Which linear any qualities could represent the given table of values

The answer is (B) y ≤ –2x + 3 .

Step-by-step explanation: We are to select the correct linear inequality that could represent the table of values given in the question.

If (x, y) = (-4, -1), then

\begin{gathered}y=-1,-2x+3=-2\times(-4)+3=8+3=11.\end{gathered}

y=−1,

−2x+3=−2×(−4)+3=8+3=11.

So, y < -2x + 3.

If (x, y) = (-2, 4), then

\begin{gathered}y=4,-2x+3=-2\times(-2)+3=4+3=7.\end{gathered}

y=4,

−2x+3=−2×(−2)+3=4+3=7.

So, y < -2x + 3.

If (x, y) = (3, -3), then

\begin{gathered}y=-3,-2x+3=-2\times 3+3=-6+3=-3.\end{gathered}

y=−3,

−2x+3=−2×3+3=−6+3=−3.

So, y = -2x + 3.

If (x, y) = (3, -4), then

\begin{gathered}y=-4,-2x+3=-2\times 3+3=-6+3=-3.\end{gathered}

y=−4,

−2x+3=−2×3+3=−6+3=−3.

So, y < -2x + 3.

Therefore, we have y is either less than or equal to (-2x + 3).

Thus, y ≤ –2x + 3 and (B) is the correct option.

, and

for

step-by-step explanation:

we can analyze the pattern at which the terms increase. let us denote , , , so that we can call

note the following:

the pattern: to get the next term value, we just add the next term number (e.g. 1st, 2nd, to the previous term value (term values are the given numbers in the question, 1, 3, 6, 10, 15,

from , to get to , we add

from , to get to , we add

etc.

we need to form a recursive formula, a formula that has the next term as a function of the previous term. so a possible recursive formula can be

, and for

126 maybe. sorry