Answer:
m<ACB = 59degrees
Step-by-step explanation:
To get <ACB, first we need arcAB
arcAB + arcAC + arcBC = 360
arcAB + 89 + 153 = 360
arcAB+242 = 360
arcAB = 360 - 242
arcAB = 118
Also, m<ACB = 1/2arcAB
m<ACB = 1/2*118
m<ACB = 59degrees
Answer:
4, 2, 1, 2, 4
Step-by-step explanation:
-2 <= 0, so use the first equation, f(x) = (1/2)^x, to find f(-2).
f(-2) = (1/2)^-2 = 1^-2/2^-2 = 2^2/1^2 = 4/1 = 4
-1 <= 0, so use the first equation, f(x) = (1/2)^x, to find f(-1).
f(-1) = (1/2)^-1 = 1^-1/2^-1 = 2^1/1^1 = 2/1 = 2
0 <= 0, so use the first equation, f(x) = (1/2)^x, to find f(0).
f(0) = (1/2)^0 = 1^0/2^0 = 1/1 = 1
1 > 0, so use the second equation, f(x) = 2^x, to find f(1).
f(1) = 2^1 = 2
2 > 0, so use the second equation, f(x) = 2^x, to find f(2).
f(2) = 2^2 = 4
The interquartile range is 5!
Lol
Try to calculate them in your head and if that doesn't help use a calculator. I hope this helps
