True
e must be 0, j can be anything
312.5 J
The apothem of a regular polygon? well lets look at the math behind it before i recall it... you can scroll down to the bottom of the page if you don't want to read this. the formula is on the bottom of the page * A regular polygon is made up of a sequence of isoceles triangles.. * How do we know that they are isoceles? ------1)the triangles that make up a regular polygon are congruent -------2)the radii are always congruent . the radii of a regular polygon goes from it's center to the vertices...(hint:think of a circle's radius) * due to the fact that you have isoceles triangles they have to be made by angle bisectors through the regular polygon otherwise they couldn't be congruent * okay now that we know that the triangles are isoceles we also know that the apothem is an angle bisector so it cuts the measurement of a side in half. lets use j for our the measurement of our side. * okay we got the angle measures and our apothem made two congruent triangles so now we can use trig ratios to find our apothem so the formula is a=0.5j(tan [n-2]*180/2n) where n is the # of sides and j is the measurement of a side or you can simplify that to a=0.5j(tan [n-2]*90/n) i am using degrees for my angle meausure by the way
#include<stdio.h> #include<string.h> #include<conio.h> int count=0; char arr1[20],arr2[20]; int main() { int j; arr1[0]=1; arr1[1]='\0'; int flag=3; for(j=3;j>=-1;j--) printf(" "); printf("1\n"); while(flag>=0) { for(j=3;j>=count;j--) printf(" "); arr2[0]=arr1[0]; arr2[strlen(arr1)]=arr1[strlen(arr1)-1]; for(j=1;j<=(strlen(arr1)-1);j++) arr2[j]=arr1[j]+arr1[j-1]; arr2[strlen(arr1)+1]='\0'; for(j=0;j<strlen(arr2);j++) printf("%d ",arr2[j]); for(j=0;j<=strlen(arr2);j++) arr1[j]=arr2[j]; flag--; count++; printf("\n\n"); } getch(); }
four more than j or j+4 or j--4 or 2j+(4-j) or (0.0j + 0.04) *100 BOGIF buy one get infinity free
For triangles JKL and PQR to be congruent, the corresponding sides and angles must be equal. Given that J corresponds to P, K to Q, and L to R, if the angles J, K, and L are equal to angles P, Q, and R respectively, then JKL is indeed congruent to PQR by the Angle-Angle-Angle (AAA) similarity criterion. However, congruence specifically requires that the sides are also equal, so if only angles are mentioned without side lengths, we cannot conclude congruence solely based on the angles. Thus, more information about the sides is needed for a definitive conclusion.
j+k+l=P
The bumps on the f and j keys are to find home row (aka. "asdf jkl;") without looking at the keyboard
It's a bit hard to explain in words. Suppose you have four congruent isosceles triangles: ABC, DEF, GHI, and JKL. (the vertexes are A, D, G and J). Place DEF and ABC so that the bases are touching. You now have quadrilateral ABDC ( when 2 points coincide, use the letter which comes first in the alphabet). Place GHI so that G is at B and H is at A. You now have quadrilateral AIDC (IBD is a straight line).Place JKL so that L is at D and J is at B. Now AIBKDC is the concave hexagon.
Utah, Kansas, Kentucky and Louisiana are U.S. states. No U.S. state begins with the letter j.
Angle J is congruent to angle K line KL is parellel to line Jm
These are the home row keys that belong to your index fingers. The home row keys, ASDF and JKL; are the eight keys your fingers return to after typing a word.
llombbjmmnhk fhkj khkhkjl k k l kl ; jkkkkkkfhjkk hjkjhlhk j l jkl k kjl ;l l;l;l; ;l ; l l ;l l;;l;k;;moklnhg h jkj jh k jl kl' ;l b ,hjk ljk l; kl h h sm mbanlgd m ,gjkv Cvfj,lgh lg jh l v l; hgvj lghf kl klgj kljklj jkl lk jlgjh kljklkl l k klj klj g lkj jkl ljk jkl kl klj jkl klj klj ljk kl kl klj kl jkl ljk jkl klj klj kl lk kl kl kl klj l ',nbgJLf adjhtful5ds fgd jk xq438969530734597 nv,mnmvch omfgwhatrudoibng? daf bdbnvbnbv n bn n v n b nb bn n sf nvbmnfmn mnbm'bmbm mnbv
Num-Lock. Activates a secondary "hidden" dialpad. U,I,O would be 4,5,6 J,K,L= 1,2,3 Respectively and M=0 789 UIO JKL M
e must be 0, j can be anything
k,j,j,j,
At least 320 J