what's the easiest way to multiply any two numbers?
482474
With a calculator. If you're trying to do it in your head, then it depends on how your head works. For me I multiply parts of the numbers and add them. If I'm trying to do something like 15X22 I'll multiply 10X22 and add it to 5X22. For me it's easier to break it down into easier equations.
464956
use each number once and only once...\n\n2*3+4*5=26   \n\nThis is too easy.  Give us a hard one!\n...come on...?
1118742
Practice with easy stuff that you can quickly remember.\n\nFor instance, if you want to be able to add numbers quickly, remember how to add numbers 1 through 9 to each other (ie 5 + 9 = 14; 2+8 = 10). Once you know these, try writing down double digit numbers and try to solve them in your head.  You will already know how to do the ones column.\n\nFor instance:\n   45\n+19\n-------\n64\n\nLearn easy problems first then move on to hard ones.  You can apply this to subtraction, division, and multiplication too.  It really is just practice that will make you better.
163135
CHOOSE one of the follwing depending upon your course curriculum\n\n\n//Use to following program for multiple calculations in one \n//run.\n\nvoid main()\n{\n\nint a,b; // Declare variables "a" and "b" to get two\n// numbers.\n\n  char Continue = 'Y'; // Declare Continue to loop through \n//the program for multiple calculations. and initialize it \n//to 'Y'\n\n  while(Continue == 'Y' || Continue == 'y') \n// Continue doing the folliwing if the user presses 'Y' for\n// continue.\n  {\n\n  printf("Enter First Number: ");\n  scanf("%d",&a); //scan or get first number\n\n  printf("&#92;nEnter Second Number: ");\n  scanf("%d",&b); //scan or get second number\n\n  printf("&#92;n %d %d  &#92;"DIVISION&#92;" = %f",a,b,(float)a/b); \n// Divide and display the result\n\n  printf("&#92;n %d MOD %d  &#92;"MODULUS&#92;" = %d",a,b,a%b); \n// Calculate Modulus.\n\n  printf("&#92;n&#92;n Do you want to Continue? (Y/N)  :"); \n// Prompt if youser wants to continue.\n  scanf("%s",&Continue);\n\n  printf("&#92;n &#92;n --------------------------------- &#92;n &#92;n"); \n// print saperator between multiple calculations.\n  }\n}\n\n-------------------------------------------------------------------------------------------------------\n\n// Use Following for single calculation in one run...\nvoid main()\n{\n  int a,b; // Declare variables "a" and "b" to get two\n// numbers.\n \n  printf("Enter First Number: ");\n  scanf("%d",&a); //scan or get first number\n  \n  printf("&#92;nEnter Second Number: ");\n  scanf("%d",&b); //scan or get second number\n  \n  printf("&#92;n %d %d  &#92;"DIVISION&#92;" = %f",a,b,(float)a/b); // \n  // Divide and display the result\n  \n  printf("&#92;n %d MOD %d  &#92;"MODULUS&#92;" = %d",a,b,a%b); \n  // Calculate Modulus and print.\n\n}\n\n\nNote: You can not use floats in modulus.
942101
start by seperating y, u then have y=3x/2+3 then plug in the domain, or the numbers u are givin\nu'll have 0, 1.5, 3, 6, 7.5
1123203
You can multiply and divide them easier.\n\nfifth root of 2x times 2x^20\n           -or-\n2x^(1/5) times 2x^20
698756
First conversion the numbers to binary, then make the OR operation.\n\nExp.:\n8 OR 3.\n\nis: 1000 OR 11\n\n= 1011.\n\nWhich mean: 11 (Dec.)\n\nYou can use the Windows Calc.
531205
2 + 2 = 4\n\nthats all I got
621145
With just ! (Factorial), +, -, * and /, it is impossible to get 150 using only 3 5s without the x^2 (Squared) operation (Which means x*x, in case you are in Fifth Grade), or using a '^2' as you say. If you say there is an alternate way, and there really is, then I give up and admit defeat. (I'm sure most others do.) If you are so sure, why don't you post the answer yourself?
333726
Perform operations in this order:\ni)   brackets\nii)  powers\niii) divide\niv)  multiply\nv)   add\nvi)  subtract\n\nSo work it out like this:\n\n2 + 2 (4 - 6 * 7) + 4\n= 2 + 2 (4 - 42) + 4\n          By doing multiplication 6 * 7 in brackets first.\n= 2 + 2 (-38) + 4\n          By doing subtraction 4 - 42 in brackets.\n= 2 + -76 + 4\n          By doing multiplication 2 * -38\n= 2 - 76 + 4\n          Since + and - sign in front of 76 combine into a -\n= -74 + 4\n          By doing subtraction 2 - 76\n= -70\n          By doing addition -74 + 4
215091
2.  \nOrder of operations says, since there are no brackets, you do multiplication and division first in the order they appear, 2*2 = 4, 4/2 = 2, then addition/subtraction in the order they appear, so 2+2-2 = 2.
373216
any number N1 can be converted to N2 BY\naddtion:\n  N1 + N2 + (-N2 ) \ndivision:\n 1/[(N1/N2)/N1]\nMULTIPLICATION\n N1 * N2 * N1^-1\nSUBTRACTION\n[N1-(N1-N2]
1068964
simplify expression: -2(y+3)+6(y+1)\n\nsimply distribute then add.
894472
Use a calculator.  Enter [2] [log] for the first one.
1094108
Use a calculator and raise it to the 1/3 power.\nDon't try to do it by hand...
735754
just plug the numbers in.  if y=8, and x=2, \nits 8=a(2^2)\nso 8=a(4)\nso a=2.\nhope that helps!
616629
First of all, change this:\n\n<p>Enter first numerator: <input type=text name=num1 size=20></p>\n<p>Enter first numerator: <input type=text name=num2 size=20></p>\n<p>Enter second denomenator: <input type=text name=num1 size=20></p>\n<p>Enter second denomenator: <input type=text name=num2 size=20></p>\n\n\nto this:\n\n<p>Enter first numerator: <input type=text name=num1 size=20></p>\n<p>Enter first denomenator: <input type=text name=num2 size=20></p>\n<p>Enter second numerator: <input type=text name=num3 size=20></p>\n<p>Enter second denomenator: <input type=text name=num4 size=20></p>\n\nThen based on the operation selected, implement the proper logic there (its basic math) and then your program will work.\n\nGood luck!!