1 Good turn deserves another
estimated time of departure ETD estimated time of arrival ETA
d i l a p i d a t e d
S-t-u-d-t
A-B-C-D-E-F-G.... R-U-D-T-F with me? RUDTF = Are u down to f*** It's a pick up line
1 day at a time.
To 1t pod
v = − 1/ a .d [A] /d t = − 1 /b .d [B] /d t = 1 /p. d [P]/ d t = 1/ q .d [Q]/ d t for the "balanced" equation aA + bB -> pP + qQ
One good turn deserves another
I am assuming you mean 1.41km/h. s = d/t, where s is speed, d is distance, and t is time st = d t = d/s = (1km)/(1.41km/h) = 0.7km
Use the product and chain rules: d/dt te^(-t²) = (d/dt t)e^(-t²) + t(d/dt e^(-t²)) = 1 × e^(-t²) + t × e^(-t²) × d/dt -t² = e^(-t²) (1 + t × -2t) = (1 - 2t²)e^(-t²)
to be determined
D = RT so if you travel at R = 50 mph in D = 1 mi 1 = 50*T T = 1/50 hour = 1.2 min * 60 = 72 mph D = RT so if you travel at R = 50 mph in D = 1 mi 1 = 50*T T = 1/50 hour = 1.2 min * 60 = 72 mph D = RT so if you travel at R = 50 mph in D = 1 mi 1 = 50*T T = 1/50 hour = 1.2 min * 60 = 72 mph D = RT so if you travel at R = 50 mph in D = 1 mi 1 = 50*T T = 1/50 hour = 1.2 min * 60 = 72 mph D = RT so if you travel at R = 50 mph in D = 1 mi 1 = 50*T T = 1/50 hour = 1.2 min * 60 = 72 mph
Answer:q.d. x 1 wk means every day for one week.
Find I = ∫ tan³ x dx. The solution is: I = ½ tan² x - log cos x. * * * Here is how we can obtain this result: First, let t = tan x, s = sin x, and c = cos x; then, dI = t³ dx, ds = c dx, dc = -s dx, and dt = (1 + t²) dx; and, of course, t = s / c. By algebra, t³ = t(t² + 1) - t; thus, we have dI = t³ dx = t(t² + 1) dx - t dx = t dt - t dx. Now, d (t²) = 2t dt; thus, t dt = ½ d(t²). On the other hand, we have d log c = dc / c = -s dx / c = -t dx; thus, t dx = -d log c. Combining these results, we have dI = t dt - t dx = ½ d(t²) - d log c. This integrates readily, giving I = ½ t² - log c, which is the solution we sought. * * * We may check our result, by differentiating back: dt / dx = 1 + t²; and d(t²) / dt = 2t; thus, (d/dx)(t²) = 2t dt / dx = 2t (1 + t²). Also, we have d log c / dc = 1 / c; and dc / dx = -s; whence, (d/dx)(log c) = (dc / dx) / c = -s / c = -t. Then, dI / dx = ½ (d/dx)(t²) - (d/dx)(log c) = t (1 + t²) - t = t + t³ - t = t³, re-assuring us that we have integrated correctly.
// transpose for the sparse matrix void main() { clrscr(); int a[10][10],b[10][10]; int m,n,p,q,t,col; int i,j; printf("enter the no of row and columns :\n"); scanf("%d %d",&m,&n); // assigning the value of matrix for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { printf("a[%d][%d]= ",i,j); scanf("%d",&a[i][j]); } } printf("\n\n"); //displaying the matrix printf("\n\nThe matrix is :\n\n"); for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { printf("%d\t",a[i][j]); } printf("\n"); } t=0; printf("\n\nthe non zero value matrix are :\n\n"); for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { // accepting only non zero value if(a[i][j]!=0) { t=t+1; b[t][1]=i; b[t][2]=j; b[t][3]=a[i][j]; } } } printf("a[0 %d %d %d\n",m,n,t); for(i=1;i<=t;i++) { printf("a[%d %d %d %d\n",i,b[i][1],b[i][2],b[i][3]); } a[0][1]=n; a[0][2]=m; a[0][3]=t; int s[10],u[10]; if(t>0) { for(i=1;i<=n;i++) { s[i]=0; } for(i=1;i<=t;i++) { s[b[i][2]]=s[b[i][2]]+1; } u[1]=1; for(i=2;i<=n;i++) { u[i]=u[i-1]+s[i-1]; } for(i=1;i<=t;i++) { j=u[b[i][2]]; a[j][1]=b[i][2]; a[j][2]=b[i][1]; a[j][3]=b[i][3]; u[b[i][2]]=j+1; } } printf("\n\n the fast transpose matrix \n\n"); printf("a[0 %d %d %d\n",n,m,t); for(i=1;i<=t;i++) { printf("a[%d %d %d %d\n",i,a[i][1],a[i][2],a[i][3]); } getch(); }
W = what d = do y = you w = want t = to t = talk a = about