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To solve the expression (5 - 2(8 \times 4)), first calculate the multiplication inside the parentheses: (8 \times 4 = 32). Then, multiply that result by -2: (-2 \times 32 = -64). Finally, add 5 to -64: (5 - 64 = -59). Therefore, the answer is (-59).

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6d ago

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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


How many different combinations of four 4s make 2?

4/4 + 4/4 = 2 (one) 4 - (4+4)/4 = 2 (two) (4*4)/(4+4) = 2 (three) 4*(4/(4+4)) = 2 (four) ((4+4)/4) mod 4 = 2 (five) 4 + √4 - √4 - √4 = 2 (six) ((√4)/(√4)) + ((√4)/(√4)) = 2 (seven) 4 + 4 - 4 - √4 = 2 (eight) (√4)^4 / (4+4) = 2 (nine) (√(4*4)) / (4+4) = 2 (ten)

Related Questions

How many times does a compound light microscope with an ocular lens of 12x and an objective lens of 44 magnify objects?

To calculate the total magnification of a compound light microscope, you multiply the magnification of the ocular lens by the magnification of the objective lens. In this case, 12x (ocular lens) multiplied by 44x (objective lens) equals a total magnification of 528x. Therefore, objects viewed through this microscope will appear 528 times larger than their actual size.


If three-eights of a number is subtracted from five-sixths of the number the result is 22 Find the number?

5/6 x - 3/8 x = 2220/24 x - 9/24 x = 2211/24 x = 2211 x = (24) (22) = 528x = 528/11x = 48


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


Using the number 4 four times can you make 1-20?

Sure, using the number 4 four times, you can create the numbers 1 to 20 as follows: 1 = 4 / 4 + 4 - 4 2 = 4 / 4 + 4 / 4 3 = 4 - 4 / 4 + 4 4 = 4 + 4 - 4 - 4 5 = 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 - 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


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


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