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:
Volume of a cylinder = pi*(xx)^2*yy
= x^4*pi*yy
Volume of a cone = 1/3*pi*(9x^2)^2*h
= 27pi*x^4*h
x^4*pi*y^2 = 27pi*x^4*h
h = y^2 / 27
Answer:
x = 25/3; y = -5/3
Step-by-step explanation:
You did well so far by multiplying both sides of the first equation by 2.
Now notice that you have -2y in the new first equation and 2y in the second equation. -2y and 2y add to 0.
Now add the equations to eliminate y.
2x - 2y = 20
(+) 4x + 2y = 30
-------------------------
6x = 50
6x/6 = 50/6
x = 50/6 = 25/3
2x - 2y = 20
2(25/3) - 2y = 20
50/3 - 2y = 60/3
-2y = 60/3 - 50/3
-2y = 10/3
-2y/2 = (10/3)/(-2)
y = -5/3
Solution: x = 25/3; y = -5/3
Answer:
The HA Theorem states; If the hypotenuse and an acute angle of a right triangle are congruent to the hypotenuse and an acute angle of another triangle, then the two triangles are congruent. ... They are both facing with their hypotenuses to the right, which means their right angles are to the left -- HA!
Step-by-step explanation:
Step-by-step explanation:
b ha ok like and follow please
