3x^2+11x-4=10x-1
3x^2+11x-10x-4+1=0
3x^2+x-3=0
Δ=1^1-4*3*(-3)=1+36=37
x1=(-1+V37)/6
x2=( -1-V37)/6
Answer:
0
1
Step-by-step explanation:
First question:
You are given a side, a, and its opposite angle, A. You are also given side b. Use that in the law of sines and solve for the other angle, B.




The sine function can never equal 2, so there is no triangle in this case.
Answer: no triangle
Second question:
You are given a side, b, and its opposite angle, B. You are also given side c. Use that in the law of sines and solve for the other angle, C.





One triangle exists for sure. Now we see if there is a second one.
Now we look at the supplement of angle C.
m<C = 52.5°
supplement of angle C: m<C' = 180° - 52.5° = 127.5°
We add the measures of angles B and the supplement of angle C:
m<B + m<C' = 63° + 127.5° = 190.5°
Since the sum of the measures of these two angles is already more than 180°, the supplement of angle C cannot be an angle of the triangle.
Answer: one triangle
Answer:
O is the center of the circle with radius IE(=ID=EF)
Step-by-step explanation:
Join all 3 points D, E, F, forming the triangle DEF.
Let the midpoint of EF be M and the midpoint of ED be N. (first picture)
Join point I to E, D and F.
Since IN is both an altitude and median to triangle EID, then triangle EID is an isosceles triangle, and IE=ID
similarly, we see that IE=IF.
conclusion: IE=ID=EF.