During a week:
Kevin works 3z during 5 days
kevin work (4z-7) during 1 day.
Kevin work 0 during 1 day.
z=unit time
f(z)= number of hours kevin works in one week.
f(z)=5(3z)+(4z-7)+0
f(z)=15z+4z-7
f(z)=19z-7
The number of hours kevin woirs in one week in terms of z is:
f(z)=19z-7
The second question is biased
Answer: 8
<u>Step-by-step explanation:</u>
EQ1: a + b + c = 2 --> b + c = 2 - a
EQ2: ab + bc + ac = -1 --> b + c = (-1 - bc)/a
EQ3: abc = -2 --> bc = -2/a
Set EQ1 = EQ2 and substitute bc using EQ3 to solve for "a":

Consider the solution a = 2 and plug it into EQ1 to solve for "b"
b + c = 2 - 2
b + c = 0
b = -c
Plug in a = 2, b = -c, and c = c into a³ + b³ + c³
2³ + (-c)³ + c³
= 8 - c³ + c³
= 8
Answer:
7.5
Step-by-step explanation: