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Alex17521 [72]
4 years ago
15

Find the slope of a line that is parallel and the slope of a line that is perpendicular to each line whose equation is given. Pl

ease only do number one and show me all the steps you did.

Mathematics
1 answer:
aleksley [76]4 years ago
4 0
The equations above are all in the format y = mx + c, where m = the gradient, or slope.

So, in the first question, y = 4x + 2, the gradient would be 4, because 4 is m in this equation (the number before the x).

The gradient of a line perpendicular to this line is equal to the negative reciprocal of the gradient of the line. A better way to explain it is if m = the gradient of the line and <u />m' = the gradient of the perpendicular line then:
m' = - \frac{1}{m}

So in the first question, the gradient of the perpendicular line is - \frac{1}{4}.
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100 POINTS HELP ASAP Linear Combination
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Answer:

1)   x=\dfrac12

2)   n = -3 \ \ \textsf{and} \ \ m = -4

3)  see below

4)  A:  0 = 1

Step-by-step explanation:

<u>Question 1</u>

15-0.5(4x-2)+4x=17

\implies 15-2x+1+4x=17

\implies 2x+16=17

\implies 2x=1

\implies x=\dfrac12

<u>Question 2</u>

\textsf{rearrange} \ n=m+1 :

\implies -m=-n+1

\textsf{add equations} \ -m=-n+1 \ \textsf{and} \ m=2n+2:\\

\implies0=n+3

\implies n=-3

\textsf{substitute} \ \ n=-3 \ \ \textsf{into} \ \ m=n-1:

\implies m=-3-1

\implies m=-4

<u>Question 3</u>

subtract the second equation from the first

divide both sides by -4

substitute found value for y into first equation

solve for x

<u>Question 4</u>

3j=k

k=3j+1

\textsf{rearrange} \ 3j=k :

\implies -k=-3j

\textsf{add equations}\  -k=-3j \ \ \textsf{and}\ \ k=3j+1:

\implies 0=1

Solution = A

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