D -8\4
When you simplify the fraction
it becomes 2
(Sin A/ secA) × ( sec A / cos A) = Sin A / cos A = tan A
Question should be cos A ÷ sec A
Answer:
(2, 1)
Step-by-step explanation:
A line that passes through the origin and has a slope of 1/2 can be best represented as y = 1/2x
it starts at the origin
goes 2 units to the right
1 unit above
2 units to the right + 1 unit above = (2, 1)
Final Answer: (2, 1)
The length of arc PS from the information given in the task content is; 31.5 ft.
<h3>What is the length of the arc PS?</h3>
It follows from the task content that;
- Point T is the center of the circle and line TR is the radius of the circle.
- Additionally, the angle subtended by arcPS = 180 - m<PS = 180 - 16 = 164⁰. This follows from the fact that line PR is a diameter.
On this note, the length of arc PS is;
arc PS = (164/360) × 2 × 3.14 × 11.
arc PS = 31.5ft.
Read more on length of an arc;
brainly.com/question/2005046
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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.