Math 7 Math 8 Unit 2-Expressions & Equations: 12/17-Check practice, Quiz 9.3, Quizizz Activity GAME CODE 672244 Join. Math as more than a set of mnemonics or discrete procedures. Students demonstrate deep conceptual understanding of foundational mathematics concepts by applying them to new situations, as well as writing and speaking about their understanding. Memorization: The rapid recall of arithmetic facts or mathematical procedures. Memorization is often. Mix and Freeze Practice in Class (1.3 RAP Practice) Practice-9/17- Solving Equations-Special Cases Practice- Types of Equations 9/18- Solving Equations Practice in Class Multi-Step Equations Wkst 9/19- Quiz Concept 1 Practice- 1.3 BIG Ideas Online (pg. 23 #7, 9, 10, 14-17, 19-21, 23, 25, 27-31, 33) 9/20-Quizzes Back Quiz Corrections. Hello, Below are the teacher-authored instructional unit samples in mathematics. Each of these units represents the work of a team of Colorado educators to translate one curriculum overview sample into a full instructional unit with learning experiences, teacher and student resources, assessment ideas, and differentiation options.

1.1

Soln:

A man can enter the stadium in 4 ways. Again the man can leave the stadium in 9 ways.

So, total no.of ways with which a man enters and then leaves the stadium = 4 * 9 = 36ways.

Soln:

There are 6 choices for a student to enter the hostel. There are 5 choices for a student to leave the hostel as different door is to be used.

So, total no.of ways = 6 * 5 = 30.

Soln:

There are 7 choices for 1^st son, 6 choices for 2^nd son and 5 choices for 3^rd son.

Now, by the basic principle of counting, the total number of ways of choice = 7 * 6 * 5 = 210.

Soln:

A man can go from city A to city B in 5 ways. As he has to return by a different road, so he can return from city B to city A in 4 ways.

So, total no.of ways by which a man can go from city A to city B and returns by a different road = 5 * 4 = 20 ways.

Old Unit4 Agendasmrs. Colville's Math Class 1

Soln:

A person can go from city A to city B in 5 ways. Again, he can go from city B to city C in 4 ways. So, a person can go from city A to city C in 5 * 4 = 20ways. The person has to return from C to A without driving on the same road twice, So, he can return from city C to city B in 3 ways and from city B to city A in 4 ways.

So, he can return from city C to city A in 3 * 4 = 12 ways.

So, Total no.of ways by which a person can go from city A to city C and return from city C to city A = 20 * 12 = 240 ways.

Soln:

Numbers formed should be of at least 3 digits means they may be of 3 digits, 4 digits, 5 digits or 6 digits.

There are 6 choices for digit in the units place. There are 5 and 4 choices for digits in ten and hundred’s place respectively.

So, total number of ways by which 3 digits numbers can be formed = 6.5.4 = 120

Similarly, the total no.of ways by which 4 digits numbers can be formed = 6.5.4.3 = 360.

the total no. of ways by which 5 digits numbers can be formed = 6.5.4.3.2 = 720.

The total no.of ways by which 4 digits numbers can be formed = 6.5.4.3.2.1 = 720.

So, total no.of ways by which the numbers of at least 3 digits can be formed = 120 + 360 + 720 + 720 = 1920.

Old Unit4 Agendasmrs. Colville's Math Classes

Soln:

The numbers formed must be of three digits and less than 500, so the digit in the hundred’s place should be 1,2,3 or 4. So, there are 4 choices for the digit in the hundred’s place. There are 5 choice for the digit in the ten’s place. There are 4 choices for the digit in the unit’s place.

Old Unit4 Agendasmrs. Colville's Math Class 9

So, no of ways by which 3 digits numbers les than 500 can be formed = 4.5.4 = 80.

Old Unit4 Agendasmrs. Colville's Math Classroom

Soln:

The numbers formed should be even. So, the digit in the unit’s place must be 2 or 4. So, the digit in unit’s place must be 2 or 4. So, for the digit in unit’s place, there are 2 choices. So, after fixing the digit in the unit’s place, remaining 4 figures can be arranged in P(4,4) ways.

Ie. $frac{{left( 4 right)!}}{{left( {4 - 4} right)!}}$ = $frac{{4!}}{{0!}}$ = $frac{{4{rm{*}}3{rm{*}}2{rm{*}}1}}{1}$ = 24 ways.

So, total no.of ways by which 5 even numbers can be formed = 2 * 24 = 48.

Soln:

The numbers formed must be of 4 digits. The digit in the thousand’s place must always be 4. For this, there is only one choice. After that, n = 6 – 1 = 5, r = 4 – 1 = 3. Then remaining 5 figures can be placed in remaining 3 places in:

Or, P(5,3) ways = $frac{{5!}}{{left( {5 - 3} right)!}}$ = $frac{{5!}}{{2!}}$ = $frac{{5{rm{*}}4{rm{*}}3{rm{*}}2{rm{*}}1}}{{2{rm{*}}1}}$ = 60 ways.

So, Total no.of ways by which 4 digits numbers between 4,000 and 5,000 can be formed = 1 * 60 = 60.

10.

Soln:

For the three digits numbers, there are 5 ways to fill in the 1^st place, there are 4 ways to fill in the 2^nd place and there are 3 ways to fill in the 3^rd place. By the basic principle of counting, number of three digits numbers = 5 * 4 * 3 = 60.

Again, for three digit numbers which are divisible by 5, the number in the unit place must be 5. So, the unit place can be filled up in 1 way. After filling up the unit place 4 numbers are left. Ten’s place can be filled up in 4 ways and hundredths place can be filled up in 3 ways. Then by the basic principle of counting, no.of 3 digits numbers which are divisible by 5 = 1 * 4 * 3 = 12.