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:
Step-by-step explanation:
(Assuming the correct angles are 30° and 45°)
We can use the tangent relation of the angle of elevation to find two equations, then we can use these equations to find the height of the pole.
Let's call the initial distance of the boy to the pole 'x'.
Then, with an angle of elevation of 30°, the opposite side to this angle is the height of the pole (let's call this 'h') minus the height of the boy, and the adjacent side to the angle is the distance x:
Then, with an angle of elevation of 45°, the opposite side to this angle is still the height of the pole minus the height of the boy, and the adjacent side to the angle is the distance x minus 10:
So rewriting both equations using the tangents values, we have that:
From the first equation, we have that:
Using this value of x in the second equation, we have that:
I think the answer is 229.1831. hope this helps!
Answer: 1) All parallelograms are quadrilaterals.
Step-by-step explanation:
It would be 50%. because half numbers and half letters
