All posts by mathconfidence

Get the Math and Points Jan 2017 CC Alg I Regents #6

jan-2017-cc-alg-i-6-q
We can already analyze the two graphs above for the zeros.
Notice how (2) crosses the x-axis at (-2,0) and (4,0) at the ‘dots’ on the x-axis so that’s not right as we are looking for zeros of -4 and 2 (close but opposite signs)>
Ok let’s check out (4)…
Notice how (4) crosses the x-axis at (-4,0) and (2,0)? :)

(If we had to check (1) and (3), we would put each equation into Y= and check out the graph and/or table but no need to do that this time around)

Get the Math and Points Jan 2017 CC Alg 1#1

jan-2017-cc-alg-i-1-q

We can figure out the answer to this problem by studying the signs.

To get the “-” in 16x^2 – 36, we need a “+” and a “-” so when they are multiplied together they will equal a “-“.
Answer (1) has 2 negatives that when multiplied will equal a positive
Answer (3) has 2 negatives that when multiplied will equal a positive
Answer (4) has 2 positives that when multiplied will equal a positive
So find the one that has a positive and negative!  2 points :)

January 2017 Brain Puzzler Solution

Q:  The digits one through nine are randomly arranged to make a 9 digit number.
What is the probability that the resulting number is divisible by eighteen?

A:  The number will always be divisible by 9 as the digits 1 – 9 add up to 45 (any number whose digits add up 9 or a multiple of 9 will be divisible by 9).    So every one of these 9 digit numbers will be divisible by 9.
If the number is also divisible by 2, then it will be divisible by 18.
So we need to find how often the number will be divisible by 2.
The last digit could be any number from 1-9 and 4 of those digits are even (2, 4, 6, 8) making the # divisible by 2 (and 9, see above).
So there is a 4/9 probability that the number will be divisible by 18.

BONUS: What is the probability that the resulting number is divisible by 27?
Still working on the divisibility by 27…please send comments