# Get a TI-84 Plus Calculator!!

## The BEST thing you can do to improve your SAT or ACT Math score is to get the TI-84 Plus calculator… and load it up with my programs.

It’s only \$108 today on Amazon, with free shipping! This is the best deal I’ve seen. By comparison, Walmart has it for \$119.27. Get it now: http://amzn.to/1vUA1IP

The TI-84 Plus is a great help to anyone taking the SAT or ACT. I have dozens of programs that will solve many typical problems in mere seconds. I can transfer these programs to your TI-84 Plus if you come to my office. BUT very soon they will be available for download on my website, with instructions and examples, so stay tuned!

By the way, there is no need to get any of the other fancier versions of this calculator. They won’t help you any more than this one does. But if you do want a certain color, well here are the other options:  http://amzn.to/1yi4bKd

(Yes, the TI-83 Plus is compatible with the programs, too, so that’s another option as listed in the above link, but the USB cable must be purchased separately.)

If all you need is the USB cable, here it is:

USB cable for  TI-84 Plus: http://amzn.to/1xAchur

USB cable for TI-83 Plus: http://amzn.to/1qAmCRx (Don’t buy the one above if you have the 83!!)

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# SAT – Probability

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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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# SAT Question Of The Day – estimating for hard geometry

### You’d be surprised how often you can really do this and get the right answer.  Follow me, and you’ll see lots of these!

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# ACT, Geometry – diagonal of a rectangular prism

Q: What is the (straight-line) distance between points A and B on the figure below?

Explanation:  We are finding the length of the green segment below:

This requires the three-dimensional version of the distance formula:

There is another way to do this, using a progression of two right triangles, with the green segment being the hypotenuse of the second triangle, but I find the above formula much simpler.  It is just an extension of the ‘usual’ (2-dimensional) distance formula we are so familiar with from Algebra and Geometry.

BUT if you want an even easier way, and I know you do, use my TI-84 Plus program ‘DIST3D’.

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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

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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# SAT – number properties

Q: Find the largest number less than 4000 that is the square of an integer.

Explanation:  The simplest, fastest way to do this is:

On a calculator, find the square root of 4000.  It is 63.245….  Round this down to 63.  Now, square 63.  It is 3,969.

This is the way you should be thinking for SAT problems.  If you decide instead to count down from 3,999, finding the square root of each number until you find a whole number, you will have to do this 31 times until you find the right one!  (And by the way, why 31?  Doesn’t 3999 – 3969 equal 30?  Was I wrong?  No!  This is another little SAT trick.  Always add 1 to include the first or last number, whichever one wasn’t counted in the original 30.)

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# SAT – number properties

Q: The variables a, x, and b are positive integers. If ax = 7, xb = 28, and ab = 4, what is the value of a?

Explanation:

a, x, and b are all positive integers, so the only possibilities for the variables are factors of 7, 28, and 4.

Since 7 is prime, a and x must be 7 and 1, but in which order? If x = 1, then b = 28, and then we can’t find an integer value for a from the last equation. So, x must instead be 7.

It follows that a = 1 and b = 4.

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# Geometry – polygons

Q: A hexagon has interior angles measuring (in degrees) 137, 100, 150, 114, 90, and x. What is the value of x?

Explanation:
The total measure of all the interior angles in a polygon is (n-2) * 180, where n is the number of sides of the polygon.
A hexagon has 6 sides, so n = 6. The total is (6-2) * 180 = 720.
Add up the five given angle measures: 137 + 100 + 150 + 114 + 90 = 591.
The measure of the remaining angle is 720 – 591 = 129.
My TI-84 Plus program POLYGON will find this total for you.
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# SAT – Counting and Ordering

Q: There are 11 basketball teams in a tournament. Each team plays each other team once. How many games are played?

Explanation 1: This question is really ‘How many ways are there to choose 2 teams from a total of 11 teams, if the order of the teams is not important?’ It’s a combination question. Find the nCr function on your calculator. On the TI-84 Plus it’s in MATH-PRB-3. Type in 11 nCr 2. It’s 55. (There is a longer way to do it by hand, using the formula for nCr.)

Explanation 2: Team A plays one game against each of the 10 other teams. That’s 10 games so far. Team B plays 9 other games (its game against Team A has already been counted). Team C plays 8 other games, … all the way down to the 1 last game between teams J and K. Add: 10 + 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 55.

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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?

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.