Answer:
the function is continuous from the left at x=1 and continuous from the right at x=0
Step-by-step explanation:
a function is continuous from the right , when
when x→a⁺ lim f(x)=f(a)
and from the left when
when x→a⁻ lim f(x)=f(a)
then since the functions presented are continuous , we have to look for discontinuities only when the functions change
for x=0
when x→0⁺ lim f(x)=lim e^x = e^0 = 1
when x→0⁻ lim f(x)=lim (x+4) = (0+4) = 4
then since f(0) = e^0=1 , the function is continuous from the right at x=0
for x=1
when x→1⁺ lim f(x)=lim (8-x) = (8-0) = 8
when x→1⁻ lim f(x)=lim e^x = e^1 = e
then since f(1) = e^1=e , the function is continuous from the left at x=1
Answer:
yeah guys i also want to know that questian
Find the Reference Angle (4pi)/7 4π 7 4 π 7 Since the angle 4π 7 4 π 7 is in the second quadrant, subtract 4π 7 4 π 7 from π π.
Thus, 
is your answer
23/40 can't be simplified as it has no common factors
