Line JK passes through points J(–3, 11) and K(1, –3). What is the equation of line JK in standard form? 7x + 2y = –1
7x + 2y = 1
14x + 4y = –1
14x + 4y = 1

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  (b) 7x + 2y = 1

Step-by-step explanation:

You don't need to know how to find the equation. You just need to know how to determine if a point satisfies the equation. Try one of the points and see which equation fits. (The numbers are smaller for point K, so we prefer to use that one.)

  7(1) +2(-3) = 1 ≠ -1 . . . . . tells you choice A doesn't work, and choice B does

The equation is ...

  7x +2y = 1

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Additional comment

The equations of choices C and D have coefficients with a common factor of 2. If the constant also had a factor of 2, we could say these equations are not in standard form, and we could reject them right away. Since the two points have integer values for x and y, we can reject these equations anyway: the sum of even numbers cannot be odd.

d

step-by-step explanation:

by observation, we can see that the the graph is simply the graph of y = |x| that has been shifted 4 units downwards and 3 units to the left.

we can recognize in choice d :

y = | x + 3 |   - 4 (rewriting in general form)

y = | x - (- 3) |   - 4

the "-4 " term means   that the graph has been shifted by 4 units in the negative y direction

the "-3 " term means   that the graph has been shifted by 3 units in the negative x direction

this matches our observation of the graph described above.

see below

step-by-step explanation:

x+2> 6

subtract 2 from each side

x+2-2 > 6-2

x > 4

there is an open circle at 4 since is it is greater than and the line goes to the right