arabera Anthony celaya 8 years ago
257
Honelako gehiago
The first method is simply to draw out the problem. for example, 4x3 can be drawn as 4 circles with 3 dots in each of them. Counting the dots will lead to the answer.
The second method is expanded form. for the problem 26x32 it can be written as (20+6) on top of (30+2). next step, is to multiply 20 to 30 and 2. then multiply 6 to 30 and 2. finally, add the products together and there is the answer.
the next problem is the area model. for the problem 46x34 we will draw a box with 4 evenly sized boxes inside. on top of the first box, you place the 40, on top of the second box you place the 6. on the left side of the top left box you place 30 and on the left of the bottom left box you place the 4. then you multiply the 40 with 30 and 4 and place them in the corresponding boxes. then you multiply 6 with 30 and 4 and place the products in the corresponding boxes. finally, add all of the products together.
next is the lattice method. for 37x24 you create another box with 4 boxes inside. then create diagonal lines from the top right corner of every box all the way down to the bottom left. now multiply the numbers into the boxes where they meet, and separate them between the diagonals.add all of the numbers down the diagonal slots and that will be the answer.
for 248x34, you first expand the, like 200+40+8 and 30+4. then multiply the first 3 expanded numbers to 4, then repeat with 30. add all of the products and there is your answer
for lattice multiplication, create a grid for the numbers being multiplied. if there is a 3 digit number and a 2 digit, then there should be 6 boxes with one digit outside of each. draw a diagonal line connecting all of them from top right to bottom left. connect each box and multiply them all together and place them as you would for addition lattice. add like the addition lattice
The scratch method of addition is most helpful when adding several numbers at once, or when adding numbers in base nine or below. for example, well add 14basefive and 12basefive. stack them on top of one another to start. next, keep in mind that every "5" as a sum will count as "1" in the next place value up. when adding in the ones column, 4+2=6. since 5 or 6 can exist with a base five, we will scratch the 4 off and the 2 off, and write a 1 right next to the two. we do this to transfer the 5(which equals 10 in basefive) to the next place value up.
Utilizing the Lattice Method of Addition tends to occur most often when adding 2 or more numbers that have 2 or more digits in each of them. You would begin by stacking the two numbers on top of each other as you would for the basic addition method. once stacked, you would draw one box under each digit in the usual answer section. Every box should have a diagonal line running from the top right corner of the box to the bottom left corner of the box. When adding, if the sum is a single digit then enter the single digit sum in the bottom right corner of the box under that row, and place a 0 in the top left corner. If the sum is 2 digits, the place the one value in the bottom right corner and the ten value in the top left corner. Keep in mind this method is meant to reduce confusion when it comes to carrying the one by not separating double digit sums. anyways, continue this process until all of the sums have been added, then add each number together that shares space in the same diagonal row. the result should be your answer.
For repeated subtraction, the set up begins the same way with long division. for the number outside of the house, well say its 10 and the number inside is 1000. the point is to subtract a large number that 10 easily multiplies into from 1000. so we know 10x10=100. so we subtract 100 from 1000 and get 900. then we know that 90x10=900 so you subtract 900-900=0. since it was multiplied by 10 and 90 to equate for everything, the answer is to add them up. 1000/10=100
The Colby error consists of adding stacked numbers and incorrectly switching the ONE'S PLACE VALUE number with the 10'S PLACE VALUE number when carrying the TEN'S PLACE VALUE number. And example is 19 + 13. 9+3=12. The correct step after finding 9+3 is to carry the 1 (which represents 10) and stack it on top of the tens's place of the stacked added numbers While the 2 is placed directly under the 9 and 3 which represents the answer for the one' solace. The Colby Error is when the 1 is written under the 9 and 3 while the 2 is carried over to the tens place because the student had been confused over which number should be carried
G.E.M.D.A.S. STANDS FOR grouping, exponents, multiplication, division, addition, subtraction. GEMDAS is meant to maintain the proper order of solving math problems with multiple differing factors (G1, E2, M3, D4, A5, S6) . Two exceptions to the rule are the two pairs MULTIPLICATION/DIVISION and ADDITION/SUBTRACTION. When it comes to multiplication or division, there is no specific order to do these problems in so work from left to right. The same "left to right" concept applies to addition and subtraction as well. Simply put, the numerical order that the problems should be solved in is as follows: G1 E2 (M or D)3 (A or S)4. An example is (2+3) -6 x 1...first work on grouped numbers which are (2+3), then multiplication which is 6 x 1, then subtract 5-6 which equates to -1.
To convert a base ten number to a different base it is best to utilize the upside-down division method. For example, if we are going to convert 200base ten to base five, you must first plug in the 200 on top of the upside-down division bar an approach this like a factor tree only using 5 (because that is the base we are converting to). 5 goes into 200 a total of 40 times with a remainder of zero.( the first remainder will be in the one's place of the final answer, the second will be second to last and so-on. also, even if the remainder is zero, it must still be written out.) next, 5 goes into 40 a total of 8 times with a remainder of zero. finally, 5 goes into 8 a total of 1 time with a remainder of three. due to the fact that 5 is unable to go into 1, the number 1 will go into the final answer which is 1300base five.
TO CONVERT ANY BASE NUMBER TO BASE 10, YOU FIRST WANT TO WRITE DOWN THE NUMBER YOU ARE CONVERTING (IN THIS CASE WE'LL USE 14 BASE 5) KEEP IN MIND THAT YOU WILL MULTIPLY THE NUMBER IN THE ONE'S PLACE BY 1, THE TEN'S PLACE BY 5, THE HUNDREDTHS PLACE BY 25, THE THOUSANDTHS PLACE BY 125, AND SO ON. IN THIS CASE, THE NUMBER ONLY HAS TWO DIGITS, MEANING WE WILL MULTIPLY IN THE ONE'S PLACE AND IN THE TENTH'S PLACE. ONCE YOU MULTIPLY EACH INDIVIDUAL NUMBER BY THERE CORRESPONDING MULTIPLE OF 5, YOU WILL ADD THE TWO PRODUCTS TOGETHER, AND THE SUM OF THOSE TWO WILL BE THE CONVERSION TO BASE TEN. FIRST, WE MULTIPLY 1 (IN THE TENS PLACE) BY 5, THEN WE MULTIPLY 4(IN THE ONE'S PLACE) BY 1, FINALLY WE ADD 5+4 WHICH IS 9. 14 BASE 5 CONVERTED TO BASE 10 IS 9.
AS AN EXAMPLE, WE WILL ADD 24 TO 17. FIRST, YOU SHOULD STACK THE NUMBERS ON TOP OF EACH OTHER, IT DOES NOT MATTER WHICH ONE IS ON TOP. FOR THIS PROBLEM WE WILL SAY THAT 24 IS ON TOP OF 17. NEXT, YOU ADD ALL OF THE NUMBERS ON THE FAR RIGHT COLUMN WHICH IS THE DIGIT IN THE ONE'S PLACE FOR BOTH PROBLEMS. 4+7=11 . WHEN YOU ADD WITH STACKED NUMBERS IN THE ONE'S PLACE AND GET A NUMBER WITH TWO DIGITS AS A RESULT, YOU WILL PLACE THE ONE'S PLACE DIGIT FROM THE ANSWER DIRECTLY BELOW THE STACKED NUMBERS YOU HAD JUST ADDED, THEN YOU WILL PLACE THE TEN'S PLACE DIGIT ABOVE THE OTHER TWO STACKED TEN'S PLACE DIGITS. WHEM YOU PLACE A DIGIT ABOVE OTHERS IN ITS SAME PLACE TO ADD, IT IS REFERRED TO AS "CARRYING THE NUMBER". AFTER YOU HAVE CARRIED THE 1 (WHICH REPRESENTS THE TEN'S PLACE) AND PLACED IT ABOVE THE OTHER TWO DIGITS IN THE TEN'S PLACE, YOU WILL ADD ALL OF THE NUMBERS FROM THE TEN'S PLACE TOGETHER. SEEING AS HOW THERE ARE ONLY NUMBERS WITH TWO DIGITS BEING ADDED, THE ENTIRE ANSWER WILL BE PLACED BELOW THE TEN'S PLACE DIGITS AND CONNECT WITH THE ONE'S PLACE DIGIT TO CREATE THE WHOLE ANSWER. NOW WE ADD 2+1+1 WHICH IS 4(REPRESENTING 40) AND THAT CONNECTS WITH THE OTHER DIGIT WHICH IS 1. THE ANSWER TO 24+17=41.
when subtracting numbers with a base (for example 12base five minus 4base five) it is best to initially stack them. since 4 cant subtract from 2, the 2 will have to borrow from the ten's place. remember, in base five a ten's place value of 1 equals 5 units. so when you cross out the one to the tens place, youll actually be adding 5 to the 2 which would lead to 7-4=3base five.
when adding base numbers together, it is best to stack them on top of one another as a first step. as an example, we will be adding 44base five, 33 base five and 24 base five. when adding down the column, if at any point the numbers create a five, cross them out and add a 1 to the next place value upwards. if there is a 4 +3, then cross them both out, and write a 2 where the 3 was to represent the remainder which would still exist in the ones place. after all of the numbers are added in that column, put the remainders beneath the addition line. continue the process until there is nothing left to add. keep in mind that some columns may have zero because they could have added multiple numbers to the next place value up.
