The rock will hit the ground at 2 seconds.
Explanation:
Given that the height, h, in feet of a rock above the ground is given by the equation
, where t is the time in seconds after the rock is thrown.
We need to determine the time at which the rock will hit the ground.
To determine the time, let us equate the height h = 0 in the equation
, we get,

Switch sides, we have,

Let us solve the equation using the quadratic formula.
Thus, we have,

Simplifying, we get,



Hence, the two values of t are
and 
Simplifying the values, we get,
and 
Dividing, we get,
and 
Since, t cannot negative values, we have,

Thus, the rock will hit the ground at 2 seconds.
Step-by-step explanation:
sorry jesica I don't know the answer
Answer:
401
Step-by-step explanation:
1. Approach
To solve this problem, one first has to think about the given figure in a certain way. In the figure, one can see that it is a circle attached on either side of a rectangle. To find the perimeter of the figure, one has to find the circumference of the circle and then add two sides of the rectangle to the answer
2. Circumference of the circle
The formula to find the circumference of a circle is;
(pi) or
(pi)
~ diameter times the value (pi)
Normally to find the circumference of a semicircle, one would have to divide this formula by 2, but since in this case, one has to add two congruent semicircles, so therefore, the effect of dividing the equation by two, only to multiply by two again cancels, and hence, there is no need to divide by 2.
Substitute in the values;
(78)(pi)
~ 245
3. Find the perimeter of the entire object
Now, one has to add the two additional sides of the figure, to the circumferences of the semicircles to get the final answer;
78 + 78 + 245
= 401
Answer:
ΔV = 0.36π in³
Step-by-step explanation:
Given that:
The radius of a sphere = 3.0
If the measurement is correct within 0.01 inches
i.e the change in the radius Δr = 0.01
The objective is to use differentials to estimate the error in the volume of sphere.
We all know that the volume of a sphere

The differential of V with respect to r is:

dV = 4 πr² dr
which can be re-written as:
ΔV = 4 πr² Δr
ΔV = 4 × π × (3)² × 0.01
ΔV = 0.36π in³