Answer:
A. The next 3 terms are —9, —11, —13
B. The sequence is Arithmetic
C. The sequence has a common difference of —2
Common difference = T2 — T1 or = T3 — T2
=T2 — T1 = —3 —(—1) = —2 or
= T3 — T2 = —5 —(—3) = —2
D. Tn = a + d(n—1)
a = first term = —1
d = Common difference = —2
n = 27
T27 = —1 + —2(27—1)
T27 = —1 + —2(26)
T27 = —1 —52
T27 = —53
Part A;
There are many system of inequalities that can be created such that only contain points D and E in the overlapping shaded regions.
Any system of inequalities which is satisfied by (-4, 2) and (-1, 5) but is not satisfied by (1, 3), (3, 1), (3, -3) and (-3, -3) can serve.
An example of such system of equation is
x < 0
y > 0
The system of equation above represent all the points in the second quadrant of the coordinate system.The area above the x-axis and to the left of the y-axis is shaded.
Part B:It can be verified that points D and E are solutions to the system of inequalities above by substituting the coordinates of points D and E into the system of equations and see whether they are true.
Substituting D(-4, 2) into the system
we have:
-4 < 0
2 > 0
as can be seen the two inequalities above are true, hence point D is a solution to the set of inequalities.
Also, substituting E(-1, 5) into the system we have:
-1 < 0
5 > 0
as can be seen the two inequalities above are true, hence point E is a solution to the set of inequalities.
Part C:Given that chicken can only be raised in the area defined by y > 3x - 4.
To identify the farms in which chicken can be raised, we substitute the coordinates of the points A to F into the inequality defining chicken's area.
For point A(1, 3): 3 > 3(1) - 4 ⇒ 3 > 3 - 4 ⇒ 3 > -1 which is true
For point B(3, 1): 1 > 3(3) - 4 ⇒ 1 > 9 - 4 ⇒ 1 > 5 which is false
For point C(3, -3): -3 > 3(3) - 4 ⇒ -3 > 9 - 4 ⇒ -3 > 5 which is false
For point D(-4, 2): 2 > 3(-4) - 4; 2 > -12 - 4 ⇒ 2 > -16 which is true
For point E(-1, 5): 5 > 3(-1) - 4 ⇒ 5 > -3 - 4 ⇒ 5 > -7 which is true
For point F(-3, -3): -3 > 3(-3) - 4 ⇒ -3 > -9 - 4 ⇒ -3 > -13 which is true
Therefore, the farms in which chicken can be raised are the farms at point A, D, E and F.
Answer:
YZ = 44
Step-by-step explanation:
Let the intersection of XW and YZ equal A. For this problem, we can use power of a point in the scenario where two secants intersect in the interior of a circle. We get the equation:
WA * AX = ZA * AY
We can substitute the values we are given to get:
(2x - 6)(21) = (x + 1)(30)
42x - 126 = 30x + 30
12x - 126 = 30
12x = 156
x = 13
Using segment addition postulate, we can say that:
YZ = YA + AZ
And we can plug in what we know from the diagram:
YZ = 30 + x + 1
We can also plug in the value of x, 13
YZ = 30 + 13 + 1 = 44
