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.
<h3>
Answer: 216 square feet</h3>
Work Shown:
Old area = 12*15 = 180 square feet
New area = 18*22 = 396 square feet
Difference = New - Old = 396 - 180 = 216
The patio's area increased by 216 square feet.
Answer:
area = ‐11
Step-by-step explanation:
hope this is correct and helps
We have an isosceles triangle;
A=opposite angle side a.
B=opposite angle side b.
C=opposite angle side c.
A=B
Method 1:
We can divide the isosceles triangle in two right triangles,
hypotenuse=7
side=9/2=4.5
B=A=arccossine (4.5/7)=49.994799...º≈50º
C/2=90º-50º=40º ⇒ C=2*40º=80º
Answer:
a=7; A=50º
b=7; B=50º
<span>c=9; C=80º
Method 2:
Law of cosines:
a²=b²+c²-2bcCosA ⇒CosA=(a²-b²-c²)/(-2bc)
CosA=(49-49-81) / (-126)=0.642857
A=arco cos (81/126)≈50º
B=A=50º
A+B+C=180º
50º+50º+C=180º
C=180º-100º
C=80º
Answer:
</span>a=7; A=50º
b=7; B=50º
<span>c=9; C=80º</span>
