A-students taking band
B-students taking chorus
|AuB| = 250
|A| = 180
[AnB] = 60
|AuB| = |A| + |B| - |AnB|
|B| = |AuB| - |A| + |AnB| = 250 - 180 + 60 = 130
ans: 130 students in chorus only
Triangles with height
and base
, with
have area
.
Such cross sections with the base of the triangle in the disk
(a disk with radius 5) have base with length

i.e. the vertical (in the
plane) distance between the top and bottom curves describing the circle
.
So when
, the cross section at that point has base

so that the area of the cross section would be 6^2/2 = 18.
In case it's relevant, the entire solid would have volume given by the integral

Angles RZT, RZS, and TZS form triangle RZT, so they sum to 180. Thus, we have:
110+3s+8s=180
Simplifying, we have:
11s=70
Dividing by 11, we see that:
s=70/11
Therefore, RZS=3s=210/11, and TZS=8s=560/11.
Answer:
B. [3, 5]
Step-by-step explanation:
The range of a graph are the y-values that are plotted on the vertical axis (y-axis). In order words, the range values, are the possible set of output values of that are plotted on the vertical axis on the graph.
"[ ]" is usually used to denote the possible set of values.
Therefore, the range of the given graph, as we can see from the y-axis, include values from 3 to 5. Therefore, the range of the graph is [3, 5].
In the ΔIJH, the value of the cosec (I) is
.
Given ΔIJH the length of the hypotenuse is 65, the length of the base is 33, and the length of the opposite side is 56.
We have to find the value of the cosec (I).
A function of an arc or angle that is most easily represented in terms of the ratios of pairs of sides of a right-angled triangle, such as the sine, cosine, tangent, cotangent, secant, or cosecant.
We know cosec (I) = hypotenuse / opposite side
Substitute the values
cosec (I) = 65/56
= 
Hence the value of cosec (I) is
.
Learn more about trigonometric function here: brainly.com/question/24349828
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