Answer:
544.65 ohm and 179.84 ohm
Explanation:
Hello
Let
Resistor 1 (R1) and Resistor 2 (R2)
in series [tex]R_{eq} =R1+R2[/tex]
in parallel
[tex]\frac{1}{R_{eqp} } =\frac{1}{R1} +\frac{1}{R2}\\\\R1=R_{eqs} -R2\\\\R1 =724.5-R2 (equation 1)\\ R_{eqp}}=\frac{R1*R2}{R1+R2}\\\\\ 135.2 = \frac{R1*R2}{R1+R2}(equation 2)\\\\\\\\replacing 1 in \ 2\\\\probema data[/tex]
[tex]135.2=\frac{(724.5-R2)(R2)}{724.5-R2+R2} }\\135.2=\frac{(-R_{2} ^{2}+724.5R2) }{724.5}\\ 135.2*724.5=-R_{2} ^{2}+724.5R_{2}\\\\R_{2} ^{2} -724.5R_{2} +97952.4 =0\\R_{2} =\frac{724.5 \±\sqrt{(-724.5^{2})-4(1)(97952.4) } }{2(1)}\\R_{2} =\frac{724.5\±\sqrt{(133092.25) } }{2(1)}\\\\R_{2} =\frac{724.5+364.81810}{2} \\R_{2} =544.65 ohm\\\\R_{2} =\frac{724.5-364.81810}{2} =179.84 ohm[/tex]\\\\
let R2=544.65 and replace in equation 1
R1=724.5-544.65
R1=179.85
so, the resistors are 544.65 ohm and 179.84 ohm
Have a great day
