Answer: The degree of the first term.
Explanation:
The function:

The left and right ends would be indicated when x is changed to -x. When this is substituted, the change is indicated by the first term because only the degree of first term is odd.
Let the left hand side be donated by -x.
Then,

Hence, the correct option is the degree of the first term indicates the left and right end points of the function.
Answer:
<em>The final speed of the second package is twice as much as the final speed of the first package.</em>
Explanation:
<u>Free Fall Motion</u>
If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

And the distance traveled downwards is:

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

Replacing into the first equation:

Rationalizing:

Let's call v1 the final speed of the package dropped from a height H. Thus:

Let v2 be the final speed of the package dropped from a height 4H. Thus:

Taking out the square root of 4:

Dividing v2/v1 we can compare the final speeds:

Simplifying:

The final speed of the second package is twice as much as the final speed of the first package.
Answer:
5.78amps
Explanation:
Given data
Time t= 57 seconds
Charge Q= 330C
Current I= ??
The expression for the electric current is given as
Q= It
Substituting we have
330= I*57
I= 330/57
I=5.78 amps
Hence the current is 5.78amps
ITS C
This element tends to lose 2 electrons to become a 2+ ion, is the correct statement regarding the element calcium. Calcium has 2 electrons in its outer shell and it is easier to lose them than it is to gain enough to become stable. When stable it has 2 more protons than electrons forming a 2+ ion.