Take a look at the attachment below. It fills in for the attachment that wasn't provided in the question -
An oblique pyramid is one that has a top not aligned with the base. Due to this, the height of the pyramid connects with two vertices at its ends to form a right angle present outside the pyramid, knowing that it is always perpendicular to the base. There is no difference between the calculations of the volume of an oblique pyramid and a pyramid however -
![\\Base Area = 2 cm * 2 cm = 4 cm^2,\\Volume ( Pyramid ) = 1 / 3 * ( Base Area ) * ( Height ),\\Volume = 1 / 3 * ( 4 ) * ( 3.75 ),\\-------------------------\\Volume = 5 cm^3](https://tex.z-dn.net/?f=%5C%5CBase%20Area%20%3D%202%20cm%20%2A%202%20cm%20%3D%204%20cm%5E2%2C%5C%5CVolume%20%28%20Pyramid%20%29%20%3D%201%20%2F%203%20%2A%20%28%20Base%20Area%20%29%20%2A%20%28%20Height%20%29%2C%5C%5CVolume%20%3D%201%20%2F%203%20%2A%20%28%204%20%29%20%2A%20%28%203.75%20%29%2C%5C%5C-------------------------%5C%5CVolume%20%3D%205%20cm%5E3)
<u><em>And thus, you're solution is 5 cm^3, or in other words option b!</em></u>
Answer:
Oh this is easy it is 6 so yea I'm right
Answer:
its c .Both equations have the same potential solutions, but equation A might have extraneous solutions.
Step-by-step explanation:
just took the test
Answer:
Width=20 feet
Since Length=Width=20 feet, the rectangle is a Square.
Step-by-step explanation:
Area, ![A=40w - w^2](https://tex.z-dn.net/?f=A%3D40w%20-%20w%5E2)
To determine the width of the rectangle that gives the maximum area, we take the derivative of A and solve for its critical point.
![A'=40 - 2w\\$When A'=0\\40-2w=0\\40=2w\\w=20 feet](https://tex.z-dn.net/?f=A%27%3D40%20-%202w%5C%5C%24When%20A%27%3D0%5C%5C40-2w%3D0%5C%5C40%3D2w%5C%5Cw%3D20%20feet)
The width of the rectangle that gives the maximum area =20 feet.
Perimeter of a rectangle=2(l+w)
Perimeter of the rectangle=80 feet
2(l+w)=80
2l+2(20)=80
2l=80-40
2l=40
l=20 feet
Since the length and width are equal, the special type of rectangle that produces this maximum area is a Square.
<span>You are given the waiting times between a subway departure schedule and the arrival of a passenger that are uniformly distributed between 0 and 6 minutes. You are asked to find the probability that a randomly selected passenger has a waiting time greater than 3.25 minutes.
Le us denote P as the probability that the randomly selected passenger has a waiting time greater than 3.25 minutes.
P = (length of the shaded region) x (height of the shaded region)
P = (6 - 3.25) * (0.167)
P = 2.75 * 0.167
P = 0.40915
P = 0.41</span><span />