To represent the number 68 using multiplication, we can use the following equations: 2 x 34, 4 x 17, 17 x 4, 34 x 2. These equations demonstrate different ways to multiply numbers to arrive at the value of 68.
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I don't understand your question but I can assume you're using whole numbers less than 68 and greater than 1 to find multiplicand pairs that equal 68. Given that, 68 breaks down into the integers 2, 2, and 17 (multiplied together they equal 68). So anyone combination of these three numbers will give you your axb=68 equations. So in this example you can have 2x34 or 4x17 besides the obvious 1*68
the lower number for the denominator greater the number is.
Using a calculator, you can find out that 50*3/50+17/2 is equal to 11.5
the highest number you can count up to using 10 bits is 1029 using binary
Call the number of 20 notes x Call the number of 50 notes y Then we are given X+y=98 and x+10 = 5*(y-34) rearrange both equations so they both equal zero x+y-98=0 and x+10-5y+170=0 We can now say x+y-98 = x+10-5y+170 Subtract x from both sides y-98 = 10-5y+170 = 180-5y 6y=180+98 y= 278/6 = 46.3333 x=98-y = 51.6666 Basically we are using simultaneous equations. We could have found x independently by making both equations = y so y= 98-x and y =(-170-10)/-5 +x = 180/5 + x/5 = 36 +x/5 so 98-x = 36 + x/5 490-5x = 180 +x x = 310/6 = 51.6666 Unfortunately this is a bad question because notes usually come in whole numbers. However, if you substitute the number 98 with the number 96, it works out perfectly.
They are used for working out equations where the numbers you are working with are not physically possible, but we just imagine they are, such as the square root of a negative number In engineering, especially Electrical Engineering, using complex numbers to represent signals (rather than sines and/or cosines) make comparing and working with signals easier.