Alan

Interestingly, a staircase diagram shows this function is cyclic once it reaches x = -1:

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(-4)³ = -64

-64 + (-3) = -67

-67*(-4) = 268

268 - (-8) = 276

276 + (-3) = 273

The side of each square must be of length 4 ( to give areas of 16). The length of HC is therefore 2, as is that of BH.

Imagine a line from H to D.  Triangle HCD has area (1/2)*2*4 = 4.  Similarly triangle HDE has area 4. Hence area of polygon HCDE is 8.

Area of polygon ABHED is therefore 16 - 8 = 8

Add this to the area of square DEFG to get a total area of 16 + 8 = 24 for polygon ABHFGD.

Start by setting: \(u=\sqrt{11-x^2}\\ .\\u-\frac{2}{u}=1\\\)

Rearrange:  \(u^2-u-2=0\\.\\(u-2)(u+1)=0\)

So u = 2 or u = -1.  However, the original square root must be positive, so only the first value of u is valid.

\(\sqrt{11-x^2}=2\\.\\11-x^2 = 4\\.\\x^2=7\\.\\x=\pm \sqrt7 \)

You do just subtract the two.  The uppercase Greek letter \(\Delta\)  is often used to indicate the change in something; so \(\Delta d\)  means the change in d.

For the force applied at right-angles to the lever arm the torque is just force x distance from pivot, so here the torque is 30 N-m

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Use the  ^y√x key (2nd row, 4'th column), or type in sqrt(x,y) for y'th root of x.

Actually, I gave answers for the first part for two cases; one when the two masses are not the same and one when they are (I could, of course, have left out the second possibility if I'd noticed that the question specified M1 > M2!)

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Yes!

(i.e. both did, independently).

Gravitational forces etc: