Question

Write a formula that shows the dependence of: the length of the side (a) of a cube on the surface area (S) of the cube

Answer 1

Answer:

Let a = side length of a cube

Let S = surface area of a cube

Area of a square = a²

Since a cube has 6 square sides:  S = 6a²

To make a the subject:

S = 6a²

Divide both sides by 6:

$\sf \implies \dfrac{S}{6}=a^2$

Square root both sides:

$\sf \implies a=\sqrt{\dfrac{S}{6}}$

(positive square root only as distance is positive)

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$\sf x=-3-\sqrt{2} \implies (x+[3+\sqrt{2}])=0$

$\sf x=-3+\sqrt{2} \implies (x+[3-\sqrt{2}])=0$

Therefore,

$\sf y=a(x+[3+\sqrt{2}]) (x+[3-\sqrt{2}])$  for some constant a

Given the y-intercept is at (0, -5)

$\sf \implies a(0+3+\sqrt{2}) (0+3-\sqrt{2})=-5$

$\sf \implies a(3+\sqrt{2}) (3-\sqrt{2})=-5$

$\sf \implies a(9-2)=-5$

$\sf \implies 7a=-5$

$\sf \implies a=-\dfrac57$

Substituting found value of a into the equation and simplifying:

$\sf y=-\dfrac57(x+[3+\sqrt{2}]) (x+[3-\sqrt{2}])$

$\sf \implies y=-\dfrac57(x^2+6x+7)$

$\sf \implies y=-\dfrac57x^2-\dfrac{30}{7}x-5$