7: since they went down further you go further from zero: -28-40=-68 or D.
8: they lost 6 and then 8 so 6+8=14 but they lost them so -14 or A :)
Recall that converting from Cartesian to polar coordinates involves the identities

As a function in polar coordinates,

depends on

, so you can write

.
Differentiating the identities with respect to

gives

The slope of the tangent line to

is given by

Given

, you have

. So the tangent line to

has a slope of

When

, the tangent line has slope

This line is tangent to the point

which in Cartesian coordinates is equivalent to

, so the equation of the tangent line is

In polar coordinates, this line has equation


The tangent line passes through the y-axis when

, so the y-intercept is

.
The vector from this point to the point of tangency on

is given by the difference of the vector from the origin to the y-intercept (which I'll denote

) and the vector from the origin to the point of tangency (denoted by

). In the attached graphic, this corresponds to the green arrow.

The angle between this vector and the vector pointing to the point of tangency is what you're looking for. This is given by



The second problem is just a matter of computing the second derivative of

with respect to

and plugging in

.



Answer: The team won 47 games.
Step-by-step explanation: The total games played was 84, and that means all games added together, but victories and losses. If the games they lost is called X, and the team won 10 more games than it lost, then it means the games they won is X + 10.
To find the value of wins and losses, we can now express both properly as;
X + (X +10) = 84
2X + 10 = 84
Subtract 10 from both sides of the equation
2X = 74
Divide both sides of the equation by 2
X = 37.
If the number of games won is X + 10, then games won is derived as
Wins = X + 10
Wins = 37 + 10
Wins = 47
The team won 47 games
Answer:6060 foot-pounds
Step-by-step explanation:
Given
Weight of cable(w')=0.6 pound per foot
depth of well=60 foot
weight of wrench=7 Pound
Now work done in raising monkey with wrench
foot-pounds
Now work done in raising rope
robot rises by climbing up the cable with one end of the cable still attached thus robot support half the weight of cable so work done


foot-pounds
foot-pounds
AT this day camp, the greatest number of campers each group can have is 14 campers.