You isolated an enveloped RNA virus. Purified RNA is not capable of .pdf

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Without computing eigenvalues, show that the matrices all have the same eigenvalues. [1 2 3 4], [10 4 -12 -5], [19 -11 24 -14] Solution HEre we find that on solving that the value of all matrices are same as : |A|= 1(4) -3(2) = 4-6= -2 |B|= -5(10) -(-12)(4)= -50+48 = -2 |C|=19(-14) - (-11)(24) = -266+264 =-2 So as there values are same, their eigan values will also be the same..

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You isolated an enveloped RNA virus. Purified RNA is not capable of supporting translation in
cellfree systems. Further analysis indicates that this virus replication is not inhibited by
actinomycin D, a strong inhibitor of DNA-dependent RNA polymerase. Based on this
information, discuss and fully explain the mechanism by which enveloped RNA replicates in
infected cells.
Solution
The virus has RNA as a genetic material by which it replicates inside the infected cells. It is not
capable of producing its own machinery for translation of information from RNA into proteins.
So it depends on the host for this machinery through which it produces new virus particles in the
host cells.
When a cell-free system is used, it does not have the proteins and enzymes from the transcription
and translation process, due to which this enveloped RNA virus does not survive.
Actinomycin D inhibits the classical RNA polymerase which uses DNA as a template in the
cells, while the enveloped RNA replicates through an RNA-dependent RNA polymerase which
uses an RNA as a template.
On whole, the enveloped virus uses the host cell machinery such as DNA polymerase, nucleotide
pool for its replication and survival in the infected host.

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I need to fill-in //TODO\'s in .cpp file and in .h file Could someone help me at least with few of them to give me an idea how deal with it. ***SinglyLinkedList.cpp #include #include #include \"SinglyLinkedList.h\" void test_constructor() { SinglyLinkedList lst = {100, 200, 300, 400, 500}; assert(*lst.at(0) == 100); assert(*lst.at(1) == 200); assert(*lst.at(2) == 300); assert(*lst.at(3) == 400); assert(*lst.at(4) == 500); assert(lst.size() == 5); } void test_remove() { SinglyLinkedList lst = {100, 200, 300, 400, 500}; lst.remove(2); assert(*lst.at(0) == 100); assert(*lst.at(1) == 200); assert(*lst.at(2) == 400); assert(*lst.at(3) == 500); assert(lst.size() == 4); } void test_insert() { // TODO } void test_push_back() { // TODO } void test_push_front() { // TODO } void test_append() { // TODO } void test_sum() { // TODO } int main() { test_constructor(); test_remove(); test_insert(); test_push_back(); test_push_front(); test_append(); test_sum(); } ***SinglyLinkedList.h #include #include template class SinglyLinkedList { // Nested class representing the nodes in the list. class SinglyLinkedListNode { public: // The value stored in this node. T value; // The next node in the sequence. SinglyLinkedListNode *next; SinglyLinkedListNode(T value) : value(value), next(nullptr) {} SinglyLinkedListNode(T value, SinglyLinkedListNode *next) : value(value), next(next) {} // Return the size (length) of the linked list. std::size_t size(); }; SinglyLinkedListNode *head; SinglyLinkedListNode *tail; public: // Constructs a new SinglyLinkedList from an initializer_list of type T[]. // This is mostly for convenience, especially when testing. SinglyLinkedList(std::initializer_list items) : head(nullptr), tail(nullptr) { if (items.size() == 0) { return; } // initializer_lists were designed to be used iteratively, // so thats what we do. // Can you think of how to write this recursively? auto it = items.begin(); while (it != items.end()) { this->push_back(*it++); } } // Return a pointer to the value at the given index. // If the index is larger than the size of the list, // return a nullptr. // // ASIDE: We will cover exceptions later. T* at(std::size_t i); // Pushes a new node onto the back of the list. void push_back(T value); // Pushes a new node onto the front of the list. void push_front(T value); // Return the size (length) of the linked list. std::size_t size(); // Remove the specified node from the list. void remove(std::size_t i); // Insert the value at the index. void insert(std::size_t i, T value); // Append the given list to this one. void append(SinglyLinkedList list); }; template T* SinglyLinkedList::at(std::size_t i) { // TODO } template void SinglyLinkedList::push_back(T value) { // TODO // Make sure that this is a O(1) operation! } template void SinglyLinkedList::push_front(T value) { // TODO // Make sure that this is a O(1) operation! } template void SinglyLinkedList::remove(std::size_t i) { // TODO // Don\'t forget to not only unlink the node, b.

I need to fill-in TODOs in .cpp file and in .h file Could some.pdf

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how important is Negative Emotionality to an accounting career? please explain in detail Solution The negative emotional attitude is found to be an important for accounting job because such emotions like, embarrassment, anxiety, frustration, anger and guilt, would lead to discrete links towards rigorous assessment and testing, feedback on employee’s performance, personal study and satisfaction. The functional characteristics have to be performed by an Accountant while on his/her job. Through all these characters, the person is derived to report and promote various strategies to achieve the objectives and goals. The negative emotions within the academic accounting environments would enhance the performance and motivation on the job under various situations. To investigate and finding the evidences are motivated by the negative nature of Accounting Jobs and its functions. Thus, it is always important to have negative emotionality in an accounting career. ==============================================.

how important is Negative Emotionality to an accounting career plea.pdf

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How do I know whether miscellaneous expense goes on top or bottom of.pdf

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Given a 1024 by 1024 RAM block, answer the following questions a) If we used that to make an 2-bit wide memory i) How many 2-bit locations would there be? ii) How many address bits would we need to address those locations? iii) How many of those address bits would go to the decoder? The MUX? b) If we used that to make a 16-bit wide memory [3] i) How many 16-bit locations would there be? ii) How many address bits would we need to address those locations? iii) How many of those address bits would go to the decoder? The MUX? Solution 2) a) Given Ram = 1024 bytes = 8192 bits i) If 2 bits = 1 location then 1024 bytes = 8192 bits = 8192 / 2 locations = 4096 locations ii) 1 location = 1 binary representation Therefore, 2n represents the locations . 4096 locations <= 2n where n is no of bits required 4096 locations <= 212 => 12 bits of address bits required iii) Out of 12 address bits ; 11 bits goes to the MUX b) i) If 16 bits = 1 location then 1024 bytes = 8192 bits = 8192 / 16 locations = 512 locations ii) 1 location = 1 binary representation Therefore, 2n represents the locations . 512 locations <= 2n where n is no of bits required 512 locations <= 29 => 9 bits of address bits required iii) out of 9 address bits ; 8 bits are given to decoder.

Given a 1024 by 1024 RAM block, answer the following questions a) If.pdf

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