Alan

log xy + 3 log (x/y) + 5 log (y/x) → log xy + log (x/y)³ + log (y/x)⁵ → log (xyx³y⁵/(y³x⁵)) → log (x⁴y⁶/(y³x⁵)) → log (y³/x)

.

There is no key for absolute value on the calculator here, but you can type in abs(x) where x is the expression for which you want the absolute value.

.

If you mean:

23.675*4² → 23.675*16 → 378.8   or 379 to nearest whole number.

If you mean

(23.675*4)² → (94.7)² → 8968/09  or 8968 to nearest whole number.

. 

60 km/h → 60*1000 m/hr → 60*1000/3600 m/s → 16.67 m/s

16.67 metres covered in one second.

I should add that the formula is correct if resistance due to air drag is ignored.  For cannonballs at this speed that is almost certainly not a good assumption!

You are using the correct formula! If the result is outrageous it can only be that the input values you are using are outrageous!

Check that you are using either radians for all devices or degrees for all devices.  Excel used radians by default, whereas the windows calculator uses degrees by default.

.

I should have said "maximum" not "minimum" in the sentence below Completing the square!

Let length = L, width = W and area = A

L + 2W = 300

A = L*W

A = (300 - 2W)*W  or  A = 300W - 2W^2

Using calculus:

dA/dW = 300 - 4W  This is zero when W = 75, hence L = 300 - 2*75  or L = 150

Without calculus:

Completing the square:

A = -2(W^2 - 150W) → -2( (W - 75)^2 -75^2 )   This is a clearly a minimum when W = 75, hence L = 150.

.

I think part (b) is as follows:

.