Therefore Essay Examples

Therefore x=2π3 is a maximum value -2sin4π3 =-2sin-32=√3And x=4π3 minimum C. f(x)=x13(x+4)The first derivative =x13ddx(x+4)+x+4ddxx13=x13+(x+4)(13x23)=x13+(x3x23+43x23)f'(x)=43x23+4x133Using the product rule to get the second derivative =43ddxx-23+x13+x-23+x13ddx43=43(-23x-53+13x-23)f''x=-89x53+49x23d. The intervals on which the function increases or decreases and local minimum when the first derivative is equals to zero f' , 43x23+4x133=04x133x-1 +1=0 it follow that x =0 or-1Function decreases for x<-1 and -1<x<0Increases for x>0Local minimum x=0e. concave down or up for the function and sketch a graph f''x=-89x53+49x23 when f''x=0 therefore-492x-53-x-23=0 x=2by...

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Therefore: abs minima =1 at x=1 Abs maxima=10 at x=-2 (d) Calculate the limits of the following limx→0x+5x2 Solution Divide both numerator and denominator by x2 =lim x->0 1/x+5/x2 Since we have 0 in the denominator, the limit=0 (e) evaluate A'(x) at x = 1, 2, and 3. A(x) = -3x 2t dtSolution =d/dx(t2) At x=1; (12)-(-32) =-5 At x=2; (22)-(-32) =-2 At x=3;(32)-(-32) =12 (f) . Let A(x) represent the area bounded by the graph and the horizontal axis and vertical lines at t=0 and t=x for the graph in Fig. 25. Evaluate A(x) for x = 1, 2, 3, 4, and 5. Solution At X=1; A=1*1= 1unit2 X=2; A=1+1.5=2.5 units2 X=3; A=2.5+(1*2)= 4.5 units2 X=4; A=4.5+1.5=6 units2 X=5; A=6+1=7...

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therefore avoid penalties would be enacted through effective team selection. There would be a selection of the team that is capable of doing the crashing in the shortest time possible, while at the same time delivering quality work. The process could even entail doing away with some of the workers and replacing them with the loyal employees. There would also be an embrace of motivation tactics aiming at increasing the zeal of the workers in achieving much fast crashing. On realizing the positive turn in activities towards effecting the crashing in the correct schedule, there would be efforts of supporting the team efforts using incentives. Comment Getting estimates on the regular timing as well as...

Therefore, the basis holds true for n=0The inductive hypothesis becomes n=k≥0, k3+k+13+(k+2)3 is also divisible by 9 → k3+k+13+(k+2)3=9p…(i)By inductionn=k+1, equation (i) becomes (k+1)3+(k+2)2+(k+3)3=9p-k3+k+33…(ii) Simplifying (ii) we have =9p-k3+k+33=9p-k3+k3+9k2+9k+27=9p+9k2+k+3=9(p+k2+k+3)This implies that (k+1)3+(k+2)2+(k+3)3 is also divisible by 9 therefore, n3+(n+1)3+(n+2)3is divisible by 9 for all n≥1 d. 83 – 3n Is divisible by 5 for all n≥1Proof If n=1 then, 8n-3n= 81+31=5 this can be divided by 5. Therefore, the basis holds true for n=1The inductive hypothesis becomes n=k ≥1. It means that 8k-3k is divisible by 5 →8k-3k=5p…(i)By induction, taking n=k+1 we...

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therefore, efficient to conduct a simple data analysis procedure. The data indicates that there is an overwhelming necessity to decongest traffic at entry and checkpoints of USC. Values that were suggested that there is a challenge in the traffic scored averages that indicated a remarkable coherence of all the 20 respondents. Most of them are beyond 4.5 with a number actually having the maximum value of 5. On the other hand, prompts that asked about the efficiency of USC traffic management gave a low average indicating dissatisfaction. The query “I have easy to a parking space at my department,” generated an average of 1.75 indicating disagreements. It is, therefore, appropriate to assume that...

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