We know that 
and this is the only point when sin and cos are equal lengths. Because both 
Now if the sin of 30° is a half that would mean that cos of 60° is also a half.
Hope this helps.
<u>r3t40</u>
The greatest common factor will be (x² – xy + y²).
<h3>Greatest common factor</h3>
This is a value or expression that can divide the given expressions without leaving a remainder.
Given the following expressions
x^3+^3 and x^2 - xy + y^2
Expand x^3+y^3
x^3+y^3 =(x + y)(x² – xy + y²).
Since (x² – xy + y²) is common to both expression, hence the greatest common factor will be (x² – xy + y²).
Learn more on GCF here: brainly.com/question/902408
#SPJ1
divide the numerator and denominator by 13
=
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.
2 minutes per lap. I think
