Answer:
A = 62.35 cm²
Step-by-step explanation:
Use the area formula A = 
, where a is the side length.
Plug in the values:
A = 
A = 
A = 62.35 cm²
The fractions which cannbe formed are: 2/3, 3/2, 2/5, 5/2, 3/5 and 5/3.
Now, we find the products with the third number:
2/3*5= 10/3 = 3.33 > 2
3/2*5 = 15 >2= 7.5 > 2
2/5*3= 6/5 (rejected)
5/2*3 = 15/2 > 2
3/5*2= 6/5 (rejected)
5/3*2 = 15/2 = 7.5> 2
25, 7*4 = 28 32-28=4 and 42/6=7*3 = 21 + 4 = 25
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:
B
Step-by-step explanation:
if she practiced for 4 hours, we can go to the 4-hour mark and go upward until we see the dot. the dot lands on 3 hits every 4 hours so we can safely assume its B
