If in the triangle ABC , BF is an angle bisector and ∠ABF=41° then angle m∠BCE=8°.
Given that m∠ABF=41° and BF is an angle bisector.
We are required to find the angle m∠BCE if BF is an angle bisector.
Angle bisector basically divides an angle into two parts.
If BF is an angle bisector then ∠ABF=∠FBC by assuming that the angle is divided into two parts.
In this way ∠ABC=2*∠ABF
∠ABC=2*41
=82°
In ΔECB we got that ∠CEB=90° and ∠ABC=82° and we have to find ∠BCE.
∠BCE+∠CEB+EBC=180 (Sum of all the angles in a triangle is 180°)
∠BCE+90+82=180
∠BCE=180-172
∠BCE=8°
Hence if BF is an angle bisector then angle m∠BCE=8°.
Learn more about angles at brainly.com/question/25716982
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we are given
geometric sequence -36, 6, -1, 1/6, ...
first term is -36

now, we can find common ratio


now, we can find nth term

now, we can plug values
and we get

now, we can find 5th term , 6th term, 7th term
fifth term:


sixth term:


seventh term:


so, next terms are
, 
,
.............Answer
Answer:
10 times greater
Step-by-step explanation:
Because value of 5 in 503 is 5*100
Then the value of 5 in 53 is 5*10
So then value of 5 in 503 is 10 times greater than the value of 5 in 53
I hope this helps
I’m sorry I don’t know but I need the points:(
(side note: your formula C = 2πr^2 is incorrect, it should be C = 2πr)
Answer:
r = 9
Step-by-step explanation:
First let's find the circumference of Circle A. To use the formula we need radius. Since we have a diameter just divide by 2 to get radius. 27 / 2 = 13.5
Now plug that into C = 2πr: 2π*(13.5) = 27π
The question says that the area of circle B is 3x this value, so Circle B's area must be: 27π * 3 = 81π
Now plug that into A = πr^2 and solve for r:

(edited to correct a brain fart)