Answer: uh I want to help you but what work I don’t see your question or- did you put a picture...
Step-by-step explanation:
For odd function f(-x)= -f(x)
1) f(-x)= sin(-x) = -sin(x)=-f(x) sin is odd function
2) f(-x) = sin(-2x) = -sin 2x=-f(x)
3) f(-x)= (-x)³+1 = -x³ +1 (not an odd function)To be an odd function, it should be like this:if a function is (x³+1), to be odd f(-x) should be -(x³+1)=-x³-14) f(x)= x/(x²+1)
f(-x) = (-x)/((-x)²+1)= -x/(x²+1)=-f(x)
So, f(-x) gives
,
that means that f(x)= x/(x²+1) is an odd function.
5)f(x) = ∛(2x)
f(-x) = ∛(2*(-x)=∛(2x*(-1)) = ∛(2x)*∛(-1)=
- ∛(2x)= -f(x)
Part A;
There are many system of inequalities that can be created such that only contain points A and E in the overlapping shaded regions.
Any system of inequalities which is satisfied by (2, -3) and (3, 1) but is not satisfied by (-3, -4), (-4, 2), (2, 4) and (-2, 3) can serve.
An example of such system of equation is
y ≤ x
y ≥ -2x
The system of equation above represent all the points in the first quadrant of the coordinate system.
The area above the line y = -2x and to the right of the line y = x is shaded.
Part B:
It can be verified that points A and E are solutions to the system of inequalities above by substituting the coordinates of points A and E into the system of equations and see whether they are true.
Substituting A(2, -3) into the system we have:
-3 ≤ 2
-3 ≥ -2(2) ⇒ -3 ≥ -4
as can be seen the two inequalities above are true, hence point A is a solution to the set of inequalities.
Also, substituting E(3, 1) into the system we have:
1 ≤ 3
1 ≥ -2(3) ⇒ 1 ≥ -6
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 William can only attend a school in her designated zone and that William's zone is defined by y < −x - 1.
To identify the schools that William is allowed to attend, we substitute the coordinates of the points A to F into the inequality defining William's zone.
For point A(2, -3): -3 < -(2) - 1 ⇒ -3 < -2 - 1 ⇒ -3 < -3 which is false
For point B(-3, -4): -4 < -(-3) - 1 ⇒ -4 < 3 - 1 ⇒ -4 < 2 which is true
For point C(-4, 2): 2 < -(-4) - 1 ⇒ 2 < 4 - 1 ⇒ 2 < 3 which is true
For point D(2, 4): 4 < -(2) - 1 ⇒ 4 < -2 - 1 ⇒ 4 < -3 which is false
For point E(3, 1): 1 < -(3) - 1 ⇒ 1 < -3 - 1 ⇒ 1 < -4 which is false
For point F(-2, 3): 3 < -(-2) - 1 ⇒ 3 < 2 - 1 ⇒ 3 < 1 which is false
Therefore, the schools that Natalie is allowed to attend are the schools at point B and C.
B about 4,000 because 6,963 rounds to 7,000 and 3,098 rounds to 3,000 - 7,000-3,000= 4,000
B. $27,000
I hope helped ^-^
