find the current flowing out of the battery. ...

# Explanation: ## Step1: Calculate parallel - resistance of 35Ω and 20Ω Use the formula for parallel - resistance $R_{p}=\frac{R_1\times R_2}{R_1 + R_2}$, where $R_1 = 35\Omega$ and $R_2=20\Omega$. So $R_{p}=\frac{35\times20}{35 + 20}=\frac{700}{55}=\frac{140}{11}\Omega\approx12.73\Omega$. ## Step2: Calculate total resistance of the circuit The parallel - combination is in series with the 15Ω resistor. So the total resistance $R_{total}=R_{p}+15=\frac{140}{11}+15=\frac{140 + 165}{11}=\frac{305}{11}\Omega\approx27.73\Omega$. ## Step3: Calculate the current using Ohm's law Ohm's law is $I=\frac{V}{R}$, where $V = 10V$ and $R = R_{total}=\frac{305}{11}\Omega$. So $I=\frac{10}{\frac{305}{11}}=\frac{110}{305}=\frac{22}{61}\approx0.36A$. # Answer: $\frac{22}{61}\text{ A}$