Two black holes (the remains of exploded stars), separated by a distance of 10.0 AU (1 AU = 1.50 ✕ 10^11 m), attract one another with a gravitational force of 6.90 ✕ 10^25 N. The combined mass of the two black holes is 5.20 ✕ 10^30 kg. What is the mass of each black hole? Largest value: Smallest value:

The largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.The given parameters;distance between the two black holes, r = 10 AU = 1.5 x 10¹² mgravitational force between the two black holes, F = 6.9 x 10²⁵ N.combined mass of the two black holes = 5.20 x 10³⁰ kgThe product of the two masses is calculated from Newton's law of universal gravitational as follows;[tex]F = \frac{Gm_1m_2}{r^2} \\\\m_1m_2 = \frac{Fr^2}{G} \\\\m_1m_2 = \frac{(6.9\times 10^{25}) \times (1.5\times 10^{12})^2}{6.67\times 10^{-11}} \\\\m_1m_2 = 2.328 \times 10^{60} \ kg^2[/tex]The sum of the two masses is given as;m₁ + m₂ = 5.2 x 10³⁰ kgm₂ = 5.2 x 10³⁰ kg - m₁The first mass is calculated as follows;m₁(5.2 x 10³⁰ - m₁) = 2.328 x 10⁶⁰5.2 x 10³⁰m₁ - m₁² = 2.328 x 10⁶⁰m₁² - 5.2 x 10³⁰m₁  + 2.328 x 10⁶⁰ = 0solve the quadratic equation using formula method;a = 1, b =-  5.2 x 10³⁰, c = 2.328 x 10⁶⁰[tex]m_1 = \frac{-b \ \ +/- \ \ \sqrt{b^2 - 4ac} }{2a} \\\\m_1 = \frac{-(-5.2\times 10^{20}) \ \ +/- \ \ \sqrt{(-5.2\times 10^{20})^2 - 4(1\times 2.328\times 10^{60})} }{2(1)} \\\\m_1 = 4.7 \times 10^{30} \ kg \ \ or \ \ 4.9 \times 10^{29} \ kg[/tex]The second mass is calculated as follows;m₂ = 5.2 x 10³⁰ kg - m₁m₂ = 5.2 x 10³⁰ kg  -  4.7 x 10³⁰ kgm₂ = 5 x 10²⁹ kgorm₂ = 5.2 x 10³⁰ kg  -  4.9 x 10²⁹ kgm₂ = 4.7 x 10³⁰ kgThus, the largest mass is 4.7 x 10³⁰ kg and the smallest mass is 5 x 10²⁹ kg.Learn more here:https://brainly.com/question/9373839