A fair charges an admission fee of 3 dollars per person let c be the cost of admission in dollars afor p people write an equatio
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Two triangle are congruent when the shape and size of both the triangle are same.
The given information is the SAS case. Thus the option B is the correct option.
<h3>What is triangle congruence theorem?</h3>Two triangle are congruent when the shape and size of both the triangle are same.
Triangle congruence theorem are-
- Angle-Side-Angle theorem (AAS)- This theorem states that two triangle is congruent when two angle and one side of the triangle are respectively equal to the two angles and same side of the other triangle.
- Side-Side-Side theorem (SSS)- When the three sides of the one triangle is equal to the three sides of the other triangle respectively, then the triangle are congruent.
- Side-Angle-Side theorem (SAS)- Two sides and the included angle of are equal to the two sides and one angle of other triangle respectively.
Given information-
The distance of the Evelyn from the airplane is 200 meters.
The initial angle at takeoff of the plane is 12 degrees.
Evelyn is approximately 104 meters from the plane.
The angle of elevation of the plane is 12°.
The triangle made for this situation is attached below.
As in the shown triangle below this question. The distance of Evelyn from the initial takeoff of the plane is shown.
This distance is equal to the base of the triangle. Therefore we get the one side of the tingle.
Now the plane is flied 100 meters, which gives the value of another side of the triangle as shown in the image below attached.
Angle of elevation of the plane is 12 degrees, which gives the one angle of the triangle.
Thus we have one side equal to the 100 meters, one angle equal to the 12 degrees and another side with equal to the 200 meters.
Thus it is the case of side angle side.
Hence, the given information is the SAS case. Thus the option B is the correct option.
Learn more about the triangle congruence theorem here;
The cardinality of the union between two sets is given by
In fact, if an element is in both A and B, it is counted in both N(A) and N(B), so we're counting it twice. By subtracting the cardinality of the interserction, we're taking back one of these "extra-counted" element, and the count is correct. In your case, the numbers are
This means that the two sets cover, without repetitions, 43 of the 66 elements in the sets. So, there are
elements which are in neither A nor B