Alan

With y = m*x + c the slope, m, is m = (9 - 8)/(5 - 3) or m = 1/2.  

Use one of the points to get c:  8 = (1/2)*3 + c so c = 8 - 3/2 or c = 13/2

Therefore:  y = (1/2)*x + 13/2

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When x is very large pi/5^x is very small.

When z is very small sin(z) → z, so when x is very large xsin(pi/5^x) → x*pi/5^x

5^x increases much more rapidly than x, so x*pi/5^x → 0

Hence, in the limit as x → ∞, xsin(pi/5^x) → 0

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V =(4/3)pi.r³

dV/dt = 4.pi.r².dr/dt so

dr/dt = -10^-4

r = r0 - 10^-4t

where (4/3)pi.r0³ = 10^-6 m³

r0 = ³√(3*10^-6/4pi) m

We want the time when r = 0:

t = r0/10^-4

(the units of 10^-4 aren't given!).

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To the nearest tenth:  t_m = 0.1 seconds,  h_m = 14.9 cm.

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a).

b).

Use the first line above to help here.

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V_AB = -20V    the minus sign indicating that B is a higher voltage than A.

V_AC = -22V

Cut out the two resistances, which aren't relevant if you are only considering the voltages here (they are relevant if you want to go on and find currents of course!), then redraw what's left as a single line with A at the bottom.

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Like so?

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Original rectangle  L*W = A

New rectangle   (L-1)*(W+1) = A + x

L*W + L - W - 1 = A + x

or L - W - 1 = x   since L*W = A

If x has to be a positive integer as well then:

When L-W = 2 units,  x = 1 square unit

If x doesn't have to be an integer then there it has no smallest value (it can be infinitesimally small). 

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