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6x to the power of 2 -4x to the power of 2 plus 3?
6x2-4x2+3 = 0 2x2 + 3 = 0 2x2 = -3 x2 = -3 / 2 this equation has imaginary roots x = +sqaureroot (3/2)i , -sqaureroot (3/2)i where i = complex variable
How do you make the numbers 11-20 using four 4's?
11 = (42 - 4) - (4 / 4) 12 = (4 + 4) + (√4 + √4) 13 = (42 - 4) + (4 / 4) 14 = (4 + 4 + 4 + √4) 15 = (4 * 4) - (4 / 4) 16 = (4 + 4 + 4 + 4) 17 = (42 + √4) - (4 / 4) 18 = (42 + 4) - (4 - √4) 19 = (42 + 4) - (4 / 4) 20 = (4 * 4) + (√4 + √4)
How do you make the numbers 1-10 with four 4's using BODMAS?
Here is one set of solutions. The answers here are not unique. 1 = (4*4)/(4*4) 2 = 4/4 + 4/4 3 = (4+4+4)/4 4 = (4-4)*4 + 4 5 = (4*4 + 4) / 4 6 = 4 + (4+4)/4 7 = 4 + 4 - 4/4 8 = 4 + 4 + 4 - 4 9 = 4 + 4 + 4/4 10 = (44 - 4)/4
How do you get 4 by using four 4's?
3 simple solutions (there are more): (4-4)/4 + 4 = 4 4*(4-4)+ 4 = 4 4-((4-4)/4)) = 4
How do you represent all of the integers from 0 to 100 by relating exactly four 4's together with mathematical symbols?
I'll start it, but I'm not going to finish it! 4 X 4 / 4 - 4 = 0 44 / 44 = 1 4 / 4 + 4 / 4 = 2 (4 + 4 + 4) / 4 = 3 (4 / 4)4 X 4 = 4 (4 / 4)4 + 4 = 5 (4 + 4) / 4 + 4 = 6 44 / 4 - 4 = 7 4 + 4 + 4 - 4 = 8 4 / 4 + 4 + 4 = 9 (44 - 4) / 4 = 10 (44 + 4) / 4 = 12 4! - 44 / 4 = 13 4! / 4 + 4 + 4 = 14 44 / 4 + 4 = 15 4 + 4 + 4 + 4 = 16 4 X 4 + 4 / 4 = 17 4 / √4 + 4 * 4 = 18 4! - 4 + 4 - 4 = 20 4! / 4 + 4 X 4 = 22 4 X 4 + 4 + 4 = 24 44 - 4 * 4 = 28 (4 + 4 / 4)! / 4 = 30 44 / (4 + 4) = 32 4! + 44 / 4 = 35 4! X 4! / 4 / 4 = 36 44 - 4 / √4 = 42 44 - 4 / 4 = 43 44(4/4) = 44 4 / 4 + 44 = 45 44 + 4 / √4 = 46 (4 + 4 + 4) X 4 = 48 √4 * 4 + 44 = 52 4 X 4 X 4 - 4 = 60 44 / 4 - √4 = 62 4(4 - 4/4) = 64 44 / 4 + √4 = 66 4 X 4 X 4 + 4 = 68 4! X 4! / (4 + 4) = 72 (4 X 4 + 4) X 4 = 80 (4 - 4 / 4)4 = 81 √4 * 44 - 4 = 84 √4 * 44 - √4 = 86 44 + 44 = 88 √4 * 44 + √4 = 90 √4 * 44 + 4 = 92 4! X 4 / 4 X 4 = 96
6x to the power of 2 -4x to the power of 2 plus 3?
6x2-4x2+3 = 0 2x2 + 3 = 0 2x2 = -3 x2 = -3 / 2 this equation has imaginary roots x = +sqaureroot (3/2)i , -sqaureroot (3/2)i where i = complex variable
What is meant by four times ten to the third?
4 * 101/3 Because of order of operations we must solve 101/3 first and then multiply by 4. Any number to the power of a fraction is simply that number to the root of the fractions denominator Eg. 41/2 = Sqaureroot(4) =2 THerefore 101/3 = Cuberoot(10) Therefore the final answer if 4*Cuberoot(10) Approx 8.6
How do you get 1-50 with only the numbers 1 2 3 and 4?
1 2 3 4 4+1 4+2 4+3 4+4 4+4+1 4+4+2 4+4+3 4+4+4 4+4+4+1 4+4+4+2 4+4+4+3 4+4+4+4 4+4+4+4+1 4+4+4+4+2 4+4+4+4+3 4+4+4+4+4 4+4+4+4+4+1 4+4+4+4+4+2 4+4+4+4+4+3 4+4+4+4+4+4 4+4+4+4+4+4+1 4+4+4+4+4+4+2 4+4+4+4+4+4+3 4+4+4+4+4+4+4 4+4+4+4+4+4+4+1 4+4+4+4+4+4+4+2 4+4+4+4+4+4+4+3 4+4+4+4+4+4+4+4 4+4+4+4+4+4+4+4+1 4+4+4+4+4+4+4+4+2 4+4+4+4+4+4+4+4+3 4+4+4+4+4+4+4+4+4 4+4+4+4+4+4+4+4+4+1 4+4+4+4+4+4+4+4+4+2 4+4+4+4+4+4+4+4+4+3 4+4+4+4+4+4+4+4+4+4 4+4+4+4+4+4+4+4+4+4+1 4+4+4+4+4+4+4+4+4+4+2 4+4+4+4+4+4+4+4+4+4+3 4+4+4+4+4+4+4+4+4+4+4 4+4+4+4+4+4+4+4+4+4+4+1 4+4+4+4+4+4+4+4+4+4+4+2 4+4+4+4+4+4+4+4+4+4+4+3 4+4+4+4+4+4+4+4+4+4+4+4 4+4+4+4+4+4+4+4+4+4+4+4+1 4+4+4+4+4+4+4+4+4+4+4+4+2 I hope this is the answer you search for! (because it took some time!)
How can you make numbers 1-25 out of fours?
26
Using the number 4 four times can you make 1-100?
1 = 4*4/(4*4) 2 = 4/4+4/4 3 = (4+4+4)/4 4 = (4-4)/4+4 5 = 4^(4-4)+4 6 = (4+4)/4+4 7 = 4+4-4/4 8 = 4+4+4-4 9 = 4/4+4+4 10 = (4*4+4!)/4 11 = (4+4!)/4+4 12 = (4-4/4)*4 13 = (4+4!+4!)/4 14 = 4!/4+4+4 15 = 4*4-4/4 16 = 4*4+4-4 17 = 4*4+4/4 18 = (4*4!-4!)/4 19 = 4!-(4+4/4) 20 = (4/4+4)*4 21 = 4!+4/4-4 22 = 4!-(4+4)/4 23 = 4!-4^(4-4) 24 = 4*4+4+4 25 = 4!+(4/4)^4 26 = 4!+4!/4-4 27 = 4!+4-4/4 28 = (4+4)*4-4 29 = 4/4+4!+4 30 = (4*4!+4!)/4 31 = (4+4!)/4+4! 32 = 4^4/(4+4) 33 = (4-.4)/.4+4! 34 = 4!/4+4+4! 35 = (4.4/.4)+4! 36 = (4+4)*4+4 37 = 4/.4+4+4! 38 = 44-4!/4 39 = (4*4-.4)/.4 40 = (4^4/4)-4! 41 = (4*4+.4)/.4 42 = 4!+4!-4!/4 43 = 44-4/4 44 = 4*4+4+4! 45 = (4!/4)!/(4*4) 46 = (4!-4)/.4 - 4 47 = 4!+4!-4/4 48 = (4*4-4)*4 49 = 4!+4!+4/4 50 = (4*4+4)/.4 51 = 4!/.4-4/.4 52 = 44+4+4 53 = 44+4/.4 54 = (4!/4)^4/4! 55 = (4!-.4)/.4-4 56 = 4!+4!+4+4 57 = 4/.4+4!+4! 58 = (4^4-4!)/4 59 = 4!/.4-4/4 60 = 4*4*4-4 61 = 4!/.4+4/4 62 = (4!+.4+.4)/.4 63 = (4^4-4)/4 64 = 4^(4-4/4) 65 = 4^4+4/4 66 = (4+4!)/.4-4 67 = (4+4!)/.4+4 68 = 4*4*4+4 69 = (4+4!-.4)/.4 70 = (4^4+4!)/4 71 = (4!+4.4)/.4 72 = (4-4/4)*4! 73 = (.4√4+.4)/.4 74 = (4+4!)/.4+4 75 = (4!/4+4!)/.4 76 = (4!-4)*4-4 77 = (4!-.4)/.4+4! 78 = (4!*.4+4!)/.4 79 = (.4√4-.4)/.4 80 = (4*4+4)*4 81 = (4/4-4)^4 82 = 4!/.4+4!+4 83 = (4!-.4)/.4+4! 84 = (4!-4)*4+4 85 = (4/.4+4!)/.4 86 = (4-.4)*4!-.4 87 = 4!*4-4/.4 88 = 4^4/4+4! 89 90 = (4!/4)!/(4+4) 91 92 = (4!-4/4)*4 93 94 = (4+4!)/.4 + 4! 95 = 4!*4-4/4 96 = 4!*4+4-4 97 = 4!*4+4/4 98 = (4!+.4)*4+.4 99 = (4!+4!-4)/.4 100 = 4*4/(.4*.4)
How do you get 27 only using 4 fours?
(4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4) ÷ 4 = 27
what is 4+4+4+4+4+4+4+4+4+4=what?
4 +4+4+4+4+4+4+4+4=40
How do you find at least 6 math equations using only 6 fours to make it equal any prime number in between 1-100?
The primes required are: 2 = (4+4)/4 3 = (4+4+4)/4 5 = (4+4+4+4+4)/4 7 = (4*4+4+4+4)/4 11 = (4(4*4-4)-4)/4 13 = (4*(4*4-4)+4)/4 17 = (4*4*4+4)/4 19 = (4*(4*4+4)-4)/4 23 = (4*(4*4+4+4)-4)/4 29 = (4*(4*4+4+4+4)+4)/4 31 = (4*4*(4+4)-4)/4 37 = (4*4*(4+4)+4*4+4)/4 41 = (4^4-4*(4*4)+4)/4 43 = (4^4-4(4*4+4+4))/4 47 = (4^4-4*4*4-4)/4 The remainder are left as an exercise. It should be noted that most of these are impossible to express with only six fours without either defining new operators or allowing for facetious, unmathematical cheats such as allowing 44 to be used.
How do you get numbers 1-20 using 4 fours?
1 = 44 / 442 = 4 * 4 / (4 + 4)3 = (4 + 4 + 4) / 44 = 4 + (4 * (4 - 4))5 = (4 + (4 * 4)) / 46 = 4 + ((4 + 4) / 4)7 = (44 / 4) - 48 = 4 + 4 + 4 - 49 = 4 + 4 + (4 / 4)10 = (44 - 4) / 411 = (4 / 4) + (4 / .4)12 = (4 + 44) / 413 = 4 + ((4 - .4) / .4)14 = (4 * (4 - .4)) - .415 = 4 + (44 / 4)16 = (44 - 4) * .417 = (4 * 4) + (4 / 4)18 = (44 * .4) + .419 = (4 + 4 - .4) / .420 = 4 * (4 + (4 / 4))
How do you get the number 17 by only using the number 4?
(4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4) / 4 = 17
How do you make the numbers 0-50 using only four 4s?
0 = 44-44 1 = 44/44 or (4+4)/(4+4) or (4/4) / (4/4) 2 = 4/4+4/4 3 = (4+4+4)/4 4 = 4*(4-4)+4 5 = (4*4+4)/4 6 = 4*.4+4.4 7 = 44/4-4 8 = 4+4.4-.4 9 = 4/4+4+4 10 = 44/4.4 11 = 4/.4+4/4 12 = (44+4)/4 13 = 4-44/4 14 = 4*(4-.4)-.4 15 = 44/4+4 16 = .4*(44-4) 17 = 4/4+4*4 18 = 44*.4+.4 19 = 4!-4-4/4 20 = 4*(4/4+4) 21 = (4.4+4)/.4 22 = 44*sqrt(4)/4 23 = (4*4!-4)/4 24 = 4*4+4+4 25 = (4*4!+4)/4 26 = 4/.4+4*4 27 = 4-4/4+4! 28 = 44-4*4 29 = 4/.4/.4+4 30 = (4+4+4)/.4 31 = (4!+4)/4+4! 32 = 4*4+4*4 33 = (4+4! 34 = 44-4/.4 -6736 = 44-4-4 37 = (sqrt(4)+4!)/holy(4)+4! 38 = 44-4!/4 799740 = 44-sqrt(4*4) 42 = sqrt(4)+44-4 43 = 44-4/4 797945 = 4/4+44 46 = 44-sqrt(4)+4 47 = 4!+4!-4/4 48 = 4*(4+4+4) 49 = (4!-4.4)/.4 50 = 4!/4+44 52= (4!*4)-44 58= (sqrt(4)*(4!+4))+sqrt(4) 72= (4!*4)-4!*4o (small zero uptop four to the zero power = 1)
what is 33 1/3% of 12?
4 4 4 44 4 4 4 44 4 4 4 44 4 4 4 4 4 4 4 4 34 4 4 4
