We will have the following:
We will first find the equivalent resistances of each parallel arrangement, that is:
1st. Is just 1 resistance of 100.0 Ohm.
2nd.
![\frac{1}{R_{\text{total}}}=\frac{_{}1}{70.0\Omega}+(1)/(70.0\Omega)\Rightarrow\frac{1}{R_{\text{total}}}=(2)/(70.0\Omega)](https://img.qammunity.org/2023/formulas/physics/college/3l1lbmt3mrimi5ip4attaoaw68os0xtus8.png)
![\Rightarrow R_{\text{total}}=(70.0)/(2)\Omega\Rightarrow R_(total)=35.0\Omega](https://img.qammunity.org/2023/formulas/physics/college/f9pap2p37na24f99zhmh4yt36h9xc0wumd.png)
So, the two resistances in parallel are equivalent to 35.0 Ohm.
3rd.
![\frac{1}{R_{\text{total}}}=(1)/(40.0\Omega)+(1)/(40.0\Omega)+(1)/(40.0\Omega)\Rightarrow\frac{1}{R_{\text{total}}}=(3)/(40.0\Omega)](https://img.qammunity.org/2023/formulas/physics/college/y0nvmp01apomxagjtwnd9bk7nb03quj622.png)
![\Rightarrow R_{\text{total}}=(40.0)/(3)\Omega\Rightarrow R_{\text{total}}\approx13.3\Omega](https://img.qammunity.org/2023/formulas/physics/college/e47dkuvo91awsliw4hrk3dbwolp9t7pixh.png)
So, the three resistances are equivalent to 40.0 / 3 Ohm.
Then we find the equivalent resistance of the resulting serial resistances:
![R_{\text{FINAL}}=100.0\Omega+35.0\Omega+(40.0)/(3)\Omega\Rightarrow R_{\text{FINAL}}=(445)/(3)\Omega](https://img.qammunity.org/2023/formulas/physics/college/p12gpfh7k73v6vt3diqmecg6esitp4i6bo.png)
![\Rightarrow R_{\text{FINAL}}\approx148.3\Omega](https://img.qammunity.org/2023/formulas/physics/college/i6yezj61u77ne5o5skgb0rjx8tlkmlopxe.png)
So, the equivalent resistance of the resistors in the circuit is 445/3 Ohms, that is approximately 148.3 Ohms.