CGP Part 1

CGP Part 1
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  • That new number underneath sort of acts as a "new dividend," which you divide for again with the same divisor. The process continues until there are no more numbers to carry down from the initial dividend. The final number on top of the L is the quotient. I'll complete this problem for us, so you can see what I mean.
  • If there is a 0 remaining, then there is no remainder. If there IS a number remaining, then you would rewrite that number as a remainder by writing it as itself over the divisor, added to the quotient. This is the final quotient. Our remainder was 9, so we would write our final quotient as 893 + 9/11.
  • Let's try an example with long division. We'll use 9832 ÷ 11 = x, and solve for x. First, rewrite the problem with an L on its side. Place the dividend inside the L, and the divisor on the outside.
  • Next, determine what you can multiply the divisor by to get the first place value of the dividend where it is greater than the divisor, going from left to right. Multiply, and put the result of your multiplication statement under the dividend, aligned with your goal number. Additionally, place the other factor in multiplication above the dividend, aligned with the one's place of the target number. In this case, 9 is smaller than 11, but 98 isn't. 11*8 is the highest you can go without going over 98, so we'll write 8 above the one's place of the 98 in the dividend, and align 88 underneath the 98.
  • O.K., cool! Thank you so much Jon!
  • No problem!
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