Even numbers have problems. Maybe she was misunderstood?
Pretty explicit, actually.
More Algebra 2 problems.
January 2020, Part II
All Questions in Part II are worth 2 credits. Work must be shown. Partial credit is given.
25. For n and p > 0, is the expression (p2n1/2)8SQRT(p5n4) equivalent to p18n6SQRT(p)? Justify your answer.
Answer:
Multiply the exponents in the parentheses of the first part of the first expression by 8. Multiply the exponents in the second part by 1/2. Then add the exponents of each matching variable.
(p16n4)(p5/2n2)
= p18 1/2n6
= p18n6SQRT(p)
Yes, they are equivalent.
Note: You must include the concluding statement to get full credit. However, the concluding statement, by itself, with no work, is not worth any points.
26. Show why x - 3 is a factor of m(x) = x3 - x2 - 5x - 3. Justify your answer.
Answer:
There are several methods.
You can show that x - 3 is a factor m(x) by calculating m(3), which will give you 0 if it is a factor.
m(3) = (3)3 - (3)2 - 5(3) - 3 = 27 - 9 - 15 - 3 = 0
Since 3 is a zero of the function, x - 3 is a factor.
You can also show this by factoring or dividing the function. You can use, for example, long division or the grid method.
Look at the image below.
Put x and -3 on the left and x3 in the first box.
Divide x into x3 and you get x2. Write that on top. Now multiply x2 by -3, which has a product of -3x3. Write that in the bottom of the first column.
We have -3x2 but the original expression has -x2, so we have to add 2x2 in the top box of the second column. Repeat the division and multiplication.
When you get to the last box, you have -3, which is what we need. There is no remainder.
(x - 3) is a factor of m(x) because it divides evenly with no remainder.
Comments and questions welcome.
More Algebra 2 problems.
What do Math teachers have? Well, for one, T-shirts like that one!
Guesses to what my actual shirt says can be left in the comments.
So, if I reteach the same material year after year, I'm allowed to re-use the same puns, right? It's all new to them.
Since I said that I would explain on my blog, I guess I need to do that here.
As we see in Panel 2, above, we are Given Secant AB with point D on the circle, and AC with point E on the circle.
We want to prove that the products of these lengths are equal: (AB)(AD) = (AC)(AE)
If we draw chords CD and BE, we create triangles ABE and ACD, as shown in Panel 3.
Angles B and C both intercept the same arc, DE, and therefore they are congruent.
Angle A is congruent to itself because of the Reflexive Property.
Therefore, triangles ABE and ACD are similar.
If they are similar, then their corresponding sides are proportional.
So AB / AE = AC / AD
If we cross-multiply, we get: (AB)(AD) = (AC)(AE)
Or, in other words, "The whole line times the other part equals the other whole line times its outside part".
Actually, not very long, and could easily be included in an actual remote video, but not so much in a four- or six-panel comic page.
More Algebra 2 problems.
January 2020, Part I
All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.
21. Which value, to the nearest tenth, is the smallest solution of f(x) = g(x) if f(x) = 3sin( 1/2 x) - 1 and g(x) = x3 - 2x + 1?
(1) -3.6
(2) -2.1
(3) -1.8
(4) 1.4
Answer: (2) -2.1
Put both equations in your graphing calculator. You should be able to see the results clearly enough to pick the correct choice. However, if you do not, you can use the Calculator functions. (2nd TRACE, option 5, on TI-83 and beyond)
22. Expressed in simplest a + bi form, (7 - 3i) + (x - 2i)2 - (4i + 2x2) is
(1) (3 - x2) - (4x + 7)i
(2) (3 + 3x2) - (4x + 7)i
(3) (3 - x2) - 7i
(4) (3 + 3x2) - 7i
Answer: (1) (3 - x2) - (4x + 7)i
23. Written in simplest form, the fraction (x3 - 9x)/(9 - x2), where x =/= 3, is equivalent to
<
(1) -x
(2) x
(3) -x(x + 3) / (3 + x)
(4) x(x - 3) / (3 - x)
Answer: (1) -x
You should immediately realize that neither (3) nor (4) are in simplest term. In (3), the two binomials are the same, which results in 1. In (4), (x - 3)/(3 - x) reduce to -1.
Solve for y: (4) + y + (11) = 9
y + 15 = 9
y = -6
24. According to a study, 45% of Americans have type O blood. If a random
number generator produces three-digit values from 000 to 999, which
values would represent those having type O blood?
(1) between 000 and 045, inclusive
(2) between 000 and 444, inclusive
(3) between 000 and 449, inclusive
(4) between 000 and 450, inclusive
Answer: (3) between 000 and 449, inclusive
45% of 1,000 numbers is 450
Including 000 means that the 450th number would be 449.
Comments and questions welcome.
More Algebra 2 problems.
Or you can fast-forward through the proof-y parts and just learn the rule and answer the questions.
More Algebra 2 problems.
January 2020, Part I
All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.
16. As θ increases from -π/2 to 0 radians, the value of cos θ will
(1) decrease from 1 to 0
(2) decrease from 0 to -1
(3) increase from -1 to 0
(4) increase from 0 to 1
Answer: (4) increase from 0 to 1
Cos(-π/2) = 0, Cos(0) = 1
17. Consider the following patterns:
I. 16, -12, 9, -6.75, …
II. 1, 4, 9, 16, …
III. 6, 18, 30, 42, …
IV. 1/2, 2/3, 3/4, 4/5, …
Which pattern is geometric?
(1) I
(2) II
(3) III
(4) IV
Answer: (1) I
The terms in pattern I have a common ratio: -12/16 = -3/4; 9/-12 = -3/4
Pattern II is quadratic. The terms are squares.
Pattern III is arithmetic. Each term is 12 more than the prior term.
Pattern IV is none of these. It has neither a common ratio or common difference.
18. Consider the system below.
x + y + z = 9
x - y - z = -1
x - y + z 21
Which value is not in the solution, (x,y,z), of the system?
(1) -8
(2) -6
(3) 11
(4) 4
Answer: (1) -8
Combine the first two equations:
Combine the second and third equations:
Solve for y: (4) + y + (11) = 9
y + 15 = 9
y = -6
19. Which statement regarding polynomials and their zeros is true?
(1) f(x) = (x2 - 1)(x + a) has zeros of 1 and -a, only
(2) f(x) = x3 - ax2 + 16x - 16a has zeroes of 4 and a, only.
(3) f(x) = (x2 + 25)(x + a) has zeros of +5 and -a.
(4) f(x) = x3 - ax2 - 9x + 9a has zeros at +3 and a.
Answer: (4) f(x) = x3 - ax2 - 9x + 9a has zeros at +3 and a.
Choice (1) has factors (x + 1)(x - 1) and a zero at -1.
In choice (2), f(4) = 64 - 16a + 64 - 16a = 128 - 32a, not zero.
In choice (3), x2 + 25 cannot be factored in (x + 5)(x - 5), so they are not zeros.
Choice (4) can be factored as follows:
f(x) = x3 - ax2 - 9x + 9a
f(x) = x2(x - a) - 9(x - a)
f(x) = (x2 - 9)(x - a)
f(x) = (x + 3)(x - 3)(x - a), x = +3 or a.
20. If a solution of 2(2x - 1) = 5x2 is expressed in simplest a + bi form,
the value of b is
Answer: (2) SQRT(6)/5
x = ( -(-4) + SQRT( (-4)2 - 4(5)(2) ) / (2(5))
x = ( 4 + SQRT( 16 - 40 ) ) / (10)
x = ( 4 + SQRT( -24 ) ) / (10)
x = ( 4 + 2i SQRT( 6 ) / (10)
x = 2/5 + i SQRT( 6 ) / 5
The answer is choice (2) because they only wanted the value of b, not the entire term. Choice one includes i.
Comments and questions welcome.
More Algebra 2 problems.
More Algebra 2 problems.
January 2020, Part I
All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.
11. Consider the data in the table below
Right Handed Left Handed Male 87 13 Female 89 11
What is the probability that a randomly selected person is male given that the person is left-handed?
(1) 13/200
(2) 13/100
(3) 13/50
(4) 13/24
Answer: 4) 13/24
There are a total of 24 left handed students. Of those 13 are male. So the probability is 13/24 of selecting a male if we know the person is left-handed.
12. The function N(x) = 90(0.86)x + 69 can be used to predict the
temperature of a cup of hot chocolate in degrees Fahrenheit after
x minutes. What is the approximate average rate of change of the
temperature of the hot chocolate, in degrees per minute, over the
interval [0, 6]?
(1) -8.93
(2) -0.11
(3) 0.11
(4) 8.93
Answer: (1) -8.93
The hot chocolate is cooling, so the temperature change is negative. Choices (3) and (4) are incorrect.
Find N(0) and N(6)
N(0) = 90(0.86)0 + 69 = 90(1) + 69 = 159
N(6) = 90(0.86)6 + 69 = 90(0.405) + 69 = 105.45
The average rate of change = (105.45 - 159) / 6 = -8.925
13. A recursive formula for the sequence 40, 30, 22.5, … is
(1) gn = 40(3/4)n
(2) g1 = 40 ; gn = gn-1 - 10
(3) gn = 40(3/4)n-1
(4) g1 = 40 ; gn = (3/4)gn-1 - 10
Answer: (4) g1 = 40 ; gn = (3/4)gn-1 - 10
Choices (1) and (3) are not recursive formulas.
Choice (2) indicates that the sequence is decreasing by 10. However, that would make g3 = 20, which it does not.
Choice (4) says that each term is 3/4 of the prior term, with the first term being 40. This is correct.
14. The J & B candy company claims that 45% of the candies it produces
are blue, 30% are brown, and 25% are yellow. Each bag holds 65
candies. A simulation was run 200 times, each of sample size 65, based
on the premise that 45% of the candies are blue. The results of the
simulation are shown below.
Bonnie purchased a bag of J & B’s candy and counted 24 blue
candies. What inference can be made regarding a bag of J & B’s with
only 24 blue candies?
(1) The company is not meeting their production standard.
(2) Bonnie’s bag was a rarity and the company should not be concerned.
(3) The company should change their claim to 37% blue candies are
produced.
(4) Bonnie’s bag is within the middle 95% of the simulated data
supporting the company’s claim.
Answer: (4) Bonnie’s bag is within the middle 95% of the simulated data
supporting the company’s claim.
The mean is 29.270, the standard deviation is 3.936.
The mean - 2 SD = 21.398, which is less than 24, so it is within the middle 95% of the data.
15. Which investigation technique is most often used to determine if a
single variable has an impact on a given population?
(1) observational study
(2) random survey
(3) controlled experiment
(4) formal interview
Answer: (3) controlled experiment
A controlled experiment is the technique used most often to test if a single variable has an impact.
Comments and questions welcome.
More Algebra 2 problems.
Work smarter, not harder. No need to reinvent the wheel. And something else I can get paid a consulting fee for saying.
I didn't want to get into was a 3-2-1 Exit Ticket was, so I just said "critique".
More Algebra 2 problems.
January 2020, Part I
All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.
6. The expression (x + a)2 + 5(x + a) + 4 is equivalent to
(1) (a + 1)(a + 4)
(2) (x + 1)(x + 4)
(3) (x + a + 1)(x + a + 4)
(4) x2 + a2 + 5x + 5a + 4
Answer: 3) (x + a + 1)(x + a + 4)
Replace (x + a) with y.
This gives you y2 + 5y + 4, which factors into (y + 1)(y + 4).
Now replace y with x + a, and you get (x + a + 1)(x + a + 4)
7. Given x =/= -2, the expression (2x2 + 5x + 8)/(x + 2) is equivalent to
After n months, which function represents Chet’s net worth, R(n)?
(1) 2x2 + 9/(x + 2)
(2) 2x + 7/(x + 2)
(3) 2x + 1 + 6/(x + 2)
(4) 2x + 9 - 10/(x + 2)
Answer: 3) 2x + 1 + 6/(x + 2)
Set up the division as in the table below:
8. Which situation best describes conditional probability?
(1) finding the probability of an event occurring two or more times
(2) finding the probability of an event occurring only once
(3) finding the probability of two independent events occurring at the
same time
(4) finding the probability of an event occurring given another event
had already occurred
Answer: 4) finding the probability of an event occurring given another event
had already occurred
Choice (1) is incorrect because it doesn't have to be the same event re-occurring.
Choice (2) is simple probability.
Choice (3) is not conditional, if they are happening at the same time.
9. Which expression is not a solution to the equation 2t = SQRT(10)?
(1) 1/2 log210
(2) log2SQRT(10)
(3) log410
(4) log104
Answer: 4) log104
Choices (3) and (4) should pop out at you because there is no reason why they should be equal, so one of them should be the answer.
2t = SQRT(10) means that log2SQRT(10) = t (choice 2)
Answer: (2) {8}
More Algebra 2 problems.
SQRT(10) = 101/2, which gives you: t = log2(10)1/2
The rules of logarithms say that log2(10)1/2 = 1/2 log2(10) (choice 1)
If 2t = SQRT(10),
then 22t = 10
and 4t = 10
So t = log410. (Choice 3)
10. What is the solution set of x = SQRT(3x + 40)?
(1) {-5, 8}
(2) {8}
(3) {-4, 10}
(4) { }
First of all, you can work backward from the choices. Second, because there isn't a + symbol in front of the square root symbol, it is assumed to be the positive root. Since this is the case, choices (1) and (3) are not possible.
x2 = 3x + 40
x2 - 3x - 40 = 0
(x - 8)(x + 5) = 0
x = 8 or x = -5 (disgard)
Check: 8 = SQRT(3(8) + 40) = SQRT(64) = 8, check.
Comments and questions welcome.
More Algebra 2 problems.
January 2020, Part I
All Questions in Part I are worth 2 credits. No work need be shown. No partial credit.
1. The expression (81x8y6)1/4 is equivalent to
(1) 3x2y3/2
(2) 3x4y2
(3) 9x2y3/2
(4) 9x4y2
Answer: 1) 3x2y3/2
The fourth root (as seen in the picture) is the same as the 1/4 power (which I had to type).
The fourth root of 81 is 3. (The square root of 81 is 9. The square root of 9 is 3.)
The rule for exponents is to multiply the exponents: 8 * 1/4 = 2, 6 * 1/4 = 6/4 = 3/2.
2. Chet has $1200 invested in a bank account modeled by the function
P(n) = 1200(1.002)n, where P(n) is the value of his account, in dollars,
after n months. Chet’s debt is modeled by the function Q(n) = 100n,
where Q(n) is the value of debt, in dollars, after n months.
After n months, which function represents Chet’s net worth, R(n)?
(1) R(n) = 1200(1.002)n + 100n
(2) R(n) = 1200(1.002)12n + 100n
(3) R(n) = 1200(1.002)n - 100n
(4) R(n) = 1200(1.002)12n - 100n
Answer: 3) R(n) = 1200(1.002)n - 100n
Net worth is the amount of his savings minus his debt, so the first two choices are incorrect.
Everything in the problem is in terms of months, so there is no need to multiply the exponent by 12, which would convert n years into months.
3. Emmeline is working on one side of a polynomial identity proof used to form Pythagorean triples. Her work is shown below:
Step 1: 25x2 + (5x2 - 5)2
Step 2: 25x2 + 25x2 + 25
Step 3: 50x2 + 25
Step 4: 75x2
What statement is true regarding Emmeline’s work?
(1) Emmeline’s work is entirely correct.
(2) There is a mistake in step 2, only.
(3) There are mistakes in step 2 and step 4.
(4) There is a mistake in step 4, only.
Answer: 3) There are mistakes in step 2 and step 4.
There are two mistakes, and both Algebra 1 kinds.
In Step 2, Emmeline didn't square the binomial correctly. There should be a -50x2 term.
In Step 4, she combined two terms which are not alike.
4. Susan won $2,000 and invested it into an account with an annual
interest rate of 3.2%. If her investment were compounded monthly,
which expression best represents the value of her investment after
t years?
(1) 2000(1.003)12t
(2) 2000(1.032)t/12
(3) 2064t/12
(4) (2000(1.032)t)/12
Answer: 1) 2000(1.003)12t
The interest rate is given annually, but it is compounded monthly.
So instead of 2000(1.032)t, we need to do 2000(1.0321/12)12.
That is, we need to find the monthly interest rate by taking the 1/12 power, but then we need to multiply the number of years by 12 to get the number of months the account will be compounded.
1.0321/12 is approximately 1.003.
5.Consider the end behavior description below.
Which function satisfies the given conditions?
Answer: 3) f(x) = -x3 + 2x - 6
The conditions state, as x increases to infiinity, f(x) decreases, and as x decreases to negative infinity, f(x) increases.
Graph (2) has the opposite behavior.
Graph (4) goes to negative infinity on both ends.
Function (1) has positive exponents only and will increase on both sides.
Function (3) is a negative cubic function, so it will start high and end low.
Comments and questions welcome.
More Algebra 2 problems.
And, of course, 317 is a prime number, so the use of decimals was unavoidable without making it trivial.
Happy St. Paddy's Day!
A review or two will be forthcoming in the days ahead. I won't be breaking it down by chapters as I did with the last one. Two blog posts will probably do fine.
Why Does the Divisibility Test For 9 Work?
There's an interesting chapter on "GOESINTOS", as in divisibility by. I'll put the notation aside and use a concrete example. I like examples first, and then general rules after.
Let's take a number like 2,467. We know that it is divisible by 9 because 2 + 4 + 6 + 7 = 18 and 1 + 8 = 9. (For that matter, we know that 18 is divisible by 9.)
But why should this matter?
Note: I touched upon this in a blog post about dividing polynomials by (x - 1), but I didn't think about the sum of the digits because I was using variables. So close!
The answer lies in writing the number out in expanded notation:
This can be rewritten as:
We can leave off the 0 power exponent, since it equals 1.
If we do the binomial expansion, it gets a little crazy, and I start to regret using a four-digit example. But here goes:
This in turn becomes:
If we regroup the terms, we can put all the terms with factors of 9 up front:
Finally, we'll factor out a 9 from those initial terms:
The sum of the original digits are being added to a multiple of 9. So the only way for the original number to be divisible by 9, is if 2 + 4 + 6 + 7 adds up to a multiple of 9. Adding one multiple of 9 to another multiple of 9 will yield a sum that is also a multiple of 9:
Okay, so I said that I would prove it in general now. Well, I lied. I could, but I'm not going to because it's a bit repetitious for a blog.
Replace my digits with A, B, C, D, and you will get the same result for the sum of A + B + C + D, proving it for any four-digit number.
Replace them with a1 + a2 + ... + an, and you can prove it for any number of digits. But that requires me to code subscripts and use the Combination notation (with more subscripts to code).
However, this is also the part where I get to say:
It's not just Nines!
In fact, if you replaces the 10s with "B" (which stands for Base, which rhymes with Your Face!), you will see that (B - 1) can always be factored out.
Let's use the above example again:
Applying the same steps would lead to:
This means that if 2,467 is in base 9, we could tell if it were divisible by 8. If it were it base 8, we could tell if it were divisible by 7. Use smaller digits, and we could check base 6, 5, or 4 if we wanted to.
Do you want to? It will give you something to work out this afternoon.
