The feasible region is the region that comes from intersecting the different regions described by the equations that constrain the problem.
Let's take a look:
From this, we can conclude that the vertices of the feasible region are:
I believe the answer is 116.
23-8=15
8*15=120
sqrt2=4
the number 120-4=116
that’s what i got not sure if that is correct
It is easy the left side as to be ten times greater then your right side for example my ones place is ten times greater my tens place is ten times greater than my hundreds place and my hundreds place is ten times greater than my thousands place i hope this will help
Answer:
a = 4
Step-by-step explanation: Given that
ar=2 where a is the first term and is the common ratio.
ar=2 ............1
Also give that a/1-r = 8 where a is the first term and r is the common ratio as well.
a/1-r = 8 ......... 2
From equation 1 make r the subject
ar=2
r= 2/a...........3
Put equation 3 into 2
a/1-2/a =8
a÷1-2/a
a = 8(1-2/a)
a= 8/1-16/a
a/1= 8a-16
a^2= 8a-16
Rewriting the above equation
a^2-8a+16=0
This will eventually be
a^2-8a+16=0
(a-4)^2= 0
a =4
