At dawn and dusk, large amount of blue and violet light(shorter wavelength) has been scattered so, the light that is recieved by an observer is mostly of a longer wavelength and therefore appears to be red.

Here, f(x)=x² -6 logx-3=0

f(2)=4-6 log2-3=-0.806

f(3)=9-6 log3-3=3.1373

f(2).f(3)=-0.806*3.1373=-2.529422 which is negative.

Hence, the root lies between 2 and 3

c₀ =(2+3)/2=2.5

f(2.5)=6.25-6 log 2.5-3=0.8623

Now

From the table,

f(2.265625)=0.001928<10^-2

Therefore, the...

Here, the given equation of parabola is y²= 8x.

The equation of tangent to the parabola y²=8x is,

y= mx + 2/m

This tangent passes through the point (-2, 3)

So, 3 = -2m + 2/m

or, 3m + 2m² = 2

or, 2m²+3m - 2= 0

or, 2m² + (4 - 1)m -2 = 0

or, 2m² + 4m - m - 2 = 0

or, 2m(m + 2) - 1(m+2) = 0

or, (m + 2) (2m - 1) = 0

Either, Or,

m = -2 m = 1/2

Required angle is,

Time taken (t) = 3 sec
Velocity of sound (v) = 332 m/s
Distance travelled (d) = ?

We know,

d = v x t = 332 x 3 = 996 m

Hence, the source of thunder is about 996 m far from the boy.