The dimensions of the tire are 190/75-15.

1. Find the width of the tire.

2. Find the length of the side wall in milimeters and inches.

3. Find the diameter of the whole tire in inches and milimeters.
by Foretmason 5 hou

it would be a on edgenuit y

the diagram below is your final answer

the degrees would remain the same ( the number to the left of the decimal point.

the multiply the decimal by 60 to get the minutes and use the numbers to the left of the decimal point.

then multiply the decimal portion of that by 60 for the seconds and round off as needed.

39 .956652

the 39 remains the degrees.

the minutes = 0.956652 x 60 = 57.39912

the minutes = 57

the seconds = 0.39912 x 60 = 23.95 = 24 ( rounded to nearest tenth)

so 39.956652 = 39 degrees, 57 minutes and 24 seconds

-75.159809

the -75 remains the degrees

the minutes = 0.159809 x 60 = 9.58854

the minutes = 9

the seconds = 0.58854 x 60 = 35.31, rounded to nearest tenth = 35.3 seconds

so -75.159809 = -75 degrees, 9 minutes and 35.3 seconds.

the first coordinate is north or south, since it it positive it is north.

the second number is east or west, because it is negative it is west.

39 degrees, 57 minutes and 24 seconds n

-75 degrees, 9 minutes and 35.3 seconds w

ending amount = beginning amount / (2^n)

where "n" = number of half-lives (or elapsed time / half-life)

ending amount = 50 mg / (2 ^10 / 5.3)

ending amount = 50 mg / (2 ^ 1.8867924528)

ending amount = 50 m / 3.6981210597

ending amount = 13.5203794557 mg

or about 13.5 mg remain

source:  

152 thatbk ypi vjsl b abjb uoabr