A.) To find the maximum height, we can take the derivative of h(t). This will give us the rate at which the horse jumps (velocity) at time t.
h'(t) = -32t + 16
When the horse reaches its maximum height, its position on h(t) will be at the top of the parabola. The slope at this point will be zero because the line tangent to the peak of a parabola is a horizontal line. By setting h'(t) equal to 0, we can find the critical numbers which will be the maximum and minimum t values.
-32t + 16 = 0
-32t = -16
t = 0.5 seconds
b.) To find out if the horse can clear a fence that is 3.5 feet tall, we can plug 0.5 in for t in h(t) and solve for the maximum height.
h(0.5) = -16(0.5)^2 + 16(-0.5) = 4 feet
If 4 is the maximum height the horse can jump, then yes, it can clear a 3.5 foot tall fence.
c.) We know that the horse is in the air whenever h(t) is greater than 0.
-16t^2 + 16t = 0
-16t(t-1)=0
t = 0 and 1
So if the horse is on the ground at t = 0 and t = 1, then we know it was in the air for 1 second.
No it's over : 7.3 , unless your prompts context is close to as in getting there or close in general?
The radius of the cylinder's base is 8, so the volume of the cylinder is
(using the approximation of
).
The sphere has radius 6, so the volume of the sphere is
.
The volume of the shaded region is the difference between the two: 3617.28 - 904.32 = 2712.96. Rounded to the nearest integer gives an answer of 2713.
Answer:
Table 1,
1 16
2 8
3 4
4 2
Step-by-step explanation
The equation for this function can be written as
y = 16^1/x
This is exponential decay.
Answer:
∠ BAC = 80°
Step-by-step explanation:
The sum of the interior angles of quadrilateral ACDB = 360°
DB and DC are tangents to the circle, thus
∠ DBA = ∠ ACD = 90° ( angle between tangent/ circle at point of contact )
Thus
∠ BAC + 90° + 90° + 100° = 360°
∠ BAC + 280° = 360° ( subtract 280° from both sides )
∠ BAC = 80°
