The converson of the English to Christianity began a rich period of .pdf

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The converson of the English to Christianity began a rich period of artistic production in England, especially in the creation of a. secular mocaics, b. illuminated manuscripts, c. ecclesiastical treatises, d. gothic cathedrals Solution b. illuminated manuscripts Elaborately decorated manuscripts were prepared after the conversion. Thus b is the correct option..

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The Antoine equation often is used to describe vapor pressures: lnP = A B /(T + C) . For temperatures near the normal boiling point: a. Derive an expression that gives the behavior of Hvap as a function of temperature which is thermodynamically consistent with the Antoine equation. Solution The Antoine equation is a mathematical expression of the relation between the vapour pressure and the temperturer of pure substances. The equation was first proposed by Ch. Antoine, a french researcher, in 1888. The basic form of the equation is: lnP = A - B / ( T + C ) and it can be transformed into this temperature-explicit form: T = B ( A - lnP ) - C where: P is the absolute vapor pressure of a substance T is the temperature of the substance A, B and C are substance-specific coefficients (i.e., constants or parameters log is typically either log10 or loge A simpler form of the equation with only two coefficients is sometimes used: lnP = A - ( B \\ T ) which can be transformed to: T = B ( A - lnP ) The Antoine equation is a semi-empirical equation which expresses vapour pressure as a function of temperature. A new, rapid and highly accurate method for obtaining its three constants from experimental data is presented and applied to ethanol, water and 14 anaesthetic substances where p is the vapour pressure, T is temperature and A, B and C are component- specific constants. Usually, the Antoine equation cannot be used to describe the entire saturated vapour pressure curve from the triple point to the critical point, because it is not flexible enough. Therefore, multiple parameter sets for a single component are commonly used. A low-pressure parameter set is used to describe the vapour pressure curve up to the normal boiling point and the second set of parameters is used for the range from the normal boiling point to the critical point. The Antoine Equation is a vapor pressure equation and describes the relation between vapor pressure and temperature for pure water between 0°C and 373.946°C. For T in Kelvins and P in kPa: 273.150 < T 373.150: A = 16.5699, B = 3984.92, C = 39.724 373.150 < T 647.096: A = 16.7285, B = 4169.84, C = 28.665 Antoine(T) = exp(A B / (T + C)) For temperatures above the critical temperature (373.946°C / 647.096K), where water is a superfluid, the increase in vapor pressure appears essentially linear† with the same slope (i.e. dP/dT 235.88 kPa/°C). This behavior appears consistent with the Ideal Gas Law and statements that supercritical water behaves like an ideal gas. Now if we define Tc = 647.096 K, we can write P(T Tc) = Antoine(T) P(T > Tc) = 235.88 (T Tc) + Antoine(Tc) Hence, the behaviour of delta hvap as a function of temperature which is termodynamically consistent with the the antoine equation is derived ..

The Antoine equation often is used to describe vapor pressures lnP .pdf

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Table 19. Highest Level of Type of organization germ layers \"asymmetry (sponges) Cell Noa Cnidaria Pha An tea cnidarians-- jellyfish, hyvha. Radar 4 nerve Platyhelminthes ema toda (flatworm) Hoda Nematoda worms) bila uran 3 and have germ Germ associated with gastrulation. Note that deuterostomes product of developmental in which leads threetissue layers Solution 3. Bilateral symmetry 5.Value of body cavity: a) Pseudocoelomic fluid is rich in protein and other dissolved substances which is responsible for turgescent state of nematode body. b) Coelom makes the digestive system independent of the body wall. It can act as hydrostatic skeleton. It transports food, waste and other materials..

Table 19. Highest Level of Type of organization germ layers .pdf

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Question 12 (5 points) The reason that mitosis can result in the production of haploid gametes in alternation of generations is because Question 12 options: The sporophyte and gametophyte are both haploid Eggs and sperm are produced directly by the diploid sporophyte via meiosis Eggs and sperm are produced by the diploid gametophyte via mitosis Eggs and sperm are produced by the haploid gametophyte via mitosis Save Question 13 (5 points) In the fern life cycle, how does fertilization occur? Question 13 options: The sporangia on the undersides of the leaves produce gametes, which are flung our of the leaf and find each other on the ground to fuse The sporophyte produces sperm and eggs in gametangia. The sperm reach the eggs by swimming through water The gametophyte produces sperm and eggs in gametangia (antheridia and archegonia). The sperm reach the eggs by swimming through water There is no sexual reproduction in ferns Save Question 14 (5 points) The earliest land plants were most likely to have lived in which of the following habitats? Question 14 options: arid desert environments Salt water marshes Tropical rainforests fresh water wetlands Save The sporophyte and gametophyte are both haploid Eggs and sperm are produced directly by the diploid sporophyte via meiosis Eggs and sperm are produced by the diploid gametophyte via mitosis Eggs and sperm are produced by the haploid gametophyte via mitosis Solution Q12 Eggs and sperms are produced by haploid gametophyte by mitosis. The large, leafy fern is the diploid organism. On the undersurface of its fronds or leaves, its cells undergo meiosis to create haploid cells. However, these cells do not immediately unite with others to recreate the diploid state. Instead, they are shed as spores and germinate into small haploid organisms. Because the diploid organism creates spores, it is called the sporophyte generation of the life cycle. Upon reaching maturity, the haploid organism creates haploid egg and sperm cells (gametes) by mitosis . Because the haploid organism creates gametes , it is called the gametophyte generation of the life cycle. The male gametes (sperm) are then released and swim to the female egg. Fusion of the gametes creates the new diploid sporophyte, completing the life cycle. Q13 The gametophyte produces sperm and eggs in gametangia (antheridia and archegonia). The sperm reach the eggs by swimming through water The male organs or antheridia produce and shed numerous motile sperm or antherozoids. These antherozoids are armed with hair-like or whip-like cilia or flagellae and are able to swim through water; they do not travel great distances and are only released when free water is available. Only a continuous film of water is needed. They antherozoids are chemically attracted to the developing flask-like female organs or archegonia where fertilization of the single egg cell embedded in each archegonium takes place. The sporophyte has a double set of chromosomes (i.e. diploid) and i.

Question 12 (5 points)The reason that mitosis can result in the pr.pdf

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peripheral circuit functions built-in to the HC12 microcontroller, that are not part of the CPU itself or the memories. Access to these is via external pins on the Microcontroller IC package. What are some of these built-in circuit functions called, and what do they do? Solution The peripheral circuit communicate with the Microcontroller using built in cirucit called BUS. The bus is a set of shared communication lines that serves as the central nervous system of the computer. Data, address, and control signals are shared by all system components via the bus. Each component connected to the bus communicates information to and from the bus via its own bus controller. The data lines,address lines, and control lines allow a specific component to access data addressed to that component. The data lines are used to communicate words to and from data. registers in the various system components such as memory, CPU, and input/output (I/O) peripherals. The address lines are used to select devices on the bus or specific data locations within memory. Devices usually have a combinational logic address decoder circuit that identifies the address code and activates the device..

peripheral circuit functions built-in to the HC12 microcontroller, t.pdf

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Mendez Corporation has 10,000 shares of its $100 par value, 7 percent cumulative preferred stock outstanding and 50,000 shares of its $1 par value common stock outstanding. In Mendez’s first four years of operation, its board of directors paid the following cash dividends: 2011, none; 2012, $120,000; 2013, $140,000; 2014, $140,000. Determine the dividends per share and total cash dividends paid to the preferred and common stockholders during each of the four years. Solution The dividends per year on preferred stock = 10,000 x $ 100 x 7 % = $ 10,000 x $ 100 x 0.07 = $ 70,000 Dividend distribution: Year Total Dividend Amount Preferred Common 2011 - - - 2012 $120,000 $120,000 $0 2013 $140,000 $90,000 $50,000 2014 $140,000 $70,000 $70,000 Explanation: First dividend is paid to Preferred stock and then excess amount goes to common share holder. For cumulative preferred stock, any deficiency in dividend, not paid carries over to subsequent years and is added to that year\'s dividend. Preferred Dividends unpaid in 2011 = $70,000 Preferred Dividends on 2012 = $70,000 + $ 70,000 = $ 140,000 Preferred Dividends paid in 2012 = $ 120,000 Preferred Dividends shortage on 2012 = $ 20,000 As the Dividend was insufficient for preferred stocks, common stocks holder was not paid any dividend during 2011 & 2012. Preferred Dividends paid in 2013 = $ 70,000 + $ 20,000 = $ 90,000 Common Dividends paid in 2013 = $ 140,000 - $ 90,000 = $ 50,000 Preferred Dividends paid in 2014 = $ 70,000 Common Dividends paid in 2014 = $ 140,000 - $ 70,000 = $ 70,000 Computation of Dividends per share: 2011 2012 2013 2014 Preferred Stock (Dividends per share) $0 $12.00 $9.00 $7.00 Common Stock (Dividends per share) $0 $0 $1.00 $1.40 Explanation: Dividends per share = Dividend paid/No. of shares Year 2012: Dividends per share for Preferred Stock = $120,000/10,000 = $ 12 Dividend per share for Common Stock = $ 0 Year 2013: Dividends per share for Preferred Stock = $90,000/10,000 = $ 9 Dividends per share for Common Stock = $ 50,000/50,000 = $ 1 Year 2014: Dividends per share for Preferred Stock = $70,000 /10,000 = $ 7 Dividends per share for Common Stock = $ 70,000/50,000 = $ 1.40 Year Total Dividend Amount Preferred Common 2011 - - - 2012 $120,000 $120,000 $0 2013 $140,000 $90,000 $50,000 2014 $140,000 $70,000 $70,000.

Mendez Corporation has 10,000 shares of its $100 par value, 7 percen.pdf

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If the dpy-11 and unc-31 genes are linked on the same chromosome, how many phenotypes would the F_2 progeny show? If the two genes are on different chromosomes and therefore independently assorting, how many phenotypes would you see in the F_2 worms? List the phenotype(s). Strawberry plants have varieties that flower when the days are still short, and other varieties that flower only when the days get longer. The long-day allele is dominant to the short-day allele. You\'d like to test whether or not this gene is inherited in a Mendelian fashion. When you cross between heterozygous plants and count the offspring in the F_1 generation, this is what you find: 85 plants with no flowers on short days and 35 plants with flowers on short days What would you expect the ratio of phenotypes be? What is your Chi squared value? Show your work What does this mean? Solution 2. Linked genes tend to inherit together for the next-generations. No crossing over ororoccurs independent assortment occurs. No new characters appear. Where as gegenes are on different chromosomes, they are non linked. Crossing over or independent assortment takes place. New characters appear along with parental characters. So dpy-11 & unc-31 present on same chromosome, they are linked and give only parental characters. Only 2 phenotypes appear. If the said genes present on different chromosomes, they give 4 different phenotypes 3. a. Heterozygous long day representing as Ll. When crossing or hybridization occur between these genes Ll x Ll ----------> 3:1 is the phenotypes ratio. Three are long and one is short. Chi square value can be calculated by using the formula. (O-E)^2/E O = observed phenotypes; E = Expected phenotypes.

If the dpy-11 and unc-31 genes are linked on the same chromosome, how.pdf

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How does a wave formation influence the resultant waveSolution.pdf

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How are health services paid for? Provide a definition for the term \"third-party payer\", discuss the different payers currently operating in the market, and assess the importance of each. Please use a table format to list the information. Solution There are two main payers types: single payer and third-party payer. Single payer is when the person that obtains the health service is the same which pay for it, here relationship is direct taker and provider of health services. However, it is more common the presence of a third party payer, which is an external agent that covers the other health expenses or part of them, they are insurance companies or public social protection systems. Single Payer Third-Party Payer Type of relation with provider Direct Indirect Capability to compare Prices/Features Relatively High Almost Null Payment proportion Full Low percentage Single Payer Third-Party Payer Type of relation with provider Direct Indirect Capability to compare Prices/Features Relatively High Almost Null Payment proportion Full Low percentage.

How are health services paid for Provide a definition for the term .pdf

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Given the information regarding mitosis and meiosis, clearly and convincingly explain in your own words how an error in MITOSIS could possibly lead to a gamete containing new genetic information and thereby giving rise to a new phenotype. Use detail be comprehensive. (7 points) Solution The process of mitosis is divided into two phases: Interphase and M-phase. Interphase is further divided into G1, S and G2 phase. In G1, protein synthesis occurs for S- phase, in S phase, DNA replicates and in G2 phase, protein synthesis occurs for M-phase to start properly. Now, during S phase, the ‘2n’ nucleus converts into 4n. So, the cell entering M-phase is already ‘4n’. M-phase is the actual mitotic phase, in which the cell divides into two. It contains four phases, prophase, metaphase, anaphase and telophase. Your question is related with these four phases. In prophase, the entire DNA converts into chromosomes; the process is known as condensation of DNA. In metaphase, these chromosomes align on the equator. Now, if alignment at equator is not correct, then anaphase cannot be started due to metaphase check point. In anaphase, the aligned chromosomes move towards the opposite poles. This phase is the main phase, related with formation of gametes. The cell is already ‘4n’. ‘2n’ chromosomes move towards one end, and the rest of ‘2n’ move towards the other end. Now, suppose an error occurs in mitosis, and ‘3n’ chromosomes move towards one end, and 1n goes towards the other cell; then two gametes, ‘3n’ and n will be formed. This ‘3n’ and ‘n’ gametes are gametes having different genetic information in comparison to their ‘2n’ counterparts. Interestingly, Meiosis II will result in gamete formation is mitotic division (equational division). Haploid cells with \'2c\' amount of DNA leave Meiosis I. Now, in anaphase, suppose one chromosome\' 2c DNA migrates to one end (due to error) and no DNA migrates to the other end; then two gametes will be formed; one having all normal chromosomes + one double chromosome with 2c DNA, and the other having no DNA. Please note: normally gametes are 1n (haploid) and 1c. Suppose this gamete with double DNA (2c) fertilizes with a gamete with half DNA (1c), then the resulting cell will have “all normal chromosomes + one chromosome with ‘3c’ amount of DNA”. This condition if occurring with chromosome 21 is known as Down\' syndrome. It is important to note that errors in Mitosis result in non-disjunction; which further results in many Mendelian disorders; as well as ploidy defects. ** Hope it helps..

Given the information regarding mitosis and meiosis, clearly and con.pdf

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First, look to determine what is the earliest era that fossils have been found here in Georgia. Next, see if you can find what the oldest fossil that has been found in Georgia. Give the name of the fossil, when it was found, and where the fossil was found. Give a brief description of the organism that was fossilized. Solution Trilobites are the oldest fossils found in georgia.They have been found in Georgia’s Appalachian Plateau which is present the Coosa River in Floyd County. These fossils are considered to be arthropods,Moreover they had to shed their shells to grow.As a result the fossils found in these places usually shed shells as bits of individuals or as complete individuals or as imprints of individuals..

First, look to determine what is the earliest era that fossils have .pdf

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Figure out and write down a definition of a translation that does no.pdf

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explain why right and why wrong Indentify configuration (below) which I s not a correct representation of H and I. Solution option C is not correct one. Explanation: If we stretch all right hand fingers in the direction of current direction given then the thumb gives us into the board (into symbol) or out of the board (dot symbol). In the option C the field should be out of the board, so it is not correct one..

explain why right and why wrong Indentify configuration (below) whic.pdf

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Develop a system flowchart and then write a menu-driven C++ program that uses user-defined functions arrays, and a random number generator. Upon program execution, the screen will be cleared and the menu shown below will appear at the top of the screen and centered. The menu items are explained below. Help Smallest Largest Quit H or h ( for Help ) option will invoke a function named help() which will display a help screen. The help screen(s) should guide the user how to interact with the program, type of data to be used, and what results would the program produce. Each help screen should remain on the monitor until the user strikes any key. Once the user completes reading the help() screens, striking any key will clear the screen and the menu is displayed again. The main() function will declare an array of 60 elements. The elements are of type double. The menu options S and L, will ask the user for the actual number of elements for the array. The program must verify that the actual number of elements does not exceed 60 and is not a negative number or 0. Your code must check for these. S or s ( for Smallest ) option will invoke a function named smallest( ) which will prompt the user for the number of elements for the array to be examined using the function sizeOfArray( ) which will read and validate the desired array elements and returns it to the calling function. The program will then use the returned size of the array to fill the array using the number generator functions shown at the end of this assignment. Your program will ask the user for the range of values for the elements of the array and uses these numbers in a random generator function to fill the array. Once the array is filled, the program will call the function findSmallest( ) which will find and return the smallest number in the array to the calling function. The program will then call the function frequency(), that will compute and return the frequency of occurrence of the smallest number. The program will then display the array elements, the smallest number, and its frequency of occurrence using the function display( ), in the format shown below. The output shown is for an array of six elements with an array identifier a. a[0] = xxxx.xx a[1] = xxxx.xx a[2] = xxxx.xx a[3] = xxxx.xx a[4] = xxxx.xx a[5] = xxxx.xx Smallest no. = xxxx.xx Frequency = xx Note that for the array elements, two digits after the decimal point is required (i.e., 365.24) and the frequence of occurrence is of type integer. The function prototypes to be used are as follows: void smallest(double s[], int size); where s is the original array declared in the main() function and size is the maximum array size specified (60 elements in our case). //gets the desired array size and returns it int sizeOfArray(void); //finds the smallest number in the array and returns it via the return statement. double findSmallest(double s[], int myS); //where s is the specified array name and myS is the desired size of the array.

Develop a system flowchart and then write a menu-driven C++ program .pdf

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Description of nonaxial movement Description of other types of movement Part B - Uniaxial Movement Focus your attention on the section called uniaxial movements in parts (b) and (c) in Focus Figure 8.1 Drag and drop the terms describing a characteristic of uniaxial movement or the location in which it is found to appropriate blanks in the sentences. Solution Answer: There are four types of synovial joint movements; Uniaxial, Biaxial, Multiaxial and Nonaxial. Nonaxial movement: It is a gliding movement; no axix around which movement can occur. At a plane joint (gliding joint), the articulating surfaces of the bones are flat or slightly curved and of approximately the same size, which allows the bones to slide against each other. Example: This type of movement found between the carpal bones (intercarpal joints) of the wrist or tarsal bones (intertarsal joints) of the foot, between the clavicle and acromion of the scapula (acromioclavicular joint), and between the superior and inferior articular processes of adjacent vertebrae (zygapophysial joints). Uniaxial movement: The movement occurs around one axis; only in one plane. Example: Pivot and Hinge joints Biaxial movement: Occuring around two axes at right angle to each other. Example: Saddle and condyloid joint At a condyloid joint (ellipsoid joint), the shallow depression at the end of one bone articulates with a rounded structure from an adjacent bone or bones. One example is the radiocarpal joint of the wrist, between the shallow depression at the distal end of the radius bone and the rounded scaphoid, lunate, and triquetrum carpal bones. At a saddle joint, both of the articulating surfaces for the bones have a saddle shape, which is concave in one direction and convex in the other. The primary example is the first carpometacarpal joint, between the trapezium (a carpal bone) and the first metacarpal bone at the base of the thumb. Multiaxial movement: The movement occurs around several axes. Example: Ball and Socket joint The joint with the greatest range of motion is the ball-and-socket joint. At these joints, the rounded head of one bone (the ball) fits into the concave articulation (the socket) of the adjacent bone. The hip joint and the glenohumeral (shoulder) joint are the only ball-and-socket joints of the body..

Description of nonaxial movement Description of other types of move.pdf

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Centers for Medicare and Medicard Services is a very large purchaser of health care in the United States. As part of the American Recovery and Reinvestment Act (ARRA), \'meaningful use\' is an effort to ensure that health care organizations and providers really do make use of electronic health record systems in a meaningful way. The following represent key characteristics of meaningful use except. a. Development of legal health records. b. Use of electronic prescribing. c. Electronic exchange of information for quality of care. d. Participation in reporting quality of care measures Solution a part of the american recovery and reinvestment act (arra), \'meaningful use\' is an effort to ensure that he can care organizations and provides do really make use of each health record systems in a meaningful way. the following represents key characteristics of meaningful use except The option is A, Devolepment of legal health records It does not involve in it,.

Centers for Medicare and Medicard Services is a very large purchaser .pdf

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An alarm system is to have 4 different devices, each with its sensor. One of the sensors is to have interrupt address 83. Show and explain how that address is created using switches and how the INTA signal is used to enable a tri-state buffer/driver to place the address on the data lines. IX. On the 82C55 Parallel Programmable interface, the Control Port has the address ffffh. The control word 81h will program the lower-C-port as input and the upper-C port as output. (We don\'t care about the other ports). The address of the C-port is fffdh. Write a program to perform as follows: Address the Control Port and program the C-port as given above. Read from the lower-C port. Add 5 to the number that is read. Output the number read to the upper-C port. Return to step c. (The program will continue until operator intervention). Solution In the previous articles we saw about the architecture of 8085 microprocessor. Now we have a rough idea about how the instructions, data\'s are transferred and processed in 8085 microprocessor. In this article let us discuss in detail about the various signals involved in transferring data and executing instructions in microprocessor. The various signals in a microprocessor can be classified as Power supply and Frequency signals: Signals which aids in supplying power and generating frequency are associated with this type. Pins like Vcc and ground are classified under this type. Address signals: Signals associated with the lower order address bus and time multiplexed higher order address bus comes under this type of signals. Data Signals: Signals associated with data bus comes under this type. Control and Status Signals: Signals which are associated with timing and control unit such HOLD, RW’, WR’ etc. comes under this type of signals. Interrupt Signals: We know that signals like TRAP, RST 5.5 etc. are interrupt signals. Such signals come under this category. Serial I/O signals: These signals are used for giving serial input and output data. Signals like SID, SOD come under this category. Acknowledgement Signals: Signals like INTA’, HLDA acts as acknowledgement signal for 8085 microprocessor. Address Bus: The pins A8-A15 denote the address bus. They are used for the most significant bit of memory address. Address/Data Bus: AD0-AD7 constitutes the Address/Data bus. They are time multiplexed. These pins are used for least significant bits of address bus in the first machine clock cycle and used as data bus for second and third clock cycle. But what is a clock cycle? What is first clock cycle and second, third so on... A clock cycle is nothing but the time taken between two adjacent pulses of the oscillator. In simple words clock cycle refers to the transition between o volts to 5 volts and back to 0 volts. So the first clock cycle means the first transition of pulse from 0volts to 5 volts and then back to 0 volts. ALE: Address Latch Enable: In the previous article we saw how ALE helps in demultiplexing the lower order ad.

An alarm system is to have 4 different devices, each with its sensor..pdf

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A small aquatic crustacean called Ceriodaphnia cornuta has two distinct phenotypes, one with \"horns\" and one without.The researcher wanted to know whether the horns had any benefit in discouraging fish from eating the daphnia. So, in a large tank he put many C.cornuta in a 3 to 1 ratio with 3 unhorned to everyone with horns. He then introduced a natural predator of the daphnia, a small fish. After an hour, he removed the fish, \"sacrificed\" it, and looked at the daphnia in its belly. Of the 44 daphnia that the fish had eaten, 40 were unhorned and 4 were horned. Is there any true pattern in which phenotype was eaten? Use chi square. Solution.

A small aquatic crustacean called Ceriodaphnia cornuta has two disti.pdf

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A proof induction has two standard parts. The first establishing that a theorem is true for some small (usually almost always trivial. Next, an inductive hypothesis is the theorem is assumed to be true for all cases up to some the theorem is then shown to be true for the next value proves the theorem (as long as k is finite). As an example, we prove that the Fibonacci numbers, F_4 = 5, ..., F_i = F_i-1 + F_i-2, satisfy F_i Solution The induction proof works by verifying whether the example works for all possible values.. Let us see how it works.. For example :: Consider E(n) 12 + 22 + 32 + ... + n2 = (1/6). n.(n+1).(2n+1) Now, we need to prove that the above example is true for all possible values.. So, first we consider n = 1.. For n = 1, obviously the sequence is true.. For n=1, LHS = 12 RHS = (1/6).1.(1+1).(2+1) = 6/6 = 1 here LHS = RHS.. Now we have to prove that : if E(k) is true, then E(k+1) is true.. Let us consider E(K) is true.. We have to prove that property is valid for n = k+1 We have to prove: 12 + 22 + 32 + ... + k2 + (k+1)2 = (1/6).(k+1).(k+2)(2k+3) Left side = 12 + 22 + 32 + ... + k2 + (k+1)2 = (12 + 22 + 32 + ... + k2) + (k+1)(k+1) = (1/6). k.(k+1).(2k+1) + (k+1)(k+1) = (1/6).(k+1). [ k(2k+1) + 6(k+1)] = (1/6).(k+1).(2 k2 + 7k + 6) Right side = (1/6).(k+1).(k+2)(2k+3) = (1/6).(k+1).(2k2 + 3k + 4k + 6) Here, LHS = RHS.. I.e., this is true for E(k+1) therefore our assumption is true.. That means the equation is true for all possible values by principle of induction...

A proof induction has two standard parts. The first establishing tha.pdf

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A population has a proportion equal to 0.30. Calculate the following.pdf

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1) Compare Figures. What did Da Vinci get right, and what did he diagram incorrectly? Answer must be no less than 250 words. 2) How do Prout’s data determine that only HCl is produced in the stomach? Answer must be no less than 500 words. salivary gland pharanx liver gall bladder duodenum large intestine tongue salivary gland esophagus stomach pancreas jejunum ileum rectum anus Solution 1). During the lifetime of Leonardo Da Vinci, he made thousands of pages of notes and anatomical drawings of human body. He observed human body keenly and drawn them with an intension to understand the human body’s composition and its function. However, by the time of his death, his scientific publications were incomplete and unpublished. But with the observation of Da Vinci’s work, scientists say that they still help us to understand the human body. One of the interesting drawings of Da Vinci is, skull. He made a series of drawings of skull. In one diagram, the front teeth (half position) are cut to show what is behind the teeth. It is difficult to cut these teeth without damaging the other bones. He also left a drawing of different knives he used for cutting. Peter Abraham, Prof. of clinical anatomy at Warwick University in the UK, says that the Da Vinci is accurate and his drawings are similar to the scientific artists that work now. He described Da Vinci as a meticulous observer and experimental scientist. Da Vinci have the ability to draw perfectly what he saw. The drawings of the upper part of the body, i.e. the liver, stomach, intestine, lungs, heart, the spinal cord, embryo, etc. But Prof. Abraham says that the uterus is wrong and it indicates the uterus of a ruminant animal such as cow. He might have used the knowledge of animal dissection in imagining the human uterus because holding the female corpses is fairly difficult during those days..

1) Compare Figures. What did Da Vinci get right, and what did he dia.pdf

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