The fencing line x is the height of a rectangle triangle of base = y, hypothenuse of 9 m, so we use Pythagoras theorem to solve:
hyp^2 = height^2 + base^2
9^2 = x^2 + y^2
x^2 = 81 - y^2
we can see that x is also the height of another rectangle triangle of base = 15 - y, hypothenuse of 12 m, so we use Pythagoras theorem to solve:
hyp^2 = height^2 + base^2
12^2 = x^2 + (15 - y)^2
lets expand:
144 = x^2 + 225 - 30y + y^2
substitute x^2 from the first equation in the last:
144 = 81 - y^2 + 225 - 30y + y^2
144 = 81 + 225 - 30y
30y = -144 + 81 + 225
y = 5.4 m
substitute in the fence equation:
x^2 = 81 - y^2
x^2 = 81 - 5.4^2
x = 7.2 m that is the length of the fence
C
.) The - 3 at the end of the function means the function is shifted down 3 units. This also shifts the asymptote down 3 units.
Answer:
22 m
Step-by-step explanation:
The distance traveled is ...
d = (1/2)at^2
where "a" is the magnitude of the acceleration.
d = (1/2)(11 m/s^2)(2 s)^2
d = 22 m
The car traveled 22 meters while braking.
First, we need to solve the differential equation.

This a separable ODE. We can rewrite it like this:

Now we integrate both sides.

We get:

When we solve for y we get our solution:

To find out if we have any horizontal asymptotes we must find the limits as x goes to infinity and minus infinity.
It is easy to see that when x goes to minus infinity our function goes to zero.
When x goes to plus infinity we have the following:

When you are calculating limits like this you always look at the fastest growing function in denominator and numerator and then act like they are constants.
So our asymptote is at y=8.
X = 2
4x + 8 = 16
-8 -8
----------------
4x= 8
-- --
4 4
------------
x=2