There are many systems of equation that will satisfy the requirement for Part A.
an example is y≤(1/4)x-3 and y≥(-1/2)x-6
y≥(-1/2)x-6 goes through the point (0,-6) and (-2, -5), the shaded area is above the line. all the points fall in the shaded area, but
y≤(1/4)x-3 goes through the points (0,-3) and (4,-2), the shaded area is below the line, only A and E are in the shaded area.
only A and E satisfy both inequality, in the overlapping shaded area.
Part B. to verify, put the coordinates of A (-3,-4) and E(5,-4) in both inequalities to see if they will make the inequalities true.
for y≤(1/4)x-3: -4≤(1/4)(-3)-3
-4≤-3&3/4 This is valid.
For y≥(-1/2)x-6: -4≥(-1/2)(-3)-6
-4≥-4&1/3 this is valid as well. So Yes, A satisfies both inequalities.
Do the same for point E (5,-4)
Part C: the line y<-2x+4 is a dotted line going through (0,4) and (-2,0)
the shaded area is below the line
farms A, B, and D are in this shaded area.
Answer:
π = V/(2r²+2rh)
Step-by-step explanation:
"Solving" for something means to isolate it on one side of an equation. Because both terms on the right side of the equation have π in common, we can factor it out. Then from there, you just divide the rest of it on both sides to isolate π. It should look like this:
V=2πr^2 + 2πrh
V=π(2r² + 2rh)
V/(2r² + 2rh)=π
Perpendicular has slope that multiplies to -1 in other sloope
thsi one
2x-5y=-35
-5y=-2x-35
y=2/5x+7
slope=2/5
2/5 times -5/2=-1
y=-5/2x+b
findn b
(10,4)
(x,y)
4=-5/2(10)+b
4=-25+b
21=b
y=-5/2x+21
2y=-5x+42
5x+2y=42
Answer:
Barbie dolls
Step-by-step explanation:
