Q: What is f u c x i n g x h i t?

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Direct Substitution Th. Limf(x)=f(a) Theorem 2 If f(x)<g(x), then limf(x)<limg(x) Squeeze Theorem If f(x)<g(x)<h(x) limf(x)=limh(x)=L, then limg(x)=L Eps - Delta Definition if 0<|x-a|<d, then |f(x)-L|<Eps. Deffinition fuction is cont. if limf(x)=f(a) Theorem 5 Any polinomial and rational functions are cont. Theorem 7 Any pol, ration, root, trig, invers, exponent, and log functions are continuous. Theorem 8 limf(g(x))=f(limg(x)) Intermidiate Value Theorem If f is cont. [a,b] and N is between f(a) & f(b), then there is a C between a & b, such that f(C)=N Deriviative f'(a)=lim(h->0):{f(a+h)-f(a)}/h Rate of change R=lim(x2->x1):{f(x2)-f(x1)}/x2-x1

A b c d e f g h i j k l m n o p q r s t u v w x y z

L'Hospitals rule states that for a function f(x) = g(x)/h(x) if the limit of g'(x)/h'(x) exists and is equal to a value, then the lmit of f(x) is also equal to that value.

A,e,f,g,h,i,k,l,m,n,t,v,w,x,y,z .... but capitalized.

A,e,f,g,h,i,k,l,m,n,t,v,w,x,y,z .... but capitalized.

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a b c d e f g h i j k l m n OOOOOOOOOOOOOOO a b c d e f g h i j k l m n OOOOOOOOOOOOOOO

h i j k l m n o p q r s t u v w x y z

#include<stdio.h> #include<conio.h> #include<math.h> float f(float x) { float y; if(x!=0) //y=sin(x)-log(x)+exp(x); y=1/(1+x); else printf("answer can not be derived"); return(y); } float g(float x) { float k; k=(-2)/(pow((1+x),3)); return(k); } void main() { float i,n,k,z,s=0,err=0; float h,x,a[100],b,c; clrscr(); printf("\n enter the range of the integration"); scanf("%f%f",&b,&c); printf("\n enter n so to divide the range into n parts"); scanf("%f",&n); h=(c-b)/n; printf("\n\n%f",h); k=(f(b)+f(c)); printf("\nk=%f\n\n",k); for(i=b+h;i<c;i+=h) { z+=f(i); printf("\n%f",z); } s=(h*(k+2*z))/2; printf("\n\n the value of the integration=%f\n\n",s); for(i=b;i<=c;i+=h) err+=g(i); err=(err*(pow(h,3)))/12; printf("\n\n approximate error= %f\n\n",err); getch(); }here...

The bond stretching frequency increases with increasing bond strength. Therefore, the order of increasing bond stretching frequency is: F-H < O-H < N-H < C-H.

ITS EASY...TRY THIS OUT..TRAPEZOIDAL METHOD#include#include#includefloat valcal(float x){return (x*x*x);}int main(){float a,b,h,c,I;int n,i;printf("THE TRAPEZOIDAL RULE:\n");printf("---------------------");printf("\n\n\nEnter the two limits and the no. of divisions:\n");scanf("%f %f %d",&a, &b, &n);h=(b-a)/n;//printf("\nVALUE of h: %f\n", h);c=a;I=valcal(a)+valcal(b);//printf("\nVALUE FOR a: %f\n", valcal(a));//printf("\nVALUE FOR b: %f\n", valcal(b));for(i=1;i=b){printf("\n\nc>b\n\n");break;}//printf("\nVALUE FOR %f: is %f\n",c, valcal(c));I=I+(2*valcal(c));//printf("\nI right now is %f", I);}printf("\n\n\nThe integration of x*x*x is: %f",(h*I)/2);printf("\n\n\n");system("pause");}SIMPSON'S 1/3RD METHOD#include#include#includefloat valcal(float x){return (1/(1+x*x));}int main(){float a,b,h,c,I;int n,i;printf("THE SIMPSON'S ONE-THIRD RULE:\n");printf("------------------------------");printf("\n\n\nEnter the two limits and the no. of divisions:\n");scanf("%f %f %d",&a, &b, &n);h=(b-a)/n;//printf("\nVALUE of h: %f\n", h);c=a;I=valcal(a)+valcal(b);//printf("\nVALUE FOR a: %f\n", valcal(a));//printf("\nVALUE FOR b: %f\n", valcal(b));for(i=1;ib){printf("\n\nc>b\n\n");break;}//printf("\nVALUE FOR %f: is %f\n",c, valcal(c));if(i%2==0)I=I+(2*valcal(c));elseI=I+4*valcal(c);//printf("\nI right now is %f", I);}printf("\n\n\nThe integration of x*x*x is: %f",(h*I)/3);printf("\n\n\n");system("pause");}NEED MORE HELP...MAIL ME YOUR PROB... SEE YA

Recall that a linear transformation T:U-->V is one such that 1) T(x+y)=T(x)+T(y) for any x,y in U 2) T(cx)=cT(x) for x in U and c in R All you need to do is show that differentiation has these two properties, where the domain is C^(infinity). We shall consider smooth functions from R to R for simplicity, but the argument is analogous for functions from R^n to R^m. Let D by the differential operator. D[(f+g)(x)] = [d/dx](f+g)(x) = lim(h-->0)[(f+g)(x+h)-(f+g)(x)]/h = lim(h-->0)[f(x+h)+g(x+g)-f(x)-g(x)]/h (since (f+g)(x) is taken to mean f(x)+g(x)) =lim(h-->0)[f(x+h)-f(x)]/h + lim(h-->0)[g(x+h) - g(x)]/h since the sum of limits is the limit of the sums =[d/dx]f(x) + [d/dx]g(x) = D[f(x)] + D[g(x)]. As for ths second criterion, D[(cf)(x)]=lim(h-->0)[(cf)(x+h)-(cf)(x)]/h =lim(h-->0)[c[f(x+h)]-c[f(x)]]/h since (cf)(x) is taken to mean c[f(x)] =c[lim(h-->0)[f(x+h)-f(x)]/h] = c[d/dx]f(x) = cD[f(x)]. since constants can be factored out of limits. Therefore the two criteria hold, and if you wished to prove this for the general case, you would simply apply the same procedure to the Jacobian matrices corresponding to Df.

It's spelled F-R-I-G-H-T-E-N-I-N-G.

n n n n n n n n n n n n n n n o o o o o o o o o o o o o t t t t t t t t t t t h h h h h h h h h h h h h h i i i i i i i i i i n n n n n n n n n n g g g g g g gg g g g g g g g gg gg g

n n n n n n n n n n n n n n n o o o o o o o o o o o o o t t t t t t t t t t t h h h h h h h h h h h h h h i i i i i i i i i i n n n n n n n n n n g g g g g g gg g g g g g g g gg gg g

The real notes for jingle bell rock on the alto sax are: 1.) gg g, gg g, gbga Bb a g r, r--, 1 measure rest 5.) bb b g- bb b g#- g- f#- g- f#- g- f#- r cccc c c f f- 13.) bb b bb b bb b g- bb b g# g- f#- g- f#- g- f#- r c#c# c(n) c b--- gg 21.) a- g eg a- a Bb b(n) b b b g--- gg a- g eg a- g- f# r gggg a a a- 29.) bb b bb b bb b g- bb b gg #g g a b c e eeee cc Eb Eb- r c# c(n) c 36.) r c# c(n) c r c#- c# c(n) c c- b---- > r g b > key n= natural -= half note ---= dotted hald note ----= whole note r= rest btw, the part at 21 is a saxophone solo!

dedede

C - aring H - andsome A - rtistic N - eat G - ood guy and E - asy to mingle with