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2 years ago

What is 246 : 7 in long division

Mathematics

2 answers:

Sever21 [200]2 years ago

6 0

35
7 |245
|21
36
35
1

35 with a remainder of 1 is the answer

Mars2501 [29]2 years ago

5 0

Here we will show you step-by-step with detailed explanation how to calculate 246 divided by 7 using long division.

Before you continue, note that in the problem 246 divided by 7, the numbers are defined as follows:

246 = dividend

7 = divisor

Step 1:

Start by setting it up with the divisor 7 on the left side and the dividend 246 on the right side like this:

7 ⟌ 2 4 6

Step 2:

The divisor (7) goes into the first digit of the dividend (2), 0 time(s). Therefore, put 0 on top:

7 ⟌ 2 4 6

Step 3:

Multiply the divisor by the result in the previous step (7 x 0 = 0) and write that answer below the dividend.

7 ⟌ 2 4 6

Step 4:

Subtract the result in the previous step from the first digit of the dividend (2 - 0 = 2) and write the answer below.

7 ⟌ 2 4 6

- 0

Step 5:

Move down the 2nd digit of the dividend (4) like this:

7 ⟌ 2 4 6

- 0

2 4

Step 6:

The divisor (7) goes into the bottom number (24), 3 time(s). Therefore, put 3 on top:

0 3

7 ⟌ 2 4 6

- 0

2 4

Step 7:

Multiply the divisor by the result in the previous step (7 x 3 = 21) and write that answer at the bottom:

0 3

7 ⟌ 2 4 6

- 0

2 4

2 1

Step 8:

Subtract the result in the previous step from the number written above it. (24 - 21 = 3) and write the answer at the bottom.

0 3

7 ⟌ 2 4 6

- 0

2 4

- 2 1

Step 9:

Move down the last digit of the dividend (6) like this:

0 3

7 ⟌ 2 4 6

- 0

2 4

- 2 1

3 6

Step 10:

The divisor (7) goes into the bottom number (36), 5 time(s). Therefore put 5 on top:

0 3 5

7 ⟌ 2 4 6

- 0

2 4

- 2 1

3 6

Step 11:

Multiply the divisor by the result in the previous step (7 x 5 = 35) and write the answer at the bottom:

0 3 5

7 ⟌ 2 4 6

- 0

2 4

- 2 1

3 6

3 5

Step 12:

Subtract the result in the previous step from the number written above it. (36 - 35 = 1) and write the answer at the bottom.

0 3 5

7 ⟌ 2 4 6

- 0

2 4

- 2 1

3 6

- 3 5

You are done, because there are no more digits to move down from the dividend.

The answer is the top number and the remainder is the bottom number.

Therefore, the answer to 246 divided by 7 calculated using Long Division is:

1 Remainder

Mark as brainlist

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Let X be a set of size 20 and A CX be of size 10. (a) How many sets B are there that satisfy A Ç B Ç X? (b) How many sets B are

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(a) Number of sets B given that

A⊆B⊆C: 2¹⁰. (That is: A is a subset of B, B is a subset of C. B might be equal to C)
A⊂B⊂C: 2¹⁰ - 2. (That is: A is a proper subset of B, B is a proper subset of C. B≠C)

(b) Number of sets B given that set A and set B are disjoint, and that set B is a subset of set X: 2²⁰ - 2¹⁰.

Step-by-step explanation:

Let $x_1, x_2, \cdots, x_{20}$ denote the 20 elements of set X.

Let $x_1, x_2, \cdots, x_{10}$ denote elements of set X that are also part of set A.

For set A to be a subset of set B, each element in set A must also be present in set B. In other words, set B should also contain $x_1, x_2, \cdots, x_{10}$ .

For set B to be a subset of set C, all elements of set B also need to be in set C. In other words, all the elements of set B should come from $x_1, x_2, \cdots, x_{20}$ .

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

For each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for set B.

In case the question connected set A and B, and set B and C using the symbol ⊂ (proper subset of) instead of ⊆, A ≠ B and B ≠ C. Two possibilities will need to be eliminated: B contains all ten "maybe" elements or B contains none of the ten "maybe" elements. That leaves $2^{10} -2 = 1024 - 2 = 1022$ possibilities.

Set A and set B are disjoint if none of the elements in set A are also in set B, and none of the elements in set B are in set A.

Start by considering the case when set A and set B are indeed disjoint.

$\begin{array}{c|cccccccc}\text{Members of X} & x_1 & x_2 & \cdots & x_{10} & x_{11} & \cdots & x_{20}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set A?} & \text{Yes}&\text{Yes}&\cdots &\text{Yes}& \text{No} & \cdots & \text{No}\\[0.5em]\displaystyle\text{Member of}\atop\displaystyle\text{Set B?}& \text{No}&\text{No}&\cdots &\text{No}& \text{Maybe} & \cdots & \text{Maybe}\end{array}$ .

Set B might be an empty set. Once again, for each element that might be in set B, there are two possibilities: either the element is in set B or it is not in set B. There are ten such elements. There are thus $2^{10} = 1024$ possibilities for a set B that is disjoint with set A.

There are 20 elements in X so that's $2^{20} = 1048576$ possibilities for B ⊆ X if there's no restriction on B. However, since B cannot be disjoint with set A, there's only $2^{20} - 2^{10}$ possibilities left.