Step-by-step explanation:
b^2-4b+3=0
b²-3x-b+3=0
b(b-3)-1(b-3)=0
(b-3)(b-1)=0
either
b=3 or b=1
.
2n^2 + 7 = -4n + 5
2n²+4n+7-5=0
2n²+4n+2=0
2(n²+2n+1)=0
(n+1)²=0/2
:.n=-1
.
x - 3x^2 = 5+ 2x - x^2
0=5+ 2x - x^2-x +3x^2
0=5+x+2x²
2x²+x+5=0
comparing above equation with ax²+bx +c we get
a=2
b=1
c=5
x={-b±√(b²-4ac)}/2a ={-1±√(1²-4×2×5)}/2×1
={-1±√-39}/2
Answer:
SinL = 7/25
CosL = 24/25
TanL = 7/24
Step-by-step explanation:
Find the diagram attached.
Using SOH CAH TOA in trigonometry identity to find the sinL, cosL and TanL
Note that the hypotenuse is the longest side = 25
The opposite will be the side facing the acute angle L
Opposite = 7
Adjacent = 24
For SinL
sinL = Opposite/Hypotenuse {SOH}
SinL = 7/25
For cosL:
CosL = Adjacent/Hypotenuse{CAH}
CosL = 24/25
For tanL:
TanL = Opposite/Adjacent {TOA}
TanL = 7/24
