
$=(a^2-10a)-(b^2+6b) +16$
$=[(a^2-2(5)a+25)-25]-[(b^2+2(3)b+9)-9]+16$
$=(a-5)^2-25-(b+3)^2+9+16$
$=(a-5)^2-(b+3)^2$
It would be B. 120 you would usually round up but the closest one to 129.02083 is 130 and since there is no 130 it would be 120
Answer:
[-4,9)
Step-by-step explanation:
im like 99% sure this is right bc i just did a problem like this (:
Answer:
B and C
(I'm not sure if b and e are supposed to be the same though) if they are then, b, c, and e
minimum value: 1
lower quartile: 3
median of the data: 6
upper quartile: 8
maximum value: 14
Sorry for the late answer, i hope this will help you in the future and everyone else who reads this.