1.) x = -3...this is a vertical line with an undefined slope. A parallel line to a vertical line is also a vertical line. A vertical line is represented by x = a number....that number being the x coordinate of ur point(4,2).
so ur parallel line is : x = 4 <==
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2.) y = -3/2x + 6....slope here is -3/2. A parallel line will have the same slope.
y = mx + b
slope(m) = -3/2
(2,-1)...x = 2 and y = -1
now we sub into the formula and find b, the y int
-1 = -3/2(2) + b
-1 = -3 + b
-1 + 3 = b
2 = b
so ur parallel line is : y = -3/2x + 2 <==
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y = -x - 2....the slope here is -1. A parallel line will have the same slope
y = mx + b
slope(m) = -1
(2,-2)...x = 2 and y = -2
now we sub into the formula and find b, the y int
-2 = -1(2) + b
-2 = -2 + b
-2 + 2 = b
0 = b
so ur parallel equation is : y = -x + 0 which can be written as : y = -x <=
3,500 it is that your welcome I have the same problem right now
If
then the derivative is
Critical points occur where 
. This happens for
In the first case, we find
In the second,
So all the critical points occur at multiples of 
, or 
. (This includes all the even multiples of .)
Answer:
9.6
Step-by-step explanation:
10 = 8
12 = x
then simply solve for x
I think it is B, but I'm not fully sure.
