No, its D
Just look at the rounds mentioned and subtract the scores from higher round with lower round.
Look at A: round 2 score - round 1 score = -2?
-3 -1 = -4 change, not -2 change so it is wrong
Look at B: round 3 score - round 1 score =-1?
-2-1 =-3 change, not -1 change so it is wrong
Look at C: round 3 score - round 2 score =-1?
-2 -(-3) = 1 change, not -1 change so it is wrong
Look at D: round 3 score - round 1 score =-3?
-2-1 = -3 change, matches with -3 so it is correct.
<em>So</em><em> </em><em>the</em><em> </em><em>right</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>1</em><em>4</em><em>.</em>
<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em>
<em>H</em><em>ope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>you</em>
<em>Good</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>
It is a bit tedious to write 6 equations, but it is a straightforward process to substitute the given point values into the form provided.
For segment ab. (x1, y1) = (1, 1); (x2, y2) = (3, 4).
... x = 1 + t(3-1)
... y = 1 + t(4-1)
ab = {x=1+2t, y=1+3t}
For segment bc. (x1, y1) = (3, 4); (x2, y2) = (1, 7).
... x = 3 + t(1-3)
... y = 4 + t(7-4)
bc = {x=3-2t, y=4+3t}
For segment ca. (x1, y1) = (1, 7); (x2, y2) = (1, 1).
... x = 1 + t(1-1)
... y = 7 + t(1-7)
ca = {x=1, y=7-6t}
-2, -3, -4.
an integer is a number that isn't a fraction. so literally any natural number like 1, 0, or -1. all integers.
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