Answer:
Given a square ABCD and an equilateral triangle DPC and given a chart with which Jim is using to prove that triangle APD is congruent to triangle BPC.
From the chart, it can be seen that Jim proved that two corresponding sides of both triangles are congruent and that the angle between those two sides for both triangles are also congruent.
Therefore, the justification to complete Jim's proof is "SAS postulate"
Step-by-step explanation:
Y > 6x + 5
y < -6x + 7
(2,17)
17 > 6(2)+5 ; 17> 17 NO
(8,8)
8>6(8)+5 NO
(-2,18)
18>6(-2)+5
18>-7 YES
18< -6(-2)+7
18< 12+7 YES (-2,18) is an answer
(0,6)
6>6(0)+5
6>5 YES
6<-6(0)+7
6<7 YES (0,6) is an answer
(-2,19)
19>6(-2)+5
19>-7 NO
(2,18)
18>6(2)+5
18>17 YES
18< -6(2)+7
18< -5 NO
Answer:
y=-3/16(x-8)^2+12
Step-by-step explanation:
Refer to the vertex form equation for a parabola:
y=a(x-h)^2+k where (h,k) is the vertex.
Therefore, we have y=a(x-8)^2+12 as our equation so far. If we plug in (16,0) we can find a:
0=a(16-8)^2+12
0=64a+12
-12=64a
-12/64=a
-3/16=a
Therefore, your final equation is y=-3/16(x-8)^2+12
The answer is the last one
