Determine which protist causes each of the following diseases in huma.pdf

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Determine which protist causes each of the following diseases in humans. leishmaniasis Trypanosoma cruzi malaria Plasmodium spp. Leishmania s beaver fever Toxoplasma gondii toxoplasmosis Giardia lamblia Chagas disease Solution Additional Information: 1) Leishmaniasis: This is caused from the bites of sandflies infected with the leishmania parasite.The cutaneous form presents with skin ulcers, while the mucocutaneous form presents with ulcers of the skin, mouth, and nose 2) Malaria: It is caused from the bite of infected female anopheles mosquito. The plasmodium parasite infect RBC\'s which leads to low haemoglobin count and recurrent fever. 3) Beaver fever: Also known as giardiasis, it causes diarrhea, abdominal pain, and weight loss. Vomiting, blood in the stool, and fever are less common. 4) Toxoplasmosis: The parasite attacks the lymph nodes, causing weakening of the immune system and seizures. 5) Chagas disease: Also known as sleeping sickness, the parasite attacks almost all the organs. If left untreated, the disease can prove fatal.Name of diseaseCausative protistLeishmaniasisLeishmania spp.MalariaPlasmodium sppBeaver feverGiardia lambliaToxoplasmosisToxoplasma gondiiChagas diseaseTrypanosoma cruzi.

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Need this in JAVA We have N numbers as an array, you need to find a prefix array and a suffix array, which we can get the maximum x or value with all elements in them. Notice that for prefix[0, 1] and suffix[r, n - 1], do not intersect (I Solution // C++ program for a Trie based O(n) solution to find max // subarray XOR #include using namespace std; // Assumed int size #define INT_SIZE 32 // A Trie Node struct TrieNode { int value; // Only used in leaf nodes TrieNode *arr[2]; }; // Utility function tp create a Trie node TrieNode *newNode() { TrieNode *temp = new TrieNode; temp->value = 0; temp->arr[0] = temp->arr[1] = NULL; return temp; } // Inserts pre_xor to trie with given root void insert(TrieNode *root, int pre_xor) { TrieNode *temp = root; // Start from the msb, insert all bits of // pre_xor into Trie for (int i=INT_SIZE-1; i>=0; i--) { // Find current bit in given prefix bool val = pre_xor & (1<arr[val] == NULL) temp->arr[val] = newNode(); temp = temp->arr[val]; } // Store value at leaf node temp->value = pre_xor; } // Finds the maximum XOR ending with last number in // prefix XOR \'pre_xor\' and returns the XOR of this maximum // with pre_xor which is maximum XOR ending with last element // of pre_xor. int query(TrieNode *root, int pre_xor) { TrieNode *temp = root; for (int i=INT_SIZE-1; i>=0; i--) { // Find current bit in given prefix bool val = pre_xor & (1<arr[1-val]!=NULL) temp = temp->arr[1-val]; // If there is no prefix with opposite // bit, then look for same bit. else if (temp->arr[val] != NULL) temp = temp->arr[val]; } return pre_xor^(temp->value); } // Returns maximum XOR value of a subarray in arr[0..n-1] int maxSubarrayXOR(int arr[], int n) { // Create a Trie and insert 0 into it TrieNode *root = newNode(); insert(root, 0); // Initialize answer and xor of current prefix int result = INT_MIN, pre_xor =0; // Traverse all input array element for (int i=0; i using namespace std; // Assumed int size #define INT_SIZE 32 // A Trie Node struct TrieNode { int value; // Only used in leaf nodes TrieNode *arr[2]; }; // Utility function tp create a Trie node TrieNode *newNode() { TrieNode *temp = new TrieNode; temp->value = 0; temp->arr[0] = temp->arr[1] = NULL; return temp; } // Inserts pre_xor to trie with given root void insert(TrieNode *root, int pre_xor) { TrieNode *temp = root; // Start from the msb, insert all bits of // pre_xor into Trie for (int i=INT_SIZE-1; i>=0; i--) { // Find current bit in given prefix bool val = pre_xor & (1<arr[val] == NULL) temp->arr[val] = newNode(); temp = temp->arr[val]; } // Store value at leaf node temp->value = pre_xor; } // Finds the maximum XOR ending with last number in // prefix XOR \'pre_xor\' and returns the XOR of this maximum // with pre_xor which is maximum XOR ending with last element // of pre_xor. int query(TrieNode *root, int pre_xor) { TrieNode *temp = root; for (int i=INT_SIZE-1; i>=0; i--) { // Find current bit in given prefix bool val = pre_xor & (1<arr[1-val]!=NULL) temp = temp-.

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An infinitely long sheet of charge of width L lies in the xy -plane between x=L/2 and x=L/2 . The surface charge density is . Derive an expression for the electric field E at height z above the centerline of the sheet. (Assume that z0 .) Express your answer in terms of the variables , L , z , unit vector k^ , and appropriate constants. Solution The sheet is kept on the xy plane with a width of L along the x axis. We would consider a section of the sheet at a distance of x from the origin of thickness dx. Since it extends infinitely along the y axis. We have an infinite rod with at distance of x from origin with linear charge density of dx Now, the electic field at z due to this infinitely long strip would be dx / 2o * sqrt(z^2 + x^2) However, by symmetry we can say that the net electric field will point along the z axis as the components in the xy plane would eventually cancel out. Hence dE = z dx / 2o * (z^2 + x^2) We integrate the above to obtain: E = ( / 2o)[atan(x/z)](from -L/2 to + L/2) or, E = ( / 2o)[atan(L/2z) - atan(-L/2z)] = ( /o)[atan(L/2z)] NOTE: There are two ways of approaching the problem, either you can assume a small section of thickness dy at distance y and then proceed, or you can assume an infinitely long strip at some distance x and then go ahead. However, it needs to be mentioned that the first way gives a little tricky integral whereas the second one, as above, is fairly straightforward. The integral above is of the form dx / (a^2 + x^2) = (1/a) atan (x/a) Further, while subtracting the max and min value for the integral, the negative. you would need the relation tanx = tan-x Hence the difference of atan of negative x and postiive x ends up getting added..

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A study tracked incidence rates of a blood vessel disorder over time for individuals 50 years and older (see figure below). What is the name for the type of the time trend pattern demonstrated in this chart? (Choose the one best answer) Secular Cyclical Calendar d Influenza Although it looks like incidence of the disease has increased over the past 40+ years, the authors dispel this hypothesis. What might be one explanation for what appears to be higher incidence rates of this disease over time if the incidence rates, in fact, did not increase? The disease is being spread more rapidly due to travel. People are living longer with the disease. cThe population over 50 years has grown in comparison to the under-50 population. d People who contract the disease are now more likely to be diagnosed due to more widespread knowledge of the disease among providers. a Women were more likely to contract the disease than men. b The rates of disease dropped substantially in 1980. c The cyclical nature of the disease indicates that it is infectious. d The prevalence appears to have increased over time.a Secularb Cyclicalc Calendar d Influenza Although it looks like incidence of the disease has increased over the past 40+ years, the authors dispel this hypothesis. What might be one explanation for what appears to be higher incidence rates of this disease over time if the incidence rates, in fact, did not increase?a The disease is being spread more rapidly due to travel.b People are living longer with the disease. cThe population over 50 years has grown in comparison to the under-50 population. d People who contract the disease are now more likely to be diagnosed due to more widespread knowledge of the disease among providers. The authors of this study concluded from the data shown in this figure that this disease is probably infectious. Why did they make this conclusion? a Women were more likely to contract the disease than men. b The rates of disease dropped substantially in 1980. c The cyclical nature of the disease indicates that it is infectious. d The prevalence appears to have increased over time. Solution can you please put the figure so that the questions can be answered..

A study tracked incidence rates of a blood vessel disorder over time.pdf

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A distant galaxy was found to have a redshift of 1.8. How far away i.pdf

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