All posts by mathconfidence

Get the Math and the Points Jan 2018 CC Alg I #2

Jan 2018 CC Alg I 2
Two questions to ask:
1. Does V(x) grow or decay?
2. When putting 4(0.65)^x vs 4(1.35)^x into the Y= of the TI-83/4, does the table look like the one above?
Hopefully the student will know that the number in parentheses (the growth/decay factor) determines if it is growth or decay but through the graphing calculator, they can figure it out, get the right answer and earn 2 points!

January 2018 Brain Puzzler Solution

Q: What is the sum of the digits of the square of $111,111,111$?

A: 81
Using the standard multiplication algorithm, $111,111,111^2=12,345,678,987,654,321,$whose digit sum is 81.

Or by looking at the pattern:

$11^2 = 121 \\ 111^2 = 12321 \\ 1111^2 = 1234321$

$111,111,111^2=12,345,678,987,654,321$ whose digit sum is 81
There is a shortcut to adding these digits
Reading from left to right, we can add the first 8 digits, 1 through 8 by making 4 pairs of 9 (1 + 8, 2 + 7, 3 + 6 and 4 + 5) making 36
Reading from right to left, we can add the last 8 digits (1- 8) the same way making 36.
Leaving the middle digit of 9.

36 + 36 + 9 = 81

Get the Math and the Points! CC Alg I June 2017 Regents #10

June 2017 CC Alg I 10Can use the TI-83/4 to figure this one out!
The zeros are the values of x that make f(x) or y equal to zero.

Answer (1) looks like this:  only 1 zero at x = -3 which can be seen on both the graph and the table
Aug 2017 CC Alg I 10a EqAug 2017 CC Alg I 10aAug 2017 CC Alg I 10a Graph

Answer (2) looks like this with 3 zeros but only one of them is an integer value (both the table and the graph show that (-4,0) is a point
Aug 2017 CC Alg I 10b yAug 2017 CC Alg I 10b tableAug 2017 CC Alg I 10b graph

Let’s try answer (3):

Aug 2017 CC Alg I 10c EqAug 2017 CC Alg I 10c tableAug 2017 CC Alg I 10c graph
Here we can see that on the table when x = -3 y = 0 and also when x = 0 and x = 4, y is also equal to zero.  Looking at the graph we can see that the 3 x-intercepts are integer values of -3, 0 and 4.  That’s it!

Pay close attention to the signs!! We need to look for the factors with the opposite sign to ‘zero it out’!  This can be a bit counterintuitive as when -3 is a solution then the factor is (x + 3), as -3 makes x + 3 = 0.  When 4 is a solution, the factor must be (x – 4)  rather than  (x+4) as plugging in 4 would not result in zero.

 

December 2017 Brain Puzzler Solution

Q: For the positive integers up to and including 2017, how many of those integers have at least one zero?

A:  There are $2017$ positive integers in total to consider.
No one digit numbers have a 0 (1-9).
Below is the list of the qualifying numbers:

There are 9 two digit numbers: 10,20, 30…90 9
There are the numbers 100 – 110 and 200-210 up to 900 – 910 99
Also 120 130 140…190 and then 220 230 etc 72
Then 1000-1099 100
Then same # of #s 1000 more than b 99
Then same # of #s 1000 more than c 72
2000-2017 18
 Total  469

Another solution is to exclude the numbers that do not have a zero and subtract from 2017:
9 one digit numbers
81 two digit numbers (9 x 9 : 9 choices for each digit 1-9)
729 three digit numbers (9 x 9 x 9: 9 choices for each digit 1-9)
For the numbers between 1000- 1999, there are the same amount as for the three digit numbers as they are the same numbers with a “1” in the thousands place value.
9 + 81 + 729 + 729 = 1548
2017 – 1548 = 469

 

 

Get the Math and the Points! Common Core Algebra I Aug 2017 Regents #3

Aug 2017 CC Alg I 3

This can be easily solved by checking the answers:  we will need a + and – to create a – so answers (1) and (2) are gone.
Look at (4): If we multiply (24.5x)(24.5x) we definitely get more than 49x^2 so finding this answer is easy.

Students can also put 49x^2 – 36 into Y1 on the TI-84 and then try each of the answers into Y2 and see which creates the same graph and/or table.

That’s it, 2 more points on the Regents!!