Friday, May 24, 2019

Weird

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

(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.




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 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.





Comments and questions welcome.

More Algebra 2 problems.

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.



Comments and questions welcome.

More Algebra 2 problems.

Wednesday, May 22, 2019

Abundant

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

(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.




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.


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).



Comments and questions welcome.

More Algebra 2 problems.

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.



Comments and questions welcome.

More Algebra 2 problems.

Monday, May 20, 2019

(x, why?) Mini: Directrix

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

(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).



Comments and questions welcome.

More Algebra 2 problems.

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.



Comments and questions welcome.

More Algebra 2 problems.

Surface Area of a Cylinder

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

(C)Copyright 2019, C. Burke.

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




Come back often for more funny math and geeky comics.




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.