Answer: A △ABC such that the bisectors of ∠ABC and ∠ACB meet at a point O.
To prove :
∠BOC=90
o
+
2
1
∠A
Step-by-step explanation: In △BOC,
∠1+∠2+∠BOC=180
o
In △ABC,
∠A+∠B+∠C=180
o
∠A+2(∠1)+2(∠2)=180
o
2
∠A
+∠1+∠2=90
o
∠1+∠2=90
o
−
2
∠A
Answer:
C
Step-by-step explanation:
Answer:
b=40
c=12
Step-by-step explanation:
b) 3+7(8-6)2
10(2)2
20^2 =40
C) 4+(−3) 2 −2÷( 6/3)
4+9− 2/6/3
13-2/6/3
13-2x3/6
13-6/6
13-1
c=12
Answer:
20
Step-by-step explanation:
If the three numbers are x, x+1, and x+2, then:
x (x+1) (x+2) = 990
x (x² + 3x + 2) = 990
x³ + 3x² + 2x - 990 = 0
(x - 9) (x² + 12x + 110) = 0
Since x must be a whole number, x = 9. You can also use simple trial and error to find that the three numbers are 9, 10, 11.
The sum is:
9 + 10 + 11 = 20
Answer:
4xy
Step-by-step explanation:
20x^2 y = 4*5 x^2 y = 2*2*5*x*x*y
16*x*y^2 = 4*4*x*y^2 = 2*2*2*2*x*y*y
what is common in both
2*2*x*y
4xy
