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4 years ago

How do u teachlong division

Mathematics

2 answers:

zavuch27 [327]4 years ago

7 0

Explanation:

To do long division, follow the method pattern "DMS down".

D for divide.

M for multiply.

S for subtract.

Down is to bring down.

Use an example:

52 ÷ 2?

Divide 2 by the first digit of 52.

2 Whenever you divide, put the answer at the top.

2∫52

Multiply answer by number you are dividing by.

2∫52

4 When not dividing, put the work under the question.

Subtract.

2∫52

4

Bring down the next digit.

2∫52

4

Divide into the new difference 12.

2∫52

4

Multiply the new digit.

2∫52

4

Subtract.

2∫52

4

12

0 You stop when you get to 0. If you ran out of digits to bring down, write this as a remainder.

Mariana [72]4 years ago

3 0

Answer:

Step-by-step explanation:

1. Put into bar

8(456

2. How many times does 8 go into 45 ( 8*?=45)

8*5=40

057

8(456

-40 45-40=5

56. Bring down the 6

-56 How many times does 8 go into 56

0 8*7=56

LAST STEP:

CHECK

8*57=456

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3 years ago

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Ratling [72]

Answer:

Step-by-step explanation:

5w+10=40

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3 years ago

I need help.. i really want to go sleep.. thank you so much...

Effectus [21]

Answer:

1) True 2) False

Step-by-step explanation:

1) Given $\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$

To verify that the above equality is true or false:

Now find $\sum\limits_{k=0}^8\frac{1}{k+3}$

Expanding the summation we get

$\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{0+3}+\frac{1}{1+3}+\frac{1}{2+3}+\frac{1}{3+3}+\frac{1}{4+3}+\frac{1}{5+3}+\frac{1}{6+3}+\frac{1}{7+3}+\frac{1}{8+3}$ $\sum\limits_{k=0}^8\frac{1}{k+3}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Now find $\sum\limits_{i=3}^{11}\frac{1}{i}$

Expanding the summation we get

$\sum\limits_{i=3}^{11}\frac{1}{i}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}$

Comparing the two series we get,

$\sum\limits_{k=0}^8\frac{1}{k+3}=\sum\limits_{i=3}^{11}\frac{1}{i}$ so the given equality is true.

2) Given $\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\sum\limits_{i=1}^3\frac{3i}{i+5}$

Verify the above equality is true or false

Now find $\sum\limits_{k=0}^4\frac{3k+3}{k+6}$

Expanding the summation we get

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3(0)+3}{0+6}+\frac{3(1)+3}{1+6}+\frac{3(2)+3}{2+6}+\frac{3(3)+4}{3+6}+\frac{3(4)+3}{4+6}$

$\sum\limits_{k=0}^4\frac{3k+3}{k+6}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}+\frac{12}{8}+\frac{15}{10}$

now find $\sum\limits_{i=1}^3\frac{3i}{i+5}$

Expanding the summation we get

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3(0)}{0+5}+\frac{3(1)}{1+5}+\frac{3(2)}{2+5}+\frac{3(3)}{3+5}$

$\sum\limits_{i=1}^3\frac{3i}{i+5}=\frac{3}{6}+\frac{6}{7}+\frac{9}{8}$

Comparing the series we get that the given equality is false.

ie, $\sum\limits_{k=0}^4\frac{3k+3}{k+6}\neq\sum\limits_{i=1}^3\frac{3i}{i+5}$

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ladessa [460]

Consider the complete question is "Ben, Cam, and Justin are lumberjacks. The number of trees they chop down is given by b+2c+3j, where b is the number of hours Ben spends chopping, c is the number of hours Cam spends chopping, and j is the number of hours Justin spends chopping. How many trees do they chop down after Ben spends 8 hours chopping, Cam spends 3 hours chopping, and Justin spends 4 hours chopping?"

Given:

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