X + 5y = 6
5y = -x + 6
y = -1/5x + 6/5....the slope here is -1/5. A perpendicular line will have a negative reciprocal slope. All that means is flip the original slope and change the sign. So we flip -1/5 and make is -5/1....and we change the sign...making it 5/1 or just 5. So our perpendicular line will have a slope of 5.
y = mx + b
slope(m) = 5
(2,-2)...x = 2 and y = -2
sub and find b, the y int
-2 = 5(2) + b
-2 = 10 + b
-2 - 10 = b
-12 = b
so ur perpendicular equation is : y = 5x - 12 <=
Answer: C
<u>Step-by-step explanation:</u>

= 
= 
= 
= 
Answer:
If `r` and `R` and the respective radii of the smaller and the bigger semi-circles then the area of the shaded portion in the given figure is: (FIGURE) `pir^2\ s qdotu n i t s` (b) `piR^2-pir^2\ s qdotu n i t s` (c) `piR^2+pir^2\ s qdotu n i t s` (d) `piR^2\ s qdotu n i t s`
Step-by-step explanation:
Answer:
its the first one
Step-by-step explanation: