Y = 2x-6
Just move the y to the other side (add y so it gets pushed to the right) then subtract the six from the right to the left side.
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 is C, 25% increase. To find the increase, subtract starting value (780) from the final value (975). It equals out to be 195. Divide 195 by the starting value which turns out to be 0.25. Then, multiply 0.25 by 100 which equals out to be 25.
Answer:
36.89°
Step-by-step explanation:
tan x = 15/20=3/4
x = 36.89
Answer:
0.3277 = 32.77%
Step-by-step explanation:
If we want the probability for Sasha being late after the fifth day, we need that in the first five days she is not late, which has a probability of 1 - 0.2 = 0.8
So, multiplying the probability for each day, we have that:
P = 0.8 * 0.8 * 0.8 * 0.8 * 0.8 = 0.3277 = 32.77%
So we have a probability of 32.77% that Sasha's first delay will occur after the fifth day.
