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Vsevolod [243]
3 years ago
7

9k - X use k = 4 and x = 6​

Mathematics
2 answers:
sukhopar [10]3 years ago
6 0

Answer:

88

Step-by-step explanation:

Use the algorithm method.

8 14

9 4

-  6

8 8

2 Therefore, 94-6=8894−6=88.

8888

Nutka1998 [239]3 years ago
3 0
9(4)= 36
36-6=30

Answer is 30
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The firefighter needs to drive 3 miles in 3 minutes. How fast should the firefighter drive?​
fiasKO [112]

Answer:

60 MPH

Step-by-step explanation:

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Square root of 154m^2 = 12.5~
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3 years ago
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
The number of one-inch cubes that will fit into a box that is 6 inches wide and one inch tall is 54. how long is the box?
Nana76 [90]
54 ÷ 6 = 9

the box is 9 inches long
5 0
3 years ago
Read 2 more answers
the difference between the squares of two numbers is 32. twice the square of the first number increased by the square of the sec
melamori03 [73]

Answer:

The numbers are 6 and 2

6 squared = 36

2 squared = 4

So 36-4 = 32

And 36 x 2 = 72 + 4 = 76

7 0
3 years ago
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