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Decoding Messages We have encoded messages by assigning the numbers 1 to 26 to the letters a to z of the alphabet, respectively, and assigning 27 to a blank space. We can decode messages of this type by finding the inverse of the encoding matrix and multiplying it times the coded message. Use A−1 and the conversion table below to decode the messages in Exercises 39–44.
a b c d e f g h i j k l m n o p
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
q r s t u v w x y z blank
17 18 19 20 21 22 23 24 25 26 27
41. Decoding Messages The encoding matrix is
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Chapter 7 Solutions
College Algebra in Context, Books a la Carte Edition plus MyLab Math with Pearson eText -- Access Card Package (5th Edition)
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- Cable AB is 103 ft long and the tension in the cable is 3900 lb. 56 ft D y A B 20° 50° x C Identify the x, y, and z components of the force exerted by the cable on the anchor B. 1 F. FI 3 Fy 2 Match each of the options above to the items below. 2,120 lb 1,120 lb -3,076 lbarrow_forwardIn the given figure, P = 51 lb. 65° 25° 35° 75 lb P B Determine the required tension in cable AC, knowing that the resultant of the three forces exerted at point C of boom BC must be directed along BC. The required tension in cable AC is lb.arrow_forwardhelp on this question about Laplace transformation?arrow_forward
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