Chapter 10.1, Problem 23P

Consider the Turing machine given in lecture which decides the languageB = {w#w | w is a binary string}.Simulate the Turing machine to show that the string 1001#1001 will be accepted by the Turing machine. Show all steps.

PLEASE SHOW ME THE RIGHT ANSWER/SOLUTION SHOW ME ALL THE NEDDED STEP   13: If the perimeter of a square is shrinking at a rate of 8 inches per second, find the rate at which its area is changing when its area is 25 square inches.

Q/Find the Laurent series of (2-3) cos↓ around z = 1. 2-1

#1). A technique is given with 150 mAs is 40 kV and produces an EI value = 400. Find the new EI value, if mAs is 75 and 34 kV are used.

Q3: Answer the following: (i) Let f(z) is an analytic function in a simply connected domain S and y is a simple, closed, positively oriented contour lying in S. Prove that f, f(z)dz = 0.

DO NOT GIVE THE WRONG ANSWER  SHOW ME ALL THE NEEDED STEPS    11: A rectangle has a base that is growing at a rate of 3 inches per second and a height that is shrinking at a rate of one inch per second. When the base is 12 inches and the height is 5 inches, at what rate is the area of the rectangle changing?

please answer by showing all the dfalowing necessary step  DO NOT GIVE ME THE WRONG ANSWER  The sides of a cube of ice are melting at a rate of 1 inch per hour. When its volume is 64 cubic inches, at what rate is its volume changing?

Wendy is looking over some data regarding the strength, measured in Pascals (Pa), of some rope and how the strength relates to the number of woven strands in the rope. The data are represented by the exponential function f(x) = 2x, where x is the number of woven strands. Explain how she can convert this equation to a logarithmic function when strength is 256 Pascals.   Please type out answer

Name: Date: Bell: Unit 11: Volume & Surface Area Homework 2: Area of Sectors Directions: Find the area of each shaded sector. Round to the hundredths place. 1. GH 11 in 2. KL 20 ft H F 64 G L 119 M K 3. BA 6.5 cm 4. YZ 14.2 m B 23 X 87° Y Z 5. KL = 27.1 mm J 32 L X:360-32.1 K A-3 360 7. BD 18 cm E 136 B X=32.8 127.0 (271) A: 069.13 Amm² 19=2102.13 A-136 360.16912 A:300cm² A=96.13 6. PQ = 2.8 in P R 311° 8. WZ 5.3 km V = Z 108 W D 9. HK = 25 ft G H KO 26 X 10. SR 26 m = S 73 T R Gina Wilson (All Things Algeb