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:
80°
Step-by-step explanation:
exterior angle = sum of two remote interior angles
27 + 53 = 80
Answer:
yes
Step-by-step explanation:
Answer:distributive
Step-by-step explanation:
The distance of plane from the starting point is 69.46 km.
Given that the plane first flies 35 km in the direction N30W and then 60 km in the direction N60E.
We have to find the distance of plane from starting point.
We have to use pythagoras theorem in this which says that the square of hypotenuse is equal to sum of squares of the base and perpendicular.

From the figure we can say that P=35 km .and B=60 km
In this we have to find the value of H when P=35 and B=60

=1225+3600
=6946
H=69.46
Hence the distance of plane from starting point is 69.46 km.
Learn more about pythagoras theorem at brainly.com/question/343682
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