Mathematics for Machine Technology
7th Edition
ISBN: 9781133281450
Author: John C. Peterson, Robert D. Smith
Publisher: Cengage Learning
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Chapter 27, Problem 18A
To determine
The maximum permissible value of length
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- vector Q/Consider the real vector space R². For every X= (X/X2) ER². Let 11x11 = \xil+\x\. Show that 1.11 define a hormon R².arrow_forward12. Suppose that a, b E R and a < b. Show that the vector space C[a, b] of all continuous complex valued functions defined on [a, b], with supremum norm is a Banach space. Ilflloc: = sup f(t), t€[a,b]arrow_forwardbrayden knows that his distance from earth to pluto 5.9 times....arrow_forward
- C. Since C SNEN 2. Suppose that V and Ware vector spaces over F. Consider the cartesian product V x W, with vector addition and scalar multiplication defined by (V1.W)+(V2, W2) (V1+V2, W₁ + W₂) for every (V1, W1). (V2, W2) EV x W and e€ F. a) Show that V x W is a vector space over F. and (v, w) (cv,₁) b) Suppose that || ||v is a norm on V and || ||w is a norm on W. Show that (v, w) defines a norm on V x W. vy+wwarrow_forwardSolve the following nonlinear system using Newton's method 1 f1(x1, x2, x3)=3x₁ = cos(x2x3) - - 2 f2(x1, x2, x3) = x² - 81(x2 +0.1)² + sin x3 + 1.06 f3(x1, x2, x3) = ex1x2 +20x3 + Using x (0) X1 X2 X3 10π-3 3 = 0.1, 0.1, 0.1 as initial conditioarrow_forward2. (a) State Fermat's principle for the propagation of a light ray from point P at (x1,y1) to Q at (x2, y2), expressing the principle as a problem in Calculus of Variations. (b) Suppose c(y) is the speed of light in a medium, given by c(y) Y where a is a constant. Find the path of a light ray between the points P: (−1,3) and Q (1, 3). Is there more than one possible path? (c) Sketch the path of the light ray, and interpret what an observer at Q would see if there were a light source at P.arrow_forward
- On a given day, the sea level pressure is 1013.2 hPA. The temperature at 3,000 ft AMSL is given as minus 4°C. The temperature difference compared with the ISA is: Can you give me a step by step explanation ISA - 13°C ISA - 4°C (c)ISA - 10°C (d) ISA +10°C ISA is +15 Celsius Answer is -13Celsiusarrow_forwardanswer the question attchedarrow_forwardAlso, construct a know-show table to outline the proof of this proposition.arrow_forward
- why the know-show table below is not valid: I know something is wrong in the step p2-p5 but I don't know how to explain it. Can you explain why please.arrow_forwardConsider the following statement: For all integers a and b, if a 0 (mod 6) and b #0 (mod 6), then ab #0 (mod 6). Which of the following statements are true? (select all that apply) Original statement ✓ Contrapositive Converse Negation ☐ None of the statements are truearrow_forwardProposition: If m is an odd integer, then m + 6 is an odd integer. Proof: For m + 6 to be an odd integer, there must exist an integer n such that m+6=2n+1. Subtracting 6 from both sides, we see that m = 2n+1-6 = = 2n― 6+1 = 2(n − 3) + 1. Since the integers are closed under subtraction, then n-3 € Z. Hence, the last equation implies that m = = 2q+1 where q = n = 3. This proves - that if m is an odd integer, then m + 6 is an odd integer. Based upon the Reading assignment and the Elements of Style >>, which of the following is the most significant error in the proof? The proof does not use complete sentences The proof contains a sentence that begins with a mathematical symbol The proof uses cumbersome notation The proof contains a variable used for more than one object The proof is written backwards The proof uses an example to prove the general casearrow_forward
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