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blsea [12.9K]
3 years ago
13

How to check your work on a division problem.

Mathematics
2 answers:
Rus_ich [418]3 years ago
4 0

Answer:

By checking it with multiplication

Step-by-step explanation:

You can use multiplication to check your division answer this way.

Do the division problem.

Multiply the quotient times the divisor.

If there is a remainder, add it to the multiplication product.

Compare this answer to the dividend. They should be the same number (630 = 630).

Verizon [17]3 years ago
3 0

Answer:

Do the division problem.

Multiply the quotient times the divisor.

If there is a remainder, add it to the multiplication product.

Compare this answer to the dividend. They should be the same number

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What is meant by the term 'domain'?
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Prove A-(BnC) = (A-B)U(A-C), explain with an example​
NikAS [45]

Answer:

Prove set equality by showing that for any element x, x \in (A \backslash (B \cap C)) if and only if x \in ((A \backslash B) \cup (A \backslash C)).

Example:

A = \lbrace 0,\, 1,\, 2,\, 3 \rbrace.

B = \lbrace0,\, 1 \rbrace.

C = \lbrace0,\, 2 \rbrace.

\begin{aligned} & A \backslash (B \cap C) \\ =\; & \lbrace 0,\, 1,\, 2,\, 3 \rbrace \backslash \lbrace 0 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace \end{aligned}.

\begin{aligned}& (A \backslash B) \cup (A \backslash C) \\ =\; & \lbrace 2,\, 3\rbrace \cup \lbrace 1,\, 3 \rbrace \\ =\; & \lbrace 1,\, 2,\, 3 \rbrace\end{aligned}.

Step-by-step explanation:

Proof for [x \in (A \backslash (B \cap C))] \implies [x \in ((A \backslash B) \cup (A \backslash C))] for any element x:

Assume that x \in (A \backslash (B \cap C)). Thus, x \in A and x \not \in (B \cap C).

Since x \not \in (B \cap C), either x \not \in B or x \not \in C (or both.)

  • If x \not \in B, then combined with x \in A, x \in (A \backslash B).
  • Similarly, if x \not \in C, then combined with x \in A, x \in (A \backslash C).

Thus, either x \in (A \backslash B) or x \in (A \backslash C) (or both.)

Therefore, x \in ((A \backslash B) \cup (A \backslash C)) as required.

Proof for [x \in ((A \backslash B) \cup (A \backslash C))] \implies [x \in (A \backslash (B \cap C))]:

Assume that x \in ((A \backslash B) \cup (A \backslash C)). Thus, either x \in (A \backslash B) or x \in (A \backslash C) (or both.)

  • If x \in (A \backslash B), then x \in A and x \not \in B. Notice that (x \not \in B) \implies (x \not \in (B \cap C)) since the contrapositive of that statement, (x \in (B \cap C)) \implies (x \in B), is true. Therefore, x \not \in (B \cap C) and thus x \in A \backslash (B \cap C).
  • Otherwise, if x \in A \backslash C, then x \in A and x \not \in C. Similarly, x \not \in C \! implies x \not \in (B \cap C). Therefore, x \in A \backslash (B \cap C).

Either way, x \in A \backslash (B \cap C).

Therefore, x \in ((A \backslash B) \cup (A \backslash C)) implies x \in A \backslash (B \cap C), as required.

8 0
2 years ago
Rewrite the equation below so that it does not have fractions.
Leviafan [203]

9514 1404 393

Answer:

  8x -24 = 9y

Step-by-step explanation:

We assume your equation is ...

  2/3x -2 = 3/4y

__

Fractions can be eliminated by multiplying the equation by the least common denominator of the fractions.

  LCM(3, 4) = 12

Multiplying the equation by 12 gives you ...

  12(2/3x -2) = 12(3/4y)

  8x -24 = 9y

5 0
3 years ago
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