The bisector of angle APQ passes through O and this is illustrated below.
<h3>How to illustrate the information?</h3>
From the information given, the center is O. and the circle passes through O and cuts at K.
In this case, it should be noted that the circles are equal according to the SAS test.
Here, AOB + APQ = 180° (Linear pair)
2AOB = 180
AOB = 90.
Therefore, the bisector of angle APQ passes through O.
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I think that this is a combination problem. From the given, the 8 students are taken 3 at a time. This can be solved through using the formula of combination which is C(n,r) = n!/(n-r)!r!. In this case, n is 8 while r is 3. Hence, upon substitution of the values, we have
C(8,3) = 8!/(8-3)!3!
C(8,3) = 56
There are 56 3-person teams that can be formed from the 8 students.
Answer:
1
Step-by-step explanation:
-2(-3x-4)=14
6x + 8 = 14
6x= 14-8
6x=6
x=1
Answer:
Hence, the perimeter of the triangles are:
P=123.2727 dm
P'=102.7272 dm
Step-by-step explanation:
In two similar triangles:
The ratio of the areas of two triangle is equal to the square of their perimeters.
Let A and A' represents the area of two triangles and P and P' represents their perimeter.
Then they are related as:
We are given:
A=72 dm^2 , A'=50 dm^2
and P+P'=226 dm.-----------(1)
i.e. 
on taking square root on both the side we get:
Now putting the value of P in equation (1) we obtain:
Hence,
P=226-102.7272=123.2727
Hence, the perimeter of the triangles are:
P=123.2727 dm
P'=102.7272 dm
Answer:
I'm sorry idk
Step-by-step explanation:
