T1 and T2 can refer to the Thoracic vertebrae (part of your spine). But I don't know what 'low' would imply.
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∙ 13y agoThe SHLD (Store H&L Direct) instruction takes 5 machine cycles and 16 clock states, not including any wait states. Opcode fetch: T1, T2, T3, and TX Low order address fetch: T1, T2, T3 High order address fetch: T1, T2, T3 Store L: T1, T2, T3 Store H: T1, T2, T3
COF = h1-h4/h2-h1=T1(s1-s4)/T2-T1(s1-s4)=T1/T2-T1
T1 and T2 refer to the first and second thoraxic vertebrae. They are the 8th and 9th vertebrae, starting from the top.
The CalDigit Thunderbolt™ T1 and T2 primarily differ in that the T1 is a single-drive solution whereas the T2 is a dual-drive solution. See related links for more information.
t1:german tiger 1 t2:german tiger 2 t1:armor 69 t2:armor 89 t1:speed 14 mph t2:speed 20 mph t1:gun is 98% great t2:gun is 99% good so german tiger 2 is better
This question refers to the combined gas law: (P1V1)/T1=(P2V2)/T2, where P is pressure, V is volume, and T is temperature in Kelvins.To solve for T1, rearrange the equation to isolate T1.T1=(P1V1T2)/(P2V2)
T1= Fat- Appears Bright e.g. Grey matter = Water- Appears Dark e.g. CSF, water T2 Just opposite to T1
Let X(t) be an iid random process and hence X(t) has an identical distribution for any t i.e., distributions are identical at instants of time t1, t2...tn, so 1st order pdfs f(x1;t1), f(x2;t2)....f(xn;tn) are time invariant and further X(t1) and X(t2) are independent for any two different t1 and t2. So, f(x1, x2, . . . , xn; t1, t2, . . . , tn) = f(x1;t1)*f(x2;t2)*....*f(xn;tn) f(x1;t1), f(x2;t2).... f(xn;tn) are time invariant, therefore their product f(x1, x2, . . . , xn; t1, t2, . . . , tn) is also time invariant which is nth order pdf. So X(t) is strict sense stationary.
A T1-T2 disc herniation is a herniation that happens in the middle or lower back. This will cause extreme pain and possible numbness in the limbs.
R= R0 * [1 + rho( t2-t1 ) ] so from this equation , rho= R-R0/[R0(t2-t1)] where rho- coefficient of resisivity R-resistance at any time t R0- resistance at 00C t2-final temperature t1-initial temperature
Gay-Lussac's law states that the pressure of a gas is directly proportional to its temperature at constant volume. The formula is P1/T1 = P2/T2, where P1 and T1 are the initial pressure and temperature, and P2 is the final pressure.
#include<iostream.h> #include<stdlib.h> #include<conio.h> struct poly { int coeff; int x; int y; int z; struct poly * next; }; class polynomial { private : poly *head; public: polynomial():head(NULL) { } void getdata(); void display(); void insert(poly *prv,poly *curr,poly *p); polynomial operator + (polynomial ); }; polynomial polynomial :: operator +(polynomial px2) { polynomial px; poly *t1,*t2,*t3,*last; t1 = head; t2 = px2.head; px.head = NULL; while(t1 != NULL && t2 != NULL) { t3 = new poly; t3->next = NULL; if(t1->x t2->z) { t3->coeff = t1->coeff + t2->coeff; t3->x = t1->x; t3->y = t1->y; t3->z = t1->z; t1 = t1->next; t2 = t2->next; } elseif(t1->x > t2->x) { t3->coeff = t1->coeff; t3->x = t1->x; t3->y = t1->y; t3->z = t1->z; t1 = t1->next; } elseif(t1->x < t2->x) { t3->coeff = t2->coeff; t3->x = t2->x; t3->y = t2->y; t3->z = t2->z; t2 = t2->next; } elseif(t1->y > t2->y) { t3->coeff = t1->coeff; t3->x = t1->x; t3->y = t1->y; t3->z = t1->z; t1 = t1->next; } elseif(t1->y < t2->y) { t3->coeff = t2->coeff; t3->x = t2->x; t3->y = t2->y; t3->z = t2->z; t2 = t2->next; } elseif(t1->z > t2->z) { t3->coeff = t1->coeff; t3->x = t1->x; t3->y = t1->y; t3->z = t1->z; t1 = t1->next; } elseif(t1->z < t2->z) { t3->coeff = t2->coeff; t3->x = t2->x; t3->y = t2->y; t3->z = t2->z; t2 = t2->next; } if(px.head == NULL) px.head = t3; else last->next = t3; last = t3; } if(t1 == NULL) t3->next = t2; else t3->next = t1; return px; } void polynomial :: insert(poly *prv,poly *curr,poly *node) { if(node->x curr->z) { curr->coeff += node->coeff; delete node; } elseif((node->x > curr->x) (node->x curr->y && node->z > curr->z)) { node->next = curr; prv->next = node; } else { prv = curr; curr = curr->next; if(curr == NULL) { prv->next = node; node->next = NULL; return; } insert(prv,curr,node); } return; } void polynomial :: getdata() { int tempcoeff; poly *node; while(1) { cout << endl << "Coefficient : "; cin >> tempcoeff; if (tempcoeff==0) break; node = new poly; node->coeff = tempcoeff; cout << endl << "Power of X : "; cin >> node->x; cout << endl << "Power of Y : "; cin >> node->y; cout << endl << "Power of Z : "; cin >> node->z; if(head == NULL) { node->next = NULL; head = node; } elseif(node->x head->z) { head->coeff += node->coeff; delete node; } elseif((node->x > head->x) (node->x head->y && node->z > head->z)) { node->next = head; head = node; } elseif (head->next == NULL) { head->next = node; node->next = NULL; } else insert(head,head->next,node); } } void polynomial :: display() { poly *temp; temp = head; cout << endl << "Polynomial :: "; while(temp != NULL) { if(temp->coeff < 0) cout << " - "; cout << abs(temp->coeff); if(temp->x != 0) cout << "x^" << temp->x; if(temp->y != 0) cout << "y^" << temp->y; if(temp->z != 0) cout << "z^" << temp->z; if(temp->next->coeff > 0) cout << " + "; temp = temp->next; } cout << " = 0"; } void main() { polynomial px1,px2,px3; clrscr(); px1.getdata(); px2.getdata(); px3 = px1 + px2; px1.display(); px2.display(); px3.display(); getch(); }