According to a Pew Research Center report from 2012, the average commute time to work in California is 27.5 minutes. To investig
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Answer:
a) The tower is 90 feet tall
b) She reaches the bottom at t = 18 minutes.
c) Her speed at time t is 5 \sqrt[]{5} ft/minute
d) Her acceleration at time t is 10 ft/minute^2
Step-by-step explanation:
Consider the path described by the child as going down the tower to have the following parametrization
a) Assuming that the child is at the top of the tower when she starts going down, we have that at the initial time (t=0) we will have the value of the height of the tower. That is z = 90-5*0 = 90 ft.
b) The child reaches the bottom as soon as z =0. We want to find the value of t that does that. Then we have 0 = 90-5t, which gives us t = 18 minutes.
c) Given the parametrization we are given, the velocity of the child at time t is given by . The speed is defined as the norm of the velocity vector,
so, the speed at time t is given by
d) ON the same fashion we want to know the norm of the second derivative of .
We have that so the acceleration is given by
First of all, a bit of theory: since the area of a square is given by
where s is the length of the square. So, if we invert this function we have
.
Moreover, the diagonal of a square cuts the square in two isosceles right triangles, whose legs are the sides, so the diagonal is the hypothenuse and it can be found by
So, the diagonal is the side length, multiplied by the square root of 2.
With that being said, your function could be something like this:
double diagonalFromArea(double area) {
double side = Math.sqrt(area);
double diagonal = side * Math.sqrt(2);
return diagonal;
}
Hello,
y=5+3*cos (2(x-π/3))
The function is periodic with periode=2π.
-1<=cos (2(x-π/3))<=1
==>-1*3<=3*cos (2(x-π/3))<=3*1
==>5-3<=5+3cos(2(x-π/3))<=5+3
==>2<= y<=8
y=5+3*cos (2(x-π/3))
The function is periodic with periode=2π.
-1<=cos (2(x-π/3))<=1
==>-1*3<=3*cos (2(x-π/3))<=3*1
==>5-3<=5+3cos(2(x-π/3))<=5+3
==>2<= y<=8