Considering the angle a by cosine rule
11^2 =7 ^2 +15^2 - 2(7)(15)cos(a)
When you do the maths,
Cos(a) =153/210 =0.729
a= cos inverse of 0.729
a=43 degrees
Considering angle b
7^2=15^2 +11^2 -2(11)(15) cos(b)
This will result in cos(b) =297/330=0.9
b= cos inverse of 0.9 = 25.8 degrees
Considering angle c
15^2=7^2 +11^2 - 2(11)(7) cos(c)
Cos(c) will be = -55/154 = -0.357
c= cos inverse of -0.357=110.9
Comparing the angles a,b and c,
C is the largest size in the triangle with an angle of 110.9 degrees
Am I right please ??
Is this the full question that is stated?
Answer:
Step-by-step explanation:
a) 3x^2-12x-11=3*x^2-3*2*2*x+3*4-23=3*(x^2-2*2x+2^2)-23=3*(x-2)^2-23
so a= -2 ,b= -23
b) x1=[12+V(12^2-4*3* -11)]/6=12/6+V(144+132)/6=2+2/6V69=2+1/3V69=
2+V69/9=2+V23/3
c=2, d= 23/3
