Solve:
(3x-2)^2-4x(2x-3)>=0
(3x-2)(3x-2)-8x^2+12x>=0
9x^2-12x+4-8x^2+12x>=0
The solution: x^2+4>=0
I hope I helped you with your solution! If you could give me brainliest, I would be very thankful!
We know that
case 1)
Applying the law of sines
a/Sin A=b/Sin B
A=56°
a=12
b=14
so
a*Sin B=b*Sin A----> Sin B=b*Sin A/a---> Sin B=14*Sin 56°/12
Sin B=0.9672
B=arc sin (0.9672)------> B=75.29°-----> B=75.3°
find angle C
A+B+C=180°-----> C=180-(A+B)----> C=180-(56+75.3)----> C=48.7°
find c
a/Sin A=c/Sin C----> c=a*Sin C/Sin A----> c=12*Sin 48.7°/Sin 56°)
c=10.87-----> c=10.9
the answer Part 1)
the dimensions of the triangle N 1
are
a=12 A=56°
b=14 B=75.3°
c=10.9 C=48.7°
case 2)
A=56°
a=12
b=14
B=180-75.3----> B=104.7°
find angle C
A+B+C=180°-----> C=180-(A+B)----> C=180-(56+104.7)----> C=19.3°
find c
a/Sin A=c/Sin C----> c=a*Sin C/Sin A----> c=12*Sin 19.3°/Sin 56°)
c=4.78-----> c=4.8
the answer Part 2)
the dimensions of the triangle N 2
are
a=12 A=56°
b=14 B=104.7°
c=4.8 C=19.3°
Exactly what was your answer for a
Answer: the answer should be B
Answer: B
Step-by-step explanation:
3x - 4 = 11
add 4 to both sides so it cancels out
3x = 15
Now divide by 3
x = 5