(x+6)^(1/2)-5=x+1
(x+6)^(1/2)=x+6
((x+6)^(1/2))^2=(x+6)^2
x+6=x^2+12x+36
0=x^2+11x+30
(-11+(11^2-4(1)(30))^(1/2))/2
(-11+((1)^(1/2))/2
(-11+1)/2=-5
(-11-1)/2=-6
((-6)+6)^1/2-5=(-6)+1
(0^(1/2))-5=-5
-6 is non-extraneous
((-5)+6)^1/2-5=(-5)+1
(1^1/2)-5=-4
1-5=-4
-4=-4
-5 is non-extraneous
Hello! My work for this problem is attached below. The answer is
Vb = 4Va
9514 1404 393
Answer:
B
Step-by-step explanation:
Angles A and C are vertical angles; angles B and C are alternate interior angles. Only line 3 of the proof is in error.
The applicable description is found in choice B.
Answer:
3.75 i think c:
Step-by-step explanation:
