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.
Answer:
Step-by-step explanation:
There are only 2 solutions for none of the rooks can capture any of the other rooks: only if they are laid out in either of the 2 diagonal of the chess board
The number of possible combination for the rock to laid out in 8x8 (64 slots) chess board
possible combination
So the probability is pretty thin
Answer:
-1
Step-by-step explanation:
Step 1: Simplify both sides of the equation.
5x=6x+1
Step 2: Subtract 6x from both sides.
5x−6x=6x+1−6x
−x=1
Step 3: Divide both sides by -1.
−x
/-1=1/-1
x=-1
May I get a Brainliest?
8 (y - 7) = - 16
Divide by 8 on both sides
y - 7 = - 2
Add 7 to both sides
y = 5
Square, Rectangle and Rhombus
