0 × <span>0 = 0
0 </span>× <span>1 = 0
1 </span>× <span>1 = 1
Which means multiplication is closed under {0, 1}
</span><span>1 </span>÷ <span>1 = 1
0 </span>÷ <span>1 = 0
</span>
Division is not closed under {0, 1}
1 + 1 = 2
Addition is not closed under {0, 1}
0 - 1 = -1
Subtraction is not closed under {0, 1} either
So it's only A. Multiplication which is closed under {0, 1}
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.
Exact Surface Area = 378pi
Approximate Surface Area = 1186.92
The approximate surface area uses pi = 3.14
The units are cm^2 or "square cm".
=========================================================
Work Shown:
SA = surface area of cylinder
SA = 2*pi*r^2 + 2*pi*r*h
SA = 2*pi*7^2 + 2*pi*7*20
SA = 2*pi*49 + 2*pi*140
SA = 2*49*pi + 2*140*pi
SA = 98pi + 280pi
SA = 378pi ..... exact surface area
SA = 378*3.14
SA = 1186.92 ..... approximate surface area
----------
Side note: The diameter 14 cuts in half to get the radius r = 7
Any value which is more than 2 standard deviations away from the mean is considered to be "unusual."
2 standard deviations above the mean 52.4 mp would be 52.4+2(1.8), or 56; 2 std devs below the mean would be 52.4 - 2(1.8), or 48.8. Thus, any value larger than 56 or any value smaller than 48.8 would be "unusual."
54.8, 49.1 and 51.3 are not unusual; 56.5 is unusual, because it's greaster than 56.
