Verify that the eqn is an identity. show steps please.sec4 xsec2 x.pdf

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Verify that the eqn is an identity. show steps please. sec4 xsec2 x=tan4 x+tan2 x Solution sec4 xsec2 x=tan4 x+tan2 x LHS sec^4x -sec^2x use the identity : 1+tan^2x = sec^2x So, (sec^2x)^2 - sec^2x =(1 +tan^2x)^2 -( 1+tan^2x) = (1 +tan^2x) [ 1+tan^2x -1] =(1 +tan^2x)(tan^2x = tan^4x +tan^2x = RHS Hence it is an identity.

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Decision Tree Table for choosing present and future value equations. ltem to Number of Present or Annual or Perpetual or evaluate occurrences Future Equation to use Periodi Terminal Series na na 0 value payment:s Future Present na na Periodic Annual Periodic Vn Vo(1+i)\" Vo=Vn/(1+i)\" Vn=pl(1+i)n-1)/[(1+1)t-1)] Single Revenue Future Terminal Perpetual Vo= p/[(1+i)\'-1] Series of Equal payments Present Terminal Vo (1-1(1+)-1](1+i)) Cost PerpetualVo Annual Terminal Vo=p[(1+i)n-1)/[i(1+i)\") interest rate per year (in decimals) Vo = initial value (in time 0) V- value in time n n total number of years of compounding or discounting considered p = amount of payment occurring every t years t number of years in one multiyear group or interval of years (length of the interval) Payments occur at the end of the period or multiyear interval Please note: these equations can be used for months or weeks or any other length of time as long as the interest rate is defined at that period, n is defined as total number of periods, and t is the number of periods in one multiperiod group or interval. In this class, we will stick primarily to using the annual rates. Source: Figure 7.3 from Davis, et al. 2001 Forest Management, p. 341 Solution a. As seen from the table, Present value of perpetual periodic CF is = P/[(1+i)^t - 1]= 50000/[1.07^45 - 1] = 50000/(21.002 - 1) = $2,499.69 b. Present value of a perpetual $100 tax = P/i = 100/0.07 = $1,428.57 Present value of the cash flow = 2,499.69 - 1,428.57 = $1,071.12 c. AS the asking price of the land is $1000 and is less than the NPV of $1,071.12 you should buy it..

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Consider this experiment I have 20-sided die where each face has a .pdf

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congenital malformations. why do you think they occurSolution.pdf

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Considering the cultures used to inoculate each medium in this laboratory, how many different microbial types should you expect to see in/on each medium? Which medium was most difficult for you to transfer from? which medium was most difficult for you to inoculate? Explain your difficulties. What is the purpose of flaming in the aseptic techniques? Provide 3 (three) reasons why the use of aseptic techniques is essential when handling microbial cultures. Methoos of isolation What is a microbial colony? Can a pure culture by prepared from a mixd broth or mixed agar slant? Explain. Solution Answers are : When the cultures are inoculated onto the surface of the medium, different colonies are obtained, based on the mother culture we have chosen colonies are obtained. Until unless it is not contaminated so. Preparation of agar slants is difficult to transfer, and inoculate, because it needs to place correct angle while solidifying, less surface area. Though among the medium we can say some difficult but not so rather difficult. Nutrient broth is also difficult to transfer because more prone of contamination and determination of the bacterial growth is also difficult. Purpose of flaming during aseptic transfer :- Flaming causes direct burning of the cultures samples which stick onto the inoculating loop to avoid from contamination, which technique is called Incineration i.e., Flame sterilization. Aseptic techniques are safe transfer of culture onto the fresh medium this is very essential because :- To avoid contamination to grow unwanted cultures and spoiling the medium components. To know the perfect study of the unknown characteristics, contamination shouldn’t be seen . To avoid infecting the person who handling the sample, some may cause severe infections and death of the investigator. To obtain the pure culture techniques to study the characteristics of the cultures. Microbial colony means, group of the similar cells which shows same characteristic features , which obtained on the surface of the solid agar medium. These are varied in shapes and surface but shows same morphological and physiological characteristics. Yes, we can prepare pure culture from mixed sample either broth or solid agar, in case of broth prepare serial dilution of the sample to obtain the pure cultures. In case of the solid slant prepare quadrant streaking on the fresh agar plate..

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Are the transfermations linear? Are they an isomorphism? Why? Solution 1. T(f)=f\' T(f+g)=(f+g)\'=f\'+g\' T(cf)=(cf)\'=cf\' So, T is linear P2 is space of polynomials of degree atmost 2 so T(P2) can have polynomials of degree atmost 1 So it cannot be an isomorphism because P2 has dimension 2 and T(P2) has dimension atmost 1 2. T(f+g)=(f+g)\'+(f+g)\'\'=f\'+f\'\'+g\'+g\'\'=T(f)+T(g) T(cf)=(cf)\'+(cf)\'\'=cf\'+cf\'\'=cT(f) HEnce, T is linear But T is not an isomorphism As T maps a polynomial to a polynomial of degree stricly smaller than itself. HEnce P3 and T(P3) have different dimension so T cannot be an isomorphism..

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bessavilla bessavilla bessavilla bessavilla Solution Besavilla Review Center, formerly called Besavilla Engineering Review Center was the oldest review center (32 years) for civil engineering up to now. It was founded by Engr. Venancio I. Besavilla, Jr. at Cebu City on 1980 and expanded its operations in Manila and other key cities in the country (Baguio, Cagayan de Oro, General Santos City and Davao) Besavilla Review Center offer comprehensive review courses to Civil and Geodetic Engineering Students using innovative and updated teaching technique. It is an intensive lecture/discussion type covering basic theories, principles, problems and applications..

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ciny Univer Laboratory es and crossing over assortment of independent the processes of Explain Meiosis. 3 Compare and contrast meiosis with mitosis. Number of times Number of cell divisions Number of daughter cells Genetic relationship of daughter cells to What structures separate during anaphase (Anaphase I in Meiosis)? Name the function(s) each process performs 4. Describe the events ofspermatogenesis and oogenesis. s. Define the followin Study suggestion 1. Explain Solution Answer for Question 3.

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A population of rabbits quadruples every 2 years. If the initial number of rabbits was 25, how long will it take for the population to reach 700?? I believe I understand how to solve the logarithm, I just don\'t understand how to extract the log it from the question. Solution The population of rabbits quadruples every 2 years. Let the number of years taken to reach 700 be T. The initial population is 25. This gives an equation: 700 = 25*(4)^(T/2) => 700/25 = 4^(T/2) => 28 = 4^(T/2) take the log of both the sides => log 28 = (T/2)*log 4 => T/2 = log 28/log 4 => T = 2*(log 28/log 4) => T = 4.807 years. The population of rabbits grows to 700 from 25 in 4.807 years..

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Assembly Language: Reverse an array. Use a loop with indirect or indexed addressing to reverse the elements of an integer array in place. Do not copy the elements to any other array. Do not use stack operations. Use the SIZEOF, TYPE, LENGTHOF operators. Fibonacci Numbers. Write a code that fills the array of 15 elements with consecutive Fibonacci numbers: Fib(1) = Fib(2) = 1; Fib (n) = Fib(n-1) + Fib(n-2) for each N > 2. corrent code: Solution data barray BYTE 6 DUP (?) dwarray DWORD 3 DUP (?) TOTAL_LENGTH = $-barray bdmessage BYTE \"Today as stored in a double word\", 0dh, 0ah, 0 dwmessage BYTE 0dh, 0ah, \"Seeing bytes in a single word\", 0dh, 0ah, 0 lengthmessage BYTE 0dh, 0ah, \"total bytes used: \", 0 finalmessage BYTE 0dh, 0ah, \"Press any key to continue ...\", 0 .code main PROC CALL Crlf MOV edx, OFFSET bdmessage CALL WriteString MOV eax, DWORD PTR [barray + 2] CALL WriteDec MOV edx, OFFSET dwmessage CALL WriteString MOVZX eax, WORD PTR [dwarray + 6] CALL WriteDec MOV edx, OFFSET lengthmessage CALL WriteString MOV eax, TOTAL_LENGTH CALL WriteDec MOV edx, OFFSET finalmessage CALL WriteString CALL ReadChar INVOKE ExitProcess, 0 main ENDP END main.

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wo alternative investment proposals are under consideration for a vacant owner by Urban Development Corporation. Plan A would require an immediate investment of $120,000 and first- year expenditure for property taxes, maintenance, and insurance of $4,000, with this amount expected to increase at a rate of $1,000 per year. Plan B would have a first cost of $170,000 and total first-year expenses of $9,000, with an increase of $1,000 per year. The economic life of each project is forecast to be 10 years; and at the end of this time, only the facilities from Plan B with a value of $50,000 are expected to salvage. During the life of the project, the facility in plan A is expected to produce $34,000 annually, whereas Plan B is expected to produce $42,000 . a) Determine the rate of return of each plan. b) Determine the rate of return of the Additional investment required in Plan B compared with Plan A. c) Which plan should Urban Development select if the company uses a MARR of 12 percent? Solution Working notes: (1) For each plan, Net Annual Benefit = Annual Benefit - Annual Expense (2) For plan B, Net Annual Benefit in terminal year (year 10) is higher by $50,000 (Salvage Value). (3) Rate of Return is the Internal Rate of Return (IRR) computed using Excel function =IRR(Value 1, Value 2,.....Value n) where \"Value\" is the Net Annual Benefit, undiscounted. (4) Based on MARR of 12%, the plan choice will be made based on NPV of the plans. NPV is the sum of all cash outflows and inflows discounted at 12%. The plan with higher positive NPV should be chosen. Calculations as follow: Plan A Year Initial Cost ($) Annual Benefit ($) Annual Cost ($) Net Annual Benefit ($) Net Cash Flow ($) Discount Factor @12% Discounted Net Annual Benefit ($) (A) (B) (C) (D) = (B) - (C) (E) = (D) + (A) (F) (G) = (E) x (F) 0 -1,20,000 -1,20,000 1.0000 -1,20,000 1 34,000 4,000 30,000 30,000 0.8929 26,786 2 34,000 5,000 29,000 29,000 0.7972 23,119 3 34,000 6,000 28,000 28,000 0.7118 19,930 4 34,000 7,000 27,000 27,000 0.6355 17,159 5 34,000 8,000 26,000 26,000 0.5674 14,753 6 34,000 9,000 25,000 25,000 0.5066 12,666 7 34,000 10,000 24,000 24,000 0.4523 10,856 8 34,000 11,000 23,000 23,000 0.4039 9,289 9 34,000 12,000 22,000 22,000 0.3606 7,933 10 34,000 13,000 21,000 21,000 0.3220 6,761 IRR (%) = 18.12% NPV ($) = 29,253 Plan B Year Initial Cost ($) Annual Benefit ($) Annual Cost ($) Net Annual Benefit ($) Net Cash Flow ($) Discount Factor @12% Discounted Net Annual Benefit ($) (A) (B) (C) (D) = (B) - (C) (E) = (D) + (A) (F) (G) = (E) x (F) 0 -1,70,000 -1,70,000 1.0000 -1,70,000 1 42,000 9,000 33,000 33,000 0.8929 29,464 2 42,000 10,000 32,000 32,000 0.7972 25,510 3 42,000 11,000 31,000 31,000 0.7118 22,065 4 42,000 12,000 30,000 30,000 0.6355 19,066 5 42,000 13,000 29,000 29,000 0.5674 16,455 6 42,000 14,000 28,000 28,000 0.5066 14,186 7 42,000 15,000 27,000 27,000 0.4523 12,213 8 42,000 16,000 26,000 26,000 0.4039 10,501 9 42,000 17,000 25,000 25,000 0.3606 9,015 10 92,000 18,.

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You are searching for new species that are highly tolerant to acidic conditions and find a new single celled organism that you call WERID. The organism is ~5 um, does not have a cell wall, but your portable field microscope is broken, and you are unable to look at the internal structures. You do have portable equipment that will allow you to analyze the chemical content of the WEIRD cells along with control prokaryotic and eukaryotic cells. Out of protein, DNA, and lipid, which analysis would be most useful if you were interested in figuring out if your cells are prokaryotic or eukaryotic? Explain your answer. Solution The organism that we have discovered is single celled, and devoid of a cell wall. It is ~5 micrometers in length. Since the organism has no cell wall, we cannot reach a conclusion based or the presence/absence of peptidoglycan. Also, morphological analysis cannot be carried out since the microscope is broken. However, we do have portable equipment to analyze the chemical content of the cells: protein, DNA and lipid, and compare it with control prokaryotic and eukaryotic cells. The protein and lipid type and content may vary between prokaryotes and eukaryotes. However, the DNA has certain differences that are absolutely unique to prokaryotes and eukaryotes. For example, the ribosomes-part of the protein synthesis machinery differ between the prokaryotes and eukaryotes in that the prokaryotes have a 16S rRNA while the eukaryotes have an 18S rRNA. This difference can be exploited using our portable equipment to check for presence of genes in the WEIRD DNA that code for either of these rRNAs, and comparing them with the controls. Also, eukaryotes usually have DNA in the form of multiple chromosomes, while the prokaryotes have a single circular DNA molecule. The above differences are irrefutable, and unique to prokaryotes and eukaryotes. Therefore, our choice for identification of the WEIRD cells would be DNA analysis. P.S. It is also important to take into account that eukaryotes are usually multicellular, while prokaryotes are unicellular. Also, eukaryotic cells are generally larger in size than that of the prokaryotes..

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Why must some nutrients be absorbed by active transport rather than passive diffusion? Solution Passive transport is the absorption of molecules down their concentration gradients, which does not require energy. Active transport is the absorption of nutrients, which require energy for the process to occur. Only small molecules and water can pass through the cell membrane by passive transport. The transport of many nutrients require energy (active transport) because of their larger molecular size and hydrophilic nature..

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Who first discovered the properties of equality?? Solution Equality is a relationship between two mathematical expressions. The equality between the mathematical expressions A and B is written as A = B. The word “equality” is derived from the Latin word “aequalis” which means equal. The symbol = was used for the first time in 1557 by Robert Recorde. It was Leibiniz who first characterized its notion..

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What happens when a bacterium is placed in a HYPOTONIC solution? What about HYPERTONIC? Solution Hypertonic solution is a solution which has a higher concentrate of solute and a solution with a low concentration of dissolved solute.When a bacterial cell placed in a hypertonic solution will shrink and die due to the water content in bacteria goes to outside having lower concentration of water.This process is called plasmolysis.At the other case,process by which a bacterial cell when placed in a hypotonic solution will swell and burst by absorbing high water concentration outside into bacterial cell is called plasmoptysis..

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Why does the stack use a FILO (first in last out) scheme? Solution When a program starts executing, a certain contiguous section of memory is set aside for the program called the stack. A stack is a limited access data structure - elements can be added and removed from the stack only at the top. push adds an item to the top of the stack, pop removes the item from the top. A helpful analogy is to think of a stack of books; you can remove only the top book, also you can add a new book on the top. Hence when the elements are pushed onto the stack using push operations, the first element pushes will be at the bottom of stack and the last element pushed in will be at top of stack. Therefore when pop operation is executed , top element ( last element pushed in )of stack is popped out first and hence first element pushed in popped out last. Thereby Stack operates in LIFO (last in first out ) or FILO ( First In Last Out) Scheme.

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Trace the path followed by a sperm and by and egg from the area of their formation to the outside of the body. Solution The path followed by sperm self from formation to ejaculation: The sperm cells are produced in the testicles. They travel through the seminiferous tubules into epididymis. Epididymis stores them until the get matured. 10 day travel through West difference where the pickup the fluid from seminal vesicles prostrate gland and cowper\'s gland. This fluid and sperm cells is called semen. It is released from the body during ejaculation. SEVEN UP!! S- seminiferous tubules E- epididymis V- vas deferens E- ejaculatory duct N-nothing U-urethra P-penis path of ovum: Female is born with all the eggs that she will have in her lifetime one egg matures every month and the egg is either fertilized are discarded. The egg starts in the ovaries and moves into the fallopian tubes, from the fallopian tubes the egg travels to the uterus The egg is fertilized in the uterus or comes out the cervix and through the vagine as a period..

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The potential distribution of organism can be influenced by all of the following EXCEPT: A. salinity B. competition with other species of organisms C. environmental temperature D. availability of environmental oxygen Solution B. Competition with other species of organism. Potential distribution of an organism can be defined by its own physiological capacity of organism. It is affected by tempreture, salinity and available eviromental oxygen..

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