Answer:
x=6 y=5
Step-by-step explanation:
Step-by-step explanation:
I am not sure what your problem here is.
you understand the inequality signs ?
anyway, to get
6×f(-2) + 3×g(1)
we can calculate every part of the expression separately, and then combine all the results into one final result.
f(-2)
we look at the definition.
into what category is -2 falling ? the one with x<-2, or the one with x>=-2 ?
is -2 < -2 ? no.
is -2 >= -2 ? yes, because -2 = -2. therefore, it is also >= -2.
so, we have to use
1/3 x³
for x = -2 that is
1/3 × (-2)³ = 1/3 × -8 = -8/3
g(1)
again, we look at the definition.
into what category is 1 falling ? the one with x > 2 ? or the one with x <= 1 ?
is 1 > 2 ? no.
is 1 <= 1 ? yes, because 1=1. therefore it is also <= 1.
so we have to use
2×|x - 1| + 3
for x = 1 we get
2×0 + 3 = 3
6×f(-2) = 6 × -8/3 = 2× -8 = -16
3×g(1) = 3× 3 = 9
and so in total we get
6×f(-2) + 3×g(1) = -16 + 9 = -7
Answer:
The answer is C. (1.1)
Step-by-step explanation:
P=2 then 2^2+4=8
P=11 then 11^2+4=125
Answer:
the critical points are (0,0) , (0, 20), (12, 0) , (4,16)
Step-by-step explanation:
To consider the autonomous system


The critical points of the above system can be derived by replacing x' = o and y' = 0.
i.e.


x = 0 or 24 - 2x - y = 0 ----- (1)
Also

y( 20 -y - x) = 0
y = 0 or 20 - y - x = 0 ----- (2)
By solving (1) and (2);
we get x = 4 and y = 16
Suppose x = 0 from (2)
y = 20
Also;
if y = 0 from (1)
x = 12
Thus, the critical points are (0,0) , (0, 20), (12, 0) , (4,16)