The current / in an electric circuit varies inversely with the resistance R. If a current of 60 amperes is produced by a resistance of 9 ohms, what current will be produced by a resistance of 6 ohms? A resistance of 6 ohms will produce a current of
amperes

Step-by-step explanation:

E = IR

E = 60(9)

E = 540 V

540 = I(6)

I = 540/6

I = 90 Amperes

you didn't post any answer choices, so its not 100% clear what your teacher is looking for

however, log(15/7) is equivalent to log(15) - log(7)

i'm using the rule log(x/y) = log(x) - log(y)

let's first notice, the center of the hyperbola is at the origin, since the focus points are equidistantly 3 units away from the origin, that also means that the distance "c" from the center to either focus is 3.

it should also be notable that, the traverse axis for this hyperbola is vertical, or namely over the y-axis and therefore, the positive fraction will be the one with y-variable.

the traverse "a" component or distance from the center to either vertex is not 2, is more than 2, so we can safely put it at √5 ≈ 2.24.