The midpoint of JK¯¯¯¯¯¯¯¯is M(6, 3). One endpoint is J(14, 9). Find the coordinates of endpoint K.
1 answer:
You might be interested in
Answer:
Step-by-step explanation:
(2(6) - 3(10)^2)/(|6 - 10)|
(12 - 300)/|-4|
-288/4 = -72
answer is c
Answer:
5x maybe dont take my word
Answer:
-86
Step-by-step explanation:
-8((-11/4)-8)
((11/4-8)
8x(-43/4)
-2×43=-86
Answer:
A) v = T/2 - 5
Step-by-step explanation:
T = 10+2v
2v = T-10
v = (T-10)/2 = T/2-5
CscA = 1/sinA
CosACscA = cosA × 1/sinA = cosA/SinA = cotA
