Answer:
R = sqrt 3 * (V /( pi * h))
V = 62.8
pi = 3.14
h = 15
now we sub
R = sqrt 3 * (62.8 / (3.14 * 15)
R = sqrt 3 * (62.6 / 47.1)
R = sqrt 3 * 1.33
R = sqrt 3.99
R = 1.9 rounds to 2 inches <===
Step-by-step explanation:
Answer:
Step-by-step explanation:
we know that
In the right triangle ABC
The function sine of angle 68 degrees is equal to divide the opposite side SB by the hypotenuse AC
so
substitute the values and solve for AC
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.
The answer to the equation is 115
Hello :
the normal vector of the plane is : d' = [4,-1,5]...(vector perpendicular to the plane)<span>
d </span><span>⊥ d' because : (4)(2)+(-1)(3)+(5)(1)=0
</span>the line is perpendicular to the plane .
