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trapecia [35]
3 years ago
13

If the point (4,-2) is included in a direct variation relationship, which point also belongs in this direct variation? A. (-4,2)

B. (-4,-2) C. (2,-4) D. (-2,4)
Mathematics
1 answer:
IceJOKER [234]3 years ago
5 0

Answer:

(-4,2) belongs in this direct variation.

Step-by-step explanation:

Let the direct variation relationship is expressed by the equation y = mx ........ (1), where x and y are in direct variation and k is the variation constant.

Now, the point (4,-2) is included in the direct variation relationship, then from equation (1) we get, -2 = 4m

⇒ m =  - \frac{1}{2}

Therefore, the equation (1) becomes y = - \frac{1}{2}x

⇒ x + 2y = 0 ......... (2)

Now, from the given four options only the point (-4,2) satisfies the relation (2).

Hence, (-4,2) belongs in this direct variation. (Answer)

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3 years ago
Help please. I attempted, but I couldn't succeed.
german

Answer:

y=(3/2)x+-14

First blank: 3

Second blank:2

Last blank:-14

Step-by-step explanation:

The line form being requested is slope-intercept form, y=mx+b where m is slope and b is y-intercept.

Also perpendicular lines have opposite reciprocal slopes so the slope of the line we are looking for is the opposite reciprocal of -2/3 which is 3/2.

So the equation so far is

y=(3/2)x+b.

We know this line goes through (x,y)=(4,-8).

So we can use this point along with our equation to find b.

-8=(3/2)4+b

-8=6+b

-14=b

The line is y=(3/2)x-14.

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