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3 years ago

Which equation represents a line that is parallel to the line represented by

Mathematics

1 answer:

Sveta_85 [38]3 years ago

5 0

Answer:

y = -4x + 12

Step-by-step explanation:

For a line to be parallel, it needs to have the same slope.

Point intercept form: y - y₁ = m (x - x₁)

Where y and x are just kept as y and x, y₁ and x₁ are the parts of a coordinate, and m = slope.

So, in order for the line to be parallel, it needs to have the same slope as the first equation. The slope of the line (which is the "m" in y = mx + b) is..

4x + y = 7

y = -4x + 7

So, m, or the slope, is -4.

Plug this, along with the coordinate the problem gives you into point-intercept form.

y - y₁ = m (x - x₁)

y - 8 = -4 (x - 1)

Simplify.

y - 8 = -4x + 4

Add 8 to both sides.

y = -4x + 12

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3 years ago

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Hello there!

Carol would have sold 60 subscriptions and Jenifer would have sold 30

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2 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 $X-(A\cup B)\subset(X-A)\cap(X-B)$ .

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

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3 years ago