The absorbance of the solution at 710nm increases so that it off the scale of the spectrophotometer (values > 2). To correct for these values, we must adjust the absorbances so that the correct values can be calculated. To do so a ratio of the highest accurate absorabance can be used to determine the extinction coefficent of 710 nm. Solve for this value based on absorbances using tube 6.

(23*2)-12=x     x=34

step-by-step explanation:

its 50 miles away from the starting point

graph them on a graphing calculator, press 2nd trace. select intersect. hit enter until it says answer. ti 84 works best

step-by-step explanation:

answer: cos [   arcsin (3/5) - arccos (1/2) ] =

step-by-step explanation: × = 3/5 = 0.6--> x = 36.87 and x = 180 - 36.87 = 143.13 deg

cos y = 1/2 --> y = +- 60 deg a. cos (x - y) = cos (143.13 - 60) = cos 83,13  

=

0.12

b. cos (36.87 - 60) = cos (-23.23) = 0.92