So EGF = 137 – 32 = 105 degrees.


2. If triangle ABC is mapped onto triangle DEF after a line reflection and triangle DEF is mapped onto triangle XYZ after a translation, the relationship between triangle ABC and triangle XYZ is that they are always

Answer: (1) congruent and similar
Reflections and translations are rigid motions which preserve the shape of the object. Also, choice (2) congruent but not similar is not possible. If two triangles are congruent, they are automatically similar.


3. An isosceles right triangle whose legs measure 6 is continuously rotated about one of its legs to form a three-dimensional object. The three-dimensional object is a

Answer: (4) cone with a diameter of 12
When a right triangle is rotated about a leg, the resulting 3-D shape will be a cone. Because the base of the triangle is 6, the radius will be 6, so the diameter is twice that, or 12.


4. In regular hexagon ABCDEF shown below, AD, BE, and CF all intersect at G.

When triangle ABG is reflected over BG and then rotated 180° about point G, triangle ABG is mapped onto

Answer: (1) Triangle FEG
When it is reflected, it is mapped onto triangle BGC, the top center of the hexagon. When it is rotated 180 degrees, it maps onto FEG, the bottom center of the hexagon.


5. A right cylinder is cut perpendicular to its base. The shape of the cross section is a

Answer: (3) rectangle
A horizontal (parallel to base) cut would give a circle. A vertical (perpendicular to base) cut gives a rectangle. If you look at a can on a shelf, it appears to be a rectangle.


6. Yolanda is making a springboard to use for gymnastics. She has 8-inch-tall springs and wants to form a 16.5° angle with the base, as modeled in the diagram below.

To the nearest tenth of an inch, what will be the length of the springboard, x?

Answer: (4) 28.2
Opposite and hypotenuse mean you need to use sine.
sin 16.5 = 8 / x
x = 8 / sin 16.5 = 28.167…


7. In the diagram below of right triangle ABC, altitude BD is drawn to hypotenuse AC.

If BD = 4, AD = x – 6, and CD = x what is the length of CD?

Answer: (3) 8
According to the Right Triangle Altitude Theorem, (BD)2 = (AD)(CD)
So 42 = (x – 6)(x)
16 = x2 - 6x. At this point, you could substitute the choices, or solve.
x2 - 6x – 16 = 0
(x – 8)(x + 2) = 0
x = 8 or x = -2. Discard the negative length.


8. Rhombus STAR has vertices S(–1,2), T(2,3), A(3,0), and R(0,–1). What is the perimeter of rhombus STAR?

Answer: (4) 4 * SQRT(10)
Each side of the rhombus has the same length, so calculate the length of one side using the distance formula, and multiply the result by 4.
SQRT( (2 - -1)2 + (3-2)2 ) = SQRT( (3)2 + (1)2 )
= SQRT(9 + 1) = SQRT(10) per side.
Perimeter is 4 * SQRT(10).


9. In the diagram below of triangle HAR and triangle NTY, angles H and N are right angles, and HAR ~ NTY.

If AR = 13 and HR = 12, what is the measure of angle Y, to the nearest degree?

Answer: (1) 23°
Angle Y corresponds to angle R. AR is the hypotenuse, and HR is adjacent to R. This means use cosine to find the size of angle R, which will also be the size of angle Y.
cos R = 12/13
r = cos-1(12/13) = 22.61 = 23 degrees. Note that choice (4) is the angle if you either used sine, or if you solved for angle A. You would get choices (2) and (3) if you used tangent, and if you used tangent and the wrong angle.


10. In the diagram below, AKS, NKC, AN, and SC are drawn such that AN = SC.

Which additional statement is sufficient to prove triangle KAN = triangle KSC by AAS?

Answer: (4) AN || SC
If AN is parallel to SC, then you would get alternate interior angles, along with the vertical angles. That is enough to prove congruency by either ASA or AAS. Choice (1) would prove congruency by SSS, but wouldn’t give any more information about the angles. Choice (2) would yield SSA, which is not sufficient to prove congruency. Choice (3) would establish that the vertical angles are also right angles, but we already know that those two angles are congruent.


11. Which equation represents a line that is perpendicular to the line represented by y = (2/3)x + 1 ?

Answer: (1) 3x + 2y = 12
The negative reciprocal of 2/3 is -3/2, so choices (3) and (4) are both incorrect.
y = -(3/2)x + b
3/2x + y = b, multiply by 2 to get rid of the fraction
3x + 2y = 2b, which gives us the first equation, if 2b is replaced by 12.


12. In the diagram of ABC below, points D and E are on sides AB and CB respectively, such that DE || AC.

If EB is 3 more than DB, AB = 14, and CD = 21, what is the length of AD?

Answer: (2) 8
Be careful because you need to find the length of DB first, but they are asking for the length of AD, so you have to subtract DB from AB.
The sides are proportional, so you can set up an equation:
x / 14 = (x + 3) / 21. You can substitute the choices from this point if you want.
21x = 14x + 42
7x = 42
x = 6
AD = 14 – 6 = 8.
Notice that the four choices represent the lengths of DB, AD, BE, and CE. You might have reasoned this one out from the diagram and then checked your work to see if you were correct.


13. Quadrilateral MATH has both pairs of opposite sides congruent and parallel. Which statement about quadrilateral MATH is always true?

Answer: (4) ∠MAT = ∠MHT
Check Part IV of this exam for a coordinate geometry problem involving a quadrilateral with vertices M, A, T, and H.
Opposite sides congruent and parallel mean that this is a parallelogram. That means that the opposite angles are also congruent.
Choice (1) says that the diagonals must be congruent, which is only true in rectangles. (You could use this in Part IV)
Choice (2) says that the consecutive angles are right angles, which is only true in rectangles. (Again, Part IV)
Choice (3) would occur in rectangles as well because the two diagonals would create four isosceles triangles.


14. In the figure shown below, quadrilateral TAEO is circumscribed around circle D. The midpoint of TA is R, and HO = PE .

If AP = 10 and EO = 12, what is the perimeter of quadrilateral TAEO?

Answer: (2) 64

AP = 10 so AR =10. R is the midpoint of AT, so RT = 10, which means TH = 10.
HO = PE means that OZ = ZE, and since OE = 12, then HO = PE = OZ = PE = 6.
4 * 10 + 4 * 6 = 64


15. The coordinates of the endpoints of directed line segment ABC are A(-8,7) and C(7,-13). If AB:BC = 3:2, the coordinates of B are

Answer: (1) (1, -5)
B is 3/5 of the distance from A to C. The x-coordinates of A and C are 7 – (-8) = 15 units apart.
3/5(15) = 9, and -8 + 9 = 1, so the x-coordinate of B is 1, which is choice (1).
To check, -13 – 7 = -20, and (3/5)(-20) = -12, and 7 – 12 = -5, which is the y-coordinate of B.


16. In triangle ABC, points D and E are on sides AB and BC, respectively, such that DE || AC, and AD:DB 3:5.

If DB = 6.3 and AC = 9.4, what is the length of DE, to the nearest tenth?

Answer: (3) 5.9

If AD:DB = 3:5, then AB:DB = 8:5, which is the ratio of the larger triangle to the smaller triangle.
So 8/5 = 9.4/x
8x = (5)(9.4)
x = (5)(9.4)/8 = 5.875.


17. In the diagram below, rectangle ABCD has vertices whose coordinates are A(7,1), B(9,3), C(3,9), and D(1,7).

Which transformation will not carry the rectangle onto itself?

Answer: (3) a rotation of 180° about the point (6,6)
The rectangle will carry onto itself in a rotation about the rectangle's center. (5, 5) is the center of the rectangle. (6, 6) is not. If you rotate about (6, 6), the image will be adjacent to the original rectangle.


18. A circle with a diameter of 10 cm and a central angle of 30° is drawn below.


What is the area, to the nearest tenth of a square centimeter, of the sector formed by the 30° angle?

Answer: (2) 6.5
The radius is 5 cm. The central angle is 30 degrees, which is 1/12 of the total circle. (30/360 = 1/12)
V = (1/12)(pi)(5)2 = 6.54..


19. A child’s tent can be modeled as a pyramid with a square base whose sides measure 60 inches and whose height measures 84 inches. What is the volume of the tent, to the nearest cubic foot?

Answer: (2) 58
60 inches = 5 feet, 84 inches = 7 feet
Volume = (1/3) (Area of the base) (height)
V = (1/3) (5) (5) (7) = 58.333...


20. In the accompanying diagram of right triangle ABC, altitude BD is drawn to hypotenuse AC.

Which statement must always be true?

Answer: (2) AD/AB = AB/AC
Short leg is to hypotenuse as short leg is to hypotenuse.
The other choices do not use corresponding sides in the correct order.


21. An equation of circle O is x2 + y2 + 4x - 8y = -16. The statement that best describes circle O is the

Answer: (4) center is (-2,4) and is tangent to the y-axis
Rewrite the equation in standard form by grouping the variables and completing the square.
x2 + y2 + 4x - 8y = -16
x2 + 4x + y2 - 8y = -16
x2 + 4x + 4 + y2 - 8y + 16 = -16 + 4 + 16
(x + 2)2 + (y - 4) = 4 = 22
The center of the circle is (-2, 4) and the radius is 2, which would make it tangent to the y-axis.


22. In triangle ABC, BD is the perpendicular bisector of ADC. Based upon this information, which statements below can be proven?


I. is a median.
II. bisects ∠ABC.
III. ABC is isosceles.

Answer: (4) I, II, and III
If BD is a bisector of ADC, it's also a median.
If BD is a perpendicular bisector, than AB and AC must be congruent. (This comes in handy to know in Part II!)
IF AB = AC, the triangle is isosceles. In an isosceles triangle, the median bisects the vertex angle.


23. Triangle RJM has an area of 6 and a perimeter of 12. If the triangle is dilated by a scale factor of 3 centered at the origin, what are the area and perimeter of its image, triangle R'J'M'?

Answer: (3) area of 54 and perimeter of 36
If the image is dilated by a scale factor of 3, then the perimeter is increased by a factor of 3, but the area is increased by a factor of 32 = 9.
12 * 3 = 36, and 6 * 9 = 54.


24. If sin (2x + 7)° = cos (4x - 7)°, what is the value of x?

Answer: (2) 15
The Sine of any angle is equal to the Cosine of the complementary angle. And the sum of the two complementary angles is 90 degrees
2x + 7 + 4x - 7 = 90
6x = 90
x = 15

End of Part I

How did you do?

Questions, comments and corrections welcome.

Wednesday, May 08, 2019

Rationalize the Denominator, Part 2

(Click on the comic if you can't see the full image.)

(C)Copyright 2019, C. Burke.

At least he gave them a common denominator to start with.

Long-time readers will recognize the brown-haired girl as Bibi, from previous comics. I've been referring to the other two (in a non-canonically way) as Freedom and Serenity. Why I'm doing that is left as an exercise for the reader.




Come back often for more funny math and geeky comics.




Tuesday, May 07, 2019

Rationalize the Denominator

(Click on the comic if you can't see the full image.)

(C)Copyright 2019, C. Burke.

Who says the denominator has to be rational anyway? Other than the curriculum? And my old teachers? And the text books?




Come back often for more funny math and geeky comics.




August 2018 Common Core Geometry Regents, Parts 3 and 4 (open-ended)

The following are some of the multiple questions from the recent August 2018 New York State Geometry Regents exam.
The questions and answers to Part I can be found here.
The questions and answers to Part II can be found here.

August 2018 Geometry, Part III

Each correct answer is worth up to 4 credits. Partial credit is available. Work must be shown. Correct answers without work receive only 1 point.


32. Given: Triangle ABC, AEC, BDE with ∠ABE = ∠CBE, and ∠ADE = ∠CDE
Prove: BDE is the perpendicular bisector of AC.

Answer:
Reason (2): "The Reflexive Property" is the reason you can say a line segment is congruent to itself
Statement (4): "<BDA = <BDC" -- ADE is congruent to CDE and statement (3) establishes that the other two angles are supplementary.
Reason (6): "CPCTC" -- Corresponding Parts of Congruent Triangles are Congruent. (Definition of Congruent Polygons)
Reason (7): If two points are each equidistant from the endpoints of a segment, then those points determine the perpendicular bisector of the segment.

Explanation of reason (7): Point B must be on the perpendicular bisector of AC. Point D must be on the perpendicular bisector of AC. Since both B and D must be on that line, then line BD must be that line that bisects AC at right angles.

This is probably not the way I would have written this proof, but it's the format we're given, and we have to fill in the blanks.


33. A homeowner is building three steps leading to a deck, as modeled by the diagram below. All three step rises, HA, FG, and DE, are congruent, and all three step runs, HG, FE, and DC, are congruent. Each step rise is perpendicular to the step run it joins. The measure of ∠CAB = 36° and m∠CBA = 90°.


If each step run is parallel to AB and has a length of 10 inches, determine and state the length of each step rise, to the nearest tenth of an inch.

Determine and state the length of AC, to the nearest inch.

Answer:
Each step run is 10 inches and there are 3 of them, so AB is 30 inches.
You have given angle A which is adjacent to AB.
For the two parts of the problem, you need to find the opposite side (and divide it by 3) and the hypotenuse.
This means using tangent and cosine, or tangent and Pythagorean Theorem.

tan 36 = x / 30
x = 30 (tan 36) = 21.796...
21.796 / 3 = 7.265. Each rise is about 7.3 inches.

cos 36 = 30 / x
x = 30 / (cos 36) = 37.08..
AC is 37 inches.

Pythagorean Theorem
302 + 21.82 = AC2
900 + 475.24 = AC2
1375.24 = AC2
AC = 37.08..., or 37 inches.


34. A bakery sells hollow chocolate spheres. The larger diameter of each sphere is 4 cm. The thickness of the chocolate of each sphere is 0.5 cm. Determine and state, to the nearest tenth of a cubic centimeter, the amount of chocolate in each hollow sphere.
The bakery packages 8 of them into a box. If the density of the chocolate is 1.308 g/cm3, determine and state, to the nearest gram, the total mass of the chocolate in the box.

Answer:
The volume of the chocolate is the Volume of the outer sphere minus the Volume of the inner sphere.
The outer has a radius of 2. The inner has a radius of 2 - 0.5 = 1.5
Use the formula V = (4/3) π r3
V = (4/3) π (2)3 - (4/3) π (1.5)3
V = 33.51 - 14.14 = 19.37 = 19.4 cm3

V = m/D, so m = VD = (19.4)(1.308) = 25.3752 each.
A package of 8 would be 25.3752 * 8 = 203 grams.


Part IV

A correct answer is worth up to 6 credits. Partial credit is available.

35. The vertices of quadrilateral MATH have coordinates M(-4,2), A(-1,–3), T(9,3), and H(6,8).
Prove that quadrilateral MATH is a parallelogram.
[The use of the set of axes is optional.]
Prove that quadrilateral MATH is a rectangle.
[The use of the set of axes below is optional.]

Answer:
Note: In my opinion, this is a fairly simple 6-point question. I would've expected this to only be worth four points, or to have another part to the question.

To prove that MATH is a parallelogram, you only need to show that the opposite sides are parallel, which means that they will have the same slope.
To prove that it is a rectangle after you have already shown that it is a parallelogram, you only need to show that any two consecutive sides are perpendicular, which means the slopes are inverse reciprocals.

You don't need to use distance formula for this problem, but you can if you want to. You could show that the opposite sides are congruent. You could show that the diagonals are congruent.

Using the graph may speed up your calculations because you can just count boxes rather than do any subtraction.

Slope MA = (-3 - 2)/(-1 - -4) = -5/3
Slope AT = (3 - -3)/9 - -1) = 6/10 = 3/5
Slope TH = (8 - 3)/(6 - 9) = 5/-3 = -5/3
Slope HM = (2 - 8)/(-4 - 6) = -6/-10 = 3/5

MATH is a parallelogram because MA is parallel to TH and AT is parallel to HM because each pair of sides has the same slope.

MATH is a rectangle because (-5/3)(3/5) = -1, so sides MA and AT are perpendicular. A parallelogram with a right angle is a rectangle.

End of Exam

How did you do?
Comments and corrections welcome. (I get many of the latter!)

Monday, May 06, 2019

August 2018 Common Core Geometry Regents, Part II (open-ended)

The following are some of the multiple questions from the recent August 2018 New York State Common Core Geometry Regents exam.
The answers to Part I can be found here
The answers to Parts III and IV can be found here

August 2018 Geometry, Part II

Each correct answer is worth up to 2 credits. Partial credit can be given. Work must be shown or explained.


25. In the circle below, is a chord. Using a compass and straightedge, construct a diameter of the circle. [Leave all construction marks.]

Answer:
The diameter will be a perpendicular bisector of the chord, so you need to construct a perpendicular bisector of AB.
Open the compass more than half the width of AB. From point A, make small arcs on either side of the line. Do the same from point B without changing the width of the compass. Make points where the arcs intersect. Use the straightedge to draw the diameter. Make sure you draw the line long enough to intersect with the circle twice! You were asked for a diameter.




26. In parallelogram ABCD shown below, the bisectors of ∠ABC and ∠DCB meet at E, a point on AD.


If m∠A = 68°, determine and state m∠BEC.

Answer:
If m∠A = 68°, then m∠BCD = 68° and m∠ABC = (180 - 68)° = 112°.
Since BE and CE are bisectors, then m∠EBC = (112/2) = 56° and m∠BCE = (68/2) = 34°
We know 2 of the angles of triangle BEC, and the three of them add up to 180.
56 + 34 + m∠BEC = 180
m∠BEC = 90

A shorter, more generic version of this explanation:
Angles ABC and BCD are supplementary. They have a sum of 180 degrees.
Because BE and CE are bisectors, the sum of EBC and BCE must be half as much, which is 180/2 = 90 degrees.
Triangle BCE has a total of 180 degrees, so Angle BEC must be 180 - 90 = 90 degrees.
This will always be true regardless of the measure of angle A.


27. In circle A below, chord BC and diameter DAE intersect at F.


If mCD = 46° and mDB = 102°, what is m∠CFE?

Answer:
Note that DAE is a diameter. So the sum of arcs DB and BE is 180 degrees. That means that BE = 180 - 102 = 78 degrees.
You have a choice now, both of which use essentially the same steps, in a different order.
You can find the measure of angle BFE (and CFD), which is then supplementary to CFE, OR you can find the measure of arc CE and then find the measure of CFE.

Method 1: The measure of Angle BFE is the average of arcs CD and BE.
(46 + 78) / 2 = 124 / 2 = 62.
180 - 62 = 118.
Angle CFE is 118 degrees
Method 2: 46 + 102 + 78 + CE = 360
226 + CE = 360
CE = 134
(102 + 134) / 2 = 236 / 2 = 118
Angle CFE is 118 degrees.


28. Trapezoids ABCD and A"B"C"D" are graphed on the set of axes below.

Describe a sequence of transformations that maps trapezoid ABCD onto trapezoid A"B"C"D".

Answer:
There are multiple answers. Notice that points have rotated, so it is not a simple reflection.
Also notice that is not a simple rotation about the origin because AB is closer to the x-axis than A"B".

Some possible answers:
1: ABCD is rotated 180 degrees about the origin, followed by a translation T0, -2.

2: ABCD is reflected across y = -1, followed by a reflection over the y-axis. (This could be in either order.)

3: ABCD is reflected across the x-axis, followed by a translation 2 units down, followed by a reflection over the y-axis.

4: ABCD is rotated 180 degrees about the point (0, -1).


29. In the model below, a support wire for a telephone pole is attached to the pole and anchored to a stake in the ground 15 feet from the base of the telephone pole. Jamal places a 6-foot wooden pole under the support wire parallel to the telephone pole, such that one end of the pole is on the ground and the top of the pole is touching the support wire. He measures the distance between the bottom of the pole and the stake in the ground.


Jamal says he can approximate how high the support wire attaches to the telephone pole by using similar triangles. Explain why the triangles are similar.

Answer:
The triangles are similar because the share the angle between the wire and the ground, and they both have vertical sides which are perpendicular to the ground, creating right angles. But the AA Postulate, the triangles are similar.


30. Aliyah says that when the line 4x + 3y = 24 is dilated by a scale factor of 2 centered at the point (3,4), the equation of the dilated line is y = -(4/3)x + 16. Is Aliyah correct? Explain why. [The use of the set of axes below is optional.]

Answer:
When a line is dilated, one of two things will happen: either you will get a parallel line, or the line doesn't change because the center point is on the line itself.
First thing, check if (3, 4) is a solution to 4x + 3y = 24
4(3) + 3(4) = 24
12 + 12 = 24
24 = 24
The dilated line is the same as the original line.

For Aliyah to be correct, y = -(4/3)x + 16 must be the same line as 4x + 3y = 24.
4x + 3y = 24
3y = -4x + 24
y = -(4/3)x + 8.
Aliyah is not correct.


31. Ian needs to replace two concrete sections in his sidewalk, as modeled below. Each section is 36 inches by 36 inches and 4 inches deep. He can mix his own concrete for $3.25 per cubic foot.

How much money will it cost Ian to replace the two concrete sections?

Answer:
Convert the inches into feet by dividing by 12.
The Volume of concrete he needs is 2 * 3 * 3 * (1/3) = 6 cubic feet.
Multiply the number of cubic feet by the cost per cubic foot:
6 * 3.25 = $19.50.

End of Part II

How did you do?

Questions, comments and corrections welcome.