Precalculus with Limits: A Graphing Approach
7th Edition
ISBN: 9781305071711
Author: Ron Larson
Publisher: Brooks/Cole
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Question
Chapter 8.5, Problem 62E
To determine
To write: all possible selections of two letters that can be formed the letters.
Expert Solution & Answer
Answer to Problem 62E
The selection can be made in 10 ways.
Explanation of Solution
Given information:
Use the letters A, B, C, D, E, and F.
Calculation:
Let us consider the following letters
A, B, C, D, E, F
So, the numbers of possible selection are written below.
ABC, ABD, ABE, ABF, BCD, BCE, BCF, CDE, CDF, DEF
Therefore, the number of possible selection are
ABC, ABD, ABE, ABF, BCD, BCE, BCF, CDE, CDF, DEF
Therefore, total number of selection are 10.
Chapter 8 Solutions
Precalculus with Limits: A Graphing Approach
Ch. 8.1 - Prob. 1ECh. 8.1 - Prob. 2ECh. 8.1 - Prob. 3ECh. 8.1 - Prob. 4ECh. 8.1 - Prob. 5ECh. 8.1 - Prob. 6ECh. 8.1 - Prob. 7ECh. 8.1 - Prob. 8ECh. 8.1 - Prob. 9ECh. 8.1 - Prob. 10E
Ch. 8.1 - Prob. 11ECh. 8.1 - Prob. 12ECh. 8.1 - Prob. 13ECh. 8.1 - Prob. 14ECh. 8.1 - Prob. 15ECh. 8.1 - Prob. 16ECh. 8.1 - Prob. 17ECh. 8.1 - Prob. 18ECh. 8.1 - Prob. 19ECh. 8.1 - Prob. 20ECh. 8.1 - Prob. 21ECh. 8.1 - Prob. 22ECh. 8.1 - Prob. 23ECh. 8.1 - Prob. 24ECh. 8.1 - Prob. 25ECh. 8.1 - Prob. 26ECh. 8.1 - Prob. 27ECh. 8.1 - Prob. 28ECh. 8.1 - Prob. 29ECh. 8.1 - Prob. 30ECh. 8.1 - Prob. 31ECh. 8.1 - Prob. 32ECh. 8.1 - Prob. 33ECh. 8.1 - Prob. 34ECh. 8.1 - Prob. 35ECh. 8.1 - Prob. 36ECh. 8.1 - Prob. 37ECh. 8.1 - Prob. 38ECh. 8.1 - Prob. 39ECh. 8.1 - Prob. 40ECh. 8.1 - Prob. 41ECh. 8.1 - Prob. 42ECh. 8.1 - Prob. 43ECh. 8.1 - Prob. 44ECh. 8.1 - Prob. 45ECh. 8.1 - Prob. 46ECh. 8.1 - Prob. 47ECh. 8.1 - Prob. 48ECh. 8.1 - Prob. 49ECh. 8.1 - Prob. 50ECh. 8.1 - Prob. 51ECh. 8.1 - Prob. 52ECh. 8.1 - Prob. 53ECh. 8.1 - Prob. 54ECh. 8.1 - Prob. 55ECh. 8.1 - Prob. 56ECh. 8.1 - Prob. 57ECh. 8.1 - Prob. 58ECh. 8.1 - Prob. 59ECh. 8.1 - Prob. 60ECh. 8.1 - Prob. 61ECh. 8.1 - Prob. 62ECh. 8.1 - Prob. 63ECh. 8.1 - 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Prob. 9ECh. 8.4 - Prob. 10ECh. 8.4 - Prob. 11ECh. 8.4 - Prob. 12ECh. 8.4 - Prob. 13ECh. 8.4 - Prob. 14ECh. 8.4 - Prob. 15ECh. 8.4 - Prob. 16ECh. 8.4 - Prob. 17ECh. 8.4 - Prob. 18ECh. 8.4 - Prob. 19ECh. 8.4 - Prob. 20ECh. 8.4 - Prob. 21ECh. 8.4 - Prob. 22ECh. 8.4 - Prob. 23ECh. 8.4 - Prob. 24ECh. 8.4 - Prob. 25ECh. 8.4 - Prob. 26ECh. 8.4 - Prob. 27ECh. 8.4 - Prob. 28ECh. 8.4 - Prob. 29ECh. 8.4 - Prob. 30ECh. 8.4 - Prob. 31ECh. 8.4 - Prob. 32ECh. 8.4 - Prob. 33ECh. 8.4 - Prob. 34ECh. 8.4 - Prob. 35ECh. 8.4 - Prob. 36ECh. 8.4 - Prob. 37ECh. 8.4 - Prob. 38ECh. 8.4 - Prob. 39ECh. 8.4 - Prob. 40ECh. 8.4 - Prob. 41ECh. 8.4 - Prob. 42ECh. 8.4 - Prob. 43ECh. 8.4 - Prob. 44ECh. 8.4 - Prob. 45ECh. 8.4 - Prob. 46ECh. 8.4 - Prob. 47ECh. 8.4 - Prob. 48ECh. 8.4 - Prob. 49ECh. 8.4 - Prob. 50ECh. 8.4 - Prob. 51ECh. 8.4 - Prob. 52ECh. 8.4 - Prob. 53ECh. 8.4 - Prob. 54ECh. 8.4 - Prob. 55ECh. 8.4 - Prob. 56ECh. 8.4 - Prob. 57ECh. 8.4 - Prob. 58ECh. 8.4 - Prob. 59ECh. 8.4 - Prob. 60ECh. 8.4 - Prob. 61ECh. 8.4 - 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- The correct answer is D Could you explain and show the steps pleasearrow_forwardTaylor Series Approximation Example- H.W More terms used implies better approximation f(x) 4 f(x) Zero order f(x + 1) = f(x;) First order f(x; + 1) = f(x;) + f'(x;)h 1.0 Second order 0.5 True f(x + 1) = f(x) + f'(x)h + ƒ"(x;) h2 2! f(x+1) 0 x; = 0 x+1 = 1 x h f(x)=0.1x4-0.15x³- 0.5x2 -0.25x + 1.2 51 Taylor Series Approximation H.w: Smaller step size implies smaller error Errors f(x) + f(x,) Zero order f(x,+ 1) = f(x) First order 1.0 0.5 Reduced step size Second order True f(x + 1) = f(x) + f'(x)h f(x; + 1) = f(x) + f'(x)h + "(xi) h2 f(x,+1) O x₁ = 0 x+1=1 Using Taylor Series Expansion estimate f(1.35) with x0 =0.75 with 5 iterations (or & s= 5%) for f(x)=0.1x 0.15x³-0.5x²- 0.25x + 1.2 52arrow_forwardCould you explain this using the formula I attached and polar coorindatesarrow_forward
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