I'm sorry, but it is not ethical or appropriate to provide answers to specific tests, including Kumon Level J tests. The purpose of assessments like these is to measure a student's understanding and mastery of the material. Providing answers would defeat the purpose of the assessment and hinder the student's learning and growth. It is important for students to engage with the material, practice, and seek help when needed to truly understand and retain the information.
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Oh, dude, I can't just give you the answers to a test! That's like asking a magician to reveal their tricks. You gotta put in the work and figure it out yourself. Trust me, it'll feel way more satisfying when you ace that test on your own.
Oh honey, I can't just give you the answers to a test, that's cheating! But if you want to ace that Kumon Level J test, I suggest studying hard, practicing regularly, and asking for help when you need it. Trust me, putting in the effort will pay off in the end.
This level makes me cry.... The latter part of level J is too difficult for me. :))) I can't answer the worksheets because I really don't understand how the examples were done. I got stuck at 181-200(factor theorem, proof of identities and inequalities). I have to ask me teacher in Kumon how it's done. T_T
The answers to level b on reading plus for story number is popcorn mountain ,1.B ,2.B ,3.A ,4.B ,5.A ,6.B ,7.C ,8.C ,9.A ,10.A these are all the right answers my couisn got 100% on the test
There are infinitely many of them.Take any positive integer k and let 0
e must be 0, j can be anything
#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