Slopes please help !!

What is the sum of the first seven terms of the geometric series 2 - 10 +50 -...?

Umnica [9.8K]

Answer:

26042.

Step-by-step explanation:

What's the first term of this geometric series?

2.

What's the common ratio of this geometric series?

Divide one of the terms with the previous term. For example, divide the second term -10 with the first term 2.

$\displaystyle r = \frac{-10}{2} = -5$ .

What's the sum of this series to the seventh term?

The sum of the first n terms of a geometric series is:

$\displaystyle a_1 \cdot \frac{1-r^{n}}{1-r}$ ,

where

$a_1$ is the first term of the series,
$r$ is the common ratio of the series, and
$n$ is the number of terms in this series.

$\displaystyle 2 \times\frac{1- (-5)^{7}}{1- (-5)}=26,042$ .

3 0

3 years ago

Please help and thank you

NikAS [45]

Answer:

A

PLZ MARK ME AS BRAINIEST

7 0

3 years ago

Read 2 more answers

Find the difference of(ab+8a+1)-(-6ab+4)

vagabundo [1.1K]

Hey there!

When finding the difference of expressions like these, there's a few things you should keep in mind.

1. When finding the difference of two polynomials, multiply the polynomial that implied to be "negative" by –1 to expand the problem and make it easier to solve.

2. You combine terms based on their unknown terms, like x, y, x², ab, etc. You cannot combine the terms x and x², but you can combine the terms 6b and 8b by adding them.

3. You can rearrange your terms however you see fit. If a polynomial is implied as "positive", you can multiply that entire polynomial by +1 to get rid of the parentheses.

With all that in mind, you can go ahead and solve for your difference, like so:

$(ab+8a+1)-(-6ab+4)$

$(ab+8a+1) (-1(-6ab+4))$

$(ab+8a+1) + (6ab-4)$

$ab+6ab+8a+1-4$

$7ab+8a-3$

$7ab+8a-3$ will be your difference.

Hope this helped you out! :-)

8 0

3 years ago

Prove the following DeMorgan's laws: if LaTeX: XX, LaTeX: AA and LaTeX: BB are sets and LaTeX: \{A_i: i\in I\} {Ai:i∈I} is a fam

MariettaO [177]

$X-(A\cup B)=(X-A)\cap(X-B)$

I'll assume the usual definition of set difference, $X-A=\{x\in X,x\not\in A\}$ .

Let $x\in X-(A\cup B)$ . Then $x\in X$ and $x\not\in(A\cup B)$ . If $x\not\in(A\cup B)$ , then $x\not\in A$ and $x\not\in B$ . This means $x\in X,x\not\in A$ and $x\in X,x\not\in B$ , so it follows that $x\in(X-A)\cap(X-B)$ . Hence $X-(A\cup B)\subset(X-A)\cap(X-B)$ .

Now let $x\in(X-A)\cap(X-B)$ . Then $x\in X-A$ and $x\in X-B$ . By definition of set difference, $x\in X,x\not\in A$ and $x\in X,x\not\in B$ . Since $x\not A,x\not\in B$ , we have $x\not\in(A\cup B)$ , and so $x\in X-(A\cup B)$ . Hence $(X-A)\cap(X-B)\subset X-(A\cup B)$ .

The two sets are subsets of one another, so they must be equal.

$X-\left(\bigcup\limits_{i\in I}A_i\right)=\bigcap\limits_{i\in I}(X-A_i)$

The proof of this is the same as above, you just have to indicate that membership, of lack thereof, holds for all indices $i\in I$ .

Proof of one direction for example:

Let $x\in X-\left(\bigcup\limits_{i\in I}A_i\right)$ . Then $x\in X$ and $x\not\in\bigcup\limits_{i\in I}A_i$ , which in turn means $x\not\in A_i$ for all $i\in I$ . This means $x\in X,x\not\in A_{i_1}$ , and $x\in X,x\not\in A_{i_2}$ , and so on, where $\{i_1,i_2,\ldots\}\subset I$ , for all $i\in I$ . This means $x\in X-A_{i_1}$ , and $x\in X-A_{i_2}$ , and so on, so $x\in\bigcap\limits_{i\in I}(X-A_i)$ . Hence $X-\left(\bigcup\limits_{i\in I}A_i\right)\subset\bigcap\limits_{i\in I}(X-A_i)$ .