Alan

Solving it as an equality provides bounds on its validity.

For example:

y^2 - 4 < 0

Solve Y^2 - 4 = 0 to get Y = 2 and Y = -2

Then -2 < y < 2

Similarly, if y^2 - 4 > 0 then y < -2 and y > 2

Use arcsin(-0.35)

On the calculator on the home page here, first press the 2nd key, then the asin key ( asin is arcsin).

Remember that angles that produce negative values for sine occur in the third and fourth quadrants (I.e. Between 180 and 360 degrees).

Use:  deltaL = L*F/(A*Y)

L is length, F is force, A is cross-sectional area, Y is Young's modulus.

Also use:  F = m*a

m is mass, a is acceleration.

Make sure the units are consistent (e.g. Area in m^2 if length in metres).

I should have had the 45 and 40 the other way round.  I.e the numerator should be 45 - -40.  

For Melody:

acceleration = change in velocity/time taken

acceleration = (40 - -45)/0.0005 m/s^2

                    = 85/0.0005 m/s^2

                    = 170000 m/s^2

x^2 + y^2 = 9

Differentiate wrt x

2x + 2y*dy/dx = 0

Hence x + y = 0. (because dy/dx = 1), so y = -x

Put this in the original equation to get. 2x^2 = 9

Two solutions:  x = 3/sqrt(2) and x = -3/sqrt(2)

Corresponding values of y are y = -3/sqrt(2) and y = 3/sqrt(2) (because y = -x)

Equation of line with slope 1 is y = x + k where k is a constant.

Plug the known pairs of x and y into this equation to find the corresponding values of k.

loga(2x^2-7) = 2loga(x-1)

loga(2x^2-7) = loga(x-1)^2

2x^2 - 7 = (x - 1)^2

2x^2 - 7 = x^2 - 2x + 1

x^2 + 2x - 8 = 0

(x + 4)(x - 2)  = 0

Can you take it from here?

Does this help:

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Like so:

\(\frac{dN}{dt}=-RNp\\ \frac{1}{N}\frac{dN}{dt}=-Rp\\\frac{d \ln N}{dt}=-Rp\\ \ln N - \ln N_0 = -Rpt\\ N=N_0e^{-Rpt}\)