Answer:
All the points that lie on the line:
Step-by-step explanation:
In order to find the maximum rate of change, we need to find the gradient of the function. The gradient of a function of two variables is defined to be:
So:
Hence:
Since we need to find all the points at which the direction of fastest change of the function 
is 
. Then:
Therefore:
So, we can conclude, that all the points where the direction of fastest change of 
lie on the line:
Answer:
Step-by-step explanation:
The position function is
and if we are looking for the time(s) that the ball is 10 feet above the surface of the moon, we sub in a 10 for s(t) and solve for t:
and
and factor that however you are currently factoring quadratics in class to get
t = .07 sec and t = 18.45 sec
There are 2 times that the ball passes 10 feet above the surface of the moon, once going up (.07 sec) and then again coming down (18.45 sec).
For part B, we are looking for the time that the ball lands on the surface of the moon. Set the height equal to 0 because the height of something ON the ground is 0:
and factor that to get
t = -.129 sec and t = 18.65 sec
Since time can NEVER be negative, we know that it takes 18.65 seconds after launch for the ball to land on the surface of the moon.
It is.
To be divisible by 6, it has to be divisible by 2 and 3.
It's an even number so divisible by 2.
The sum of the digits is divisible by 3 so number is divisible by 3.
<span>Z=15+2(x+y)
Use distributive property
Z= 15 + 2x + 2y
Subtract 2y from both sides
Z - 2y= 15 + 2x
Subtract 15 from both sides
Z - 2y - 15= 2x
Divide 2 on both sides
Final Answer: Z - 2y - 15(All over 2)= x</span>
It is 1595. Put it on excel so you can better have an overview how it is done. You need 5 columns, column A-E. On column A, put 1 to 5 in rows 1 to 5. On column C, put the answers: 2, 7, 26, 101 and 400. On column D, put number 3 in all 5 rows. Back on column B, insert =sum(C1-A1) on row 1 and copy till row 5. On column E, insert =sum(B1+C1*D1) on row 1 and copy till row 5 and you will get 1595 at the end.
