Question

**Please Help, ASAP**

Answer 1

Explanation:

We need to rearrange the following formula for the values given in parenthesis.

(1) x+xy = y, (x)

taking x common in LHS,

x(1+y)=y

$x=\dfrac{y}{1+y}$

(2) x+y = xy, (x)

Subrtacting both sides by xy.

x+y-xy = xy-xy

x+y-xy = 0

x-xy=-y

x(1-y)=-y

$x=\dfrac{-y}{1-y}$

(3) x = y+xy, (x)

Subrating both sides by xy

x-xy = y+xy-xy

x(1-y)=y

$x=\dfrac{y}{1-y}$

(4) E = (1/2)mv^2-(1/2)mu^2, (u)

Subtracting both sides by (1/2)mv^2

E-(1/2)mv^2 = (1/2)mv^2-(1/2)mu^2-(1/2)mv^2

E-(1/2)mv^2 =-(1/2)mu^2

So,

$2(E-\dfrac{1}{2}mv^2)=-mu^2\\\\u^2=\dfrac{-2}{m}(E-\dfrac{1}{2}mv^2)\\\\u=\sqrt{\dfrac{-2}{m}(E-\dfrac{1}{2}mv^2)}\\\\u=\sqrt{\dfrac{2}{m}(\dfrac{1}{2}mv^2-E)}$

(5) (x^2/a^2)-(y^2/b^2) = 1, (y)

$\dfrac{x^2}{a^2}-1=\dfrac{y^2}{b^2}\\\\y^2=b^2(\dfrac{x^2}{a^2}-1)\\\\y=b\sqrt{\dfrac{x^2}{a^2}-1}$

(6) ay^2 = x^3, (y)

$y^2=\dfrac{x^3}{a}\\\\y=\sqrt{\dfrac{x^3}{a}}$

Hence, this is the required solution.