Let
. Then

lies in the second quadrant, so

So we have

and the fourth roots of
are

where
. In particular, they are




The value of the underlined digit depends on the place the number that is underlined is. For the first digit, it would be ones. The second digit is 10s. The third is 100s and the fourth is thousands and the fifth is ten thousands and the 6 is hundred thousands, and the seventh is million, and so on.
Answer:
81/70
Step-by-step explanation:
5/14+4/5=25/70+56/70=81/70
Answer:
A $60
Step-by-step explanation:
Answer:one half
Step-by-step explanation: