Question

multiple choice 1 point. simplify $\frac{\frac{x}{2}+\frac{1}{x}}{\frac{x}{2}-\frac{1}{x}};x
eq - 2,x
eq0$. $\frac{x(x + 2)}{x(x - 2)};x
eq - 2,x
eq0$ $\frac{1}{x};x
eq0$ $\frac{x+2}{x - 2};x
eq - 2$. multiple choice 1 point. the equation $r=\frac{1}{\frac{1}{r_1}+\frac{1}{r_2}}$ represents the total resistance, $r$, when two resistors whose resistances are $r_1$ and $r_2$ are connected in parallel. find the total resistance when $r_1 = x$ and $r_2=x + 1$. $\frac{2x + 1}{x(x + 1)};x
eq - 1,0$ $\frac{2x+1}{x};x
eq - 1,-\frac{1}{2},0$ $\frac{x(x + 1)}{2x + 1};x
eq - 1,0$ $\frac{2x}{2x + 1};x
eq - 1,-\frac{1}{2},0$

Explanation:

Step1: Substitute \(r_1 = x\) and \(r_2=x + 1\) into the parallel - resistance formula

The formula for the total resistance \(r\) of two resistors \(r_1\) and \(r_2\) in parallel is \(\frac{1}{r}=\frac{1}{r_1}+\frac{1}{r_2}\). Substituting \(r_1 = x\) and \(r_2=x + 1\), we get \(\frac{1}{r}=\frac{1}{x}+\frac{1}{x + 1}\).

Step2: Find a common denominator

The common denominator of \(x\) and \(x + 1\) is \(x(x + 1)\). So, \(\frac{1}{r}=\frac{x + 1+x}{x(x + 1)}=\frac{2x+1}{x(x + 1)}\).

Step3: Solve for \(r\)

Taking the reciprocal of both sides, we have \(r=\frac{x(x + 1)}{2x + 1},x
eq0,x
eq - 1,-\frac{1}{2}\).

Answer:

\(\frac{x(x + 1)}{2x + 1},x
eq0,x
eq - 1,-\frac{1}{2}\)