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:
1.5 nautical miles
Step-by-step explanation:
See the attached diagram.
The submarine starts to dive at point A and diving along AB.
If the distance traveled underwater i.e. AB = 1.7 Nautical miles and C is the point directly above its current position at B.
Then Δ ABC is a right triangle having ∠ BAC = 28° {Given}
So, using trigonometry we can write
⇒ nautical miles.
Therefore, the surface distance that the submarine travels is 1.5 nautical miles. (Answer)
