Tag Archives: Algebra 2

11.23.14
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Do this SAT problem the EASY way!

So you ‘don’t know how’ to do this SAT Question of the Day? http://sat.collegeboard.org/practice/sat-question-of-the-day?questionId=20141121&oq=1 The explanation given on the SAT site (look please!) is a good one, but ONLY if you would have thought of it! That’s the way your Algebra teacher wants you to do the problem. But DON’T SKIP IT! Do it MY WAY!
                                        

Q: A woman drove to work at an average speed of 40 miles per hour and returned along the same route at 30 miles per hour. If her total traveling time was 1 hour, what was the total number of miles in the round trip?

                          

 
There’s always a way to get rid of a choice or two. Let’s look…

 

(A) 30

(B) 30 1/7

(C) 34 2/7

(D) 35

(E) 40

Is it 30? No. Driving a distance of 30 miles in a total of 1 hour would give an overall average speed of 30 miles per hour. 30 mi/hr is the average speed for the second part of the trip, but the speed for the first part is 40 mi/hr. So, the overall average speed would be somewhere between 30 and 40, and therefore the number of miles would be between 30 and 40. (We’re talking about 1 hour, so the speeds and the distances will be the same numbers.) Cross out choice (A). Also cross out choice (E) for the same reason.

You’re down to three choices. At this point if you just guessed, you stand to gain more than you would lose. But you can do even better. The (wrong) answer they want you to fall for is (D). It’s too easy. It’s not just the average of 30 and 40, since the times spent at 30 vs 40 miles per hour are different. Cross out choice (D).

Two choices left. You were convinced above that 30 was the wrong answer. 30 1/7 is too close to 30 to make that big of a difference. The extra 1/7 of a mile wouldn’t be enough to raise the average speed to 40 for the first part of the trip. Cross out choice (B).

Done. The answer is C.

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11.07.14
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Slope

This week’s topic is:  Distance, Midpoint, and Slope.

Q:  The slope of the line passing through (-1,-3) and (7,y) is -1/2.  What is the value of y?

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Answer: -7

Algebraic solution by hand:

Use the slope formula:

 

2014-11-07_153414

Solution using my TI-84 Plus program DISTANCE:

Run DISTANCE.  How?

Use your multiple choice answer choices.  One of them is -7.  (If you don’t have choices, draw a picture.  Plot the point (-1,-3) and do a slope of -1/2 from there – down 1, right 2, down 1, right 2, … until you get to a point where x is 7.  Read the y-value of that point – it will be -7 and check it using this program.)

x1=-1

y1=-3

x2=7

y2=-7

Ignore the distance information – we don’t need it for this problem.  Press ENTER for the midpoint information, which we don’t need either.  Press ENTER once more to find that the slope is -1/2.

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11.06.14
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Distance

This week’s topic is:  Distance, Midpoint, and Slope.

Q:  For what value(s) of x is the point (x,4) exactly 5 units away from the point (6,8)?

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Answer: 3 or 9

Solution by hand:

Use the distance formula:

distance

Solution using my TI-84 Plus program DISTANCE:

Draw a picture:

distgraph

Guess that x=2, maybe.  It’s probably an integer if this is an SAT problem.  Use the answer choices if available.

Run DISTANCE.  How?

x1=2

y1=4

x2=6

y2=8

The distance is 5.66.  It’s too big – you wanted it to be 5 – so move the point a little closer to the middle.

Maybe x=3:

x1=3

y1=4

x2=6

y2=8

The distance is exactly 5.  Perfect.  (There’s also another possible answer, x=9.  If you have multiple choice answers, you will see 3 and 9 as a choice, so try both and see that they both work.  For a grid-in SAT problem, you would only have to find one answer anyway.)           :

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10.30.14
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Ratios

Q: Last month, the ratio of sunny days to rainy days was 3:2.  How many rainy days  were there last month, if the month had 30 days?  (Assume all days were either sunny or rainy.)

 

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Answer: 12

Solution by hand:

Number of sunny days = 3x

Number of rainy days = 2x

3x + 2x = 30

5x=30

x = 6

There were 2x rainy days, so there were 12 rainy days.

 

Solution using my TI-84 Plus Program RATIO:

Run RATIO.  How?

Total number of items (days) = 30

Number of categories of items = 2  (sunny and rainy)

Ratio part for category 1 (sunny) = 3

Ratio part for category 2 (rainy) = 2

Distribution is: sunny = 18 and rainy = 12.

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10.29.14
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Proportions

Q:  x is inversely proportional to w.  If x = 6 when w = 8, what is x when w = 12?

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Answer: 4

Solution by hand:

For inverse proportions, (x1)(w1) = (x2)(w2)

(6)(8) = (x)(12)

48 = 12x

x=4

Solution using my TI-84 Plus Program PROPORTN:

Run PROPORTN.  How?

x1 = 6

y1 = 8  (here, we have renamed w1 as y1)

Have y2 (really w2)

y2 = 12

Choose ‘Inverse’ proportion.

Then, x2 = 4.

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10.27.14
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Ratios

Q:  The ratio of blue cars to non-blue cars in the parking lot is 2:7.  If there are 1260 cars in the parking lot, how many are blue?

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Answer: 280

Solution by hand:

Number of blue cars = 2x

Number of non-blue cars = 7x

2x + 7x = 1260

9x=1260

x = 140

There are 2x blue cars, so there are 280 blue cars.

 

Solution using my TI-84 Plus Program RATIO:

Run RATIO.  How?

Total number of items = 1260

Number of categories of items = 2  (blue and non-blue)

Ratio part for category 1 (blue) = 2

Ratio part for category 2 (non-blue) = 7

Distribution is: blue = 280 and non-blue = 980.

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10.02.14
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SAT – Probability

Q:  What is the probability of randomly guessing the last four digits of a telephone number correctly?

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scroll down for answer

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A: 1/10,000

Explanation:  Each digit is chosen from the integers 0 through 9, which is ten possible choices for each digit.  There are four digits to choose independently, so there are:

10 x 10 x 10 x 10 = 10,000 ways to choose the four digits.

Then, the one correct sequence of 4 digits, is one out of 10,000.  The probability is 1/10,000.

Another way to arrive at the number 10,000 is as follows:

The 4-digit sequence may be anything from 0000 to 9999.  So, how many is that?  9999 – 0000 +1 = 10,000.  Why the ‘plus one’ on the end?  If you just subtract the numbers, it tells you how many spaces there are between the numbers.  Add one to include either the first number or the last number, whichever one has not been already counted.

 

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09.18.14
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Precalculus – Domain of a function

Q: What is the domain of this function?

 

Answer:

   ‘Like’ if you got it!

Explanation:

The domain of a function is a description of the x-values that may be plugged into the function. For a square root function, the restriction is that the number under the radical must not be negative. Therefore,

 

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09.12.14
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ACT , Algebra 1 – system of equations

MathPro Q&A Forum (ACT)

 Q: For what value of A will the following system of equations have infinitely many solutions?

3x – 2y = 4

Ax + 4y = -8

 

Answer: -6

Explanation: In order for the system to have infinitely many solutions, the two equations must describe the same line.  (Recall that the solution(s) of a system are the points where the graphs of the two lines meet.  For the two lines to meet in infinitely many points, they must be the same line.)

 To make the second equation the same as the first, multiply both sides of the first equation by -2:

 -2(3x – 2y) = -2(4)

-6x + 4y = -8

So, A = -6.

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09.10.14
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Algebra 2 – exponential functions

Q: In 2010, River City had a population of 500. If the population has increased by 6% each year since then, what is the population this year, in 2014?

Answer: 631

Explanation:

Alternatively,

Calcululate 6% of 500 by doing .06 * 500 = 30.

Increase 500 by 30, and you find that the population is 530 in 2011.

Now, increase 530 by 6% (530*.06, added to 530, or shortcut is to do 530*1.06).

The population is 561.8 in 2012. Don’t round this number yet – not until you get to the end of the problem.

Doing this calculation 2 more times gives you 595.508 for 2013 and 631.23… for 2014.

Now, round to the nearest whole number: 631. It’s not the most efficient way, but you get the right answer.

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