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3 years ago

10

Which formula(s) can be used to find the nth partial sum of a geometric sequence or the sum of the first n terms of a geometric

series?

1 answer:

ivolga24 [154]3 years ago

6 0

Step-by-step explanation:

An=A1. R raised to the power n-1

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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

29 — 6x=5(1 – 6x)<br> Please show me the steps

Lady bird [3.3K]

Answer:

x= -1

Step-by-step explanation:

<u><em>START</em></u> 29 - 6x = 5(1 - 6x)

1. Distribute

29 - 6x = 5 - 30x

2. Isolate the variable

~ <u>2a</u>. Add 30x to both sides

29 - 6x + 30x = 5 - 30x + 30x

~ <u>2b</u>. Subtract 29 from both sides

29 - 29 - 6x + 30x = 5 - 29 - 30x + 30x

3. Simplify

24x = -24

4. Solve

x = -1

<em><u>END</u></em>

5 0

2 years ago

How many pounds of raisins were in her supply when she started?

sp2606 [1]

Answer:

i think it's 15 because if you use 5 and you have to do it by 3 it should be 15

8 0

2 years ago

Augustus ruled the roman empire from 27 BC to AD 14.For how many years did he rule?<br><br><br>

Semmy [17]

Answer:

40 years

Step-by-step explanation:

He was also the first Roman Emperor

4 0

2 years ago

All changes saved

slamgirl [31]

Your answer would be :A

Explanation: He claims to have had 9 sales “per day” for the week. When really it was only for 2 days. That being Thursday and Tuesday! The Mode number of the sales is 9; which is only higher from most of the days.

I hope this helps. Let me know if I am wrong plz. It’s 1 in the morning for me right now. Lol Have a blessed day bud.

4 0

2 years ago