The base of a cube is parallel to the horizon. If the cube is cut by a plane to form a cross section, under what circumstance wo
uld it be possible for the cross section be a non-rectangular parallelogram?
2 answers:
This question has this set of answer choices:
a) when the plane cuts three faces of the cube, separating one corner from the others
b) when the plane passes through a pair of vertices that do not share a common face
c) when the plane is perpendicular to the base and intersects two adjacent vertical faces
d) when the plane makes an acute angle to the base and intersects three vertical faces
e) not enough information to answer the question
The right answer is the first choice: a) when the plane cuts three faces of the cube, separating one corner from the others
You can see a picture of this case in the figure attached: as you can see the cross section (in pink) is a triangle.
a) when the plane cuts three faces of the cube, separating one corner from the others
b) when the plane passes through a pair of vertices that do not share a common face
c) when the plane is perpendicular to the base and intersects two adjacent vertical faces
d) when the plane makes an acute angle to the base and intersects three vertical faces
e) not enough information to answer the question
The right answer is the first choice: a) when the plane cuts three faces of the cube, separating one corner from the others
You can see a picture of this case in the figure attached: as you can see the cross section (in pink) is a triangle.
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Answer:
Step-by-step explanation:
Hello!
You need to construct a 95% CI for the population mean of the length of engineering conferences.
The variable has a normal distribution.
The information given is:
n= 84
x[bar]= 3.94
δ= 1.28
The formula for the Confidence interval is:
x[bar]±*(δ/n)
Lower bound(Lb): 3.698
Upper bound(Ub): 4.182
Error bound: (Ub - Lb)/2 = (4.182-3.698)/2 = 0.242
I hope it helps!
The firts thig we are going to do is create tow triangles using the angles of elevation of Paul and Jose. Since the problem is not giving us their height we'll assume that the horizontal line of sight of both of them coincide with the base of the tree.
We know that Paul is 19m from the base of the tree and its elevation angle to the top of the tree is 59°. We also know that the elevation angle of Jose and the top of the tree is 43°, but we don't know the distance between Paul of Jose, so lets label that distance as
.
Now we can build a right triangle between Paul and the tree and another one between Jose and the tree as shown in the figure. Lets use cosine to find h in Paul's trianlge:
Now we can use the law of sines to find the distance between Paul and Jose:
Now that we know the distance between Paul and Jose, the only thing left is add that distance to the distance from Paul and the base of the tree:
We can conclude that Jose is 33.9m from the base of the tree.
We know that Paul is 19m from the base of the tree and its elevation angle to the top of the tree is 59°. We also know that the elevation angle of Jose and the top of the tree is 43°, but we don't know the distance between Paul of Jose, so lets label that distance as
Now we can build a right triangle between Paul and the tree and another one between Jose and the tree as shown in the figure. Lets use cosine to find h in Paul's trianlge:
Now we can use the law of sines to find the distance
Now that we know the distance between Paul and Jose, the only thing left is add that distance to the distance from Paul and the base of the tree:
We can conclude that Jose is 33.9m from the base of the tree.