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Stells [14]
3 years ago
7

Determine whether the sequence could be arithmetic. "yes" or "no". 28, 21, 15, 10, 6, ...

Mathematics
1 answer:
IRISSAK [1]3 years ago
4 0
No there is no pattern
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What is the sum of natural numbers less than 200 neither divisible by 3 nor by 5
Sveta_85 [38]

Step-by-step explanation:

The required sum 

=(1+2+3+...+199)−(3+6+9+...+198)−(5+10+15+...+195)+(15+30+45+...+195)

=2199(1+199)−266(3+198)−239(5+195)+213(15+195)

 

=199×100−33×201−39×100+13×105=10732

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3 years ago
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What is 3/5 times 20
igor_vitrenko [27]
First make 20 into a fraction, then solve; 3/5 · 20/1. 60/5 = 12. Hope that helps!
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A hiker descended 320 feet in 40 minutes. Find the hiker's average change in elevation per minute.
umka21 [38]

320 feet / 40 minutes. This will give an average of 8 feet per minute.

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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
PLZZZZ DO THE WHOLE THINGGGG<br><br> NO WORK = NO CREDIT = REPORT<br> GIVING BRAINLIEST
stich3 [128]

Answer:

Step-by-step explanation:

where is  the work

4 0
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