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5/6 = 20/24

3/8 = 9/24

20 - 9 = 11/24

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10y ago
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6mo ago

The answer to the subtraction of 5 sixes and 3 eights is 2.

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Q: What is the answer to the fraction of 5 6's and 3 8's subtractions?
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Related questions

What is the quotient of 14s 2d 7-6s 4d 5 plus 8s 2d 3 divided by 2s 2 d?

14s 2d 7 - 6s 4d 5 + 8s 2d 3/2s 2d = 14s 7 - 3s 2s 5 + 4s 3


What fraction is equivalent to 4 6s?

4 6s = 4*6 = 24. As a fraction, this can be written as 24/1.


What is the equivalent Fraction of 4 6s?

4/6 = 2/3


What is the perpendicular bisector equation joining the points of s 2s and 3s 8s on the Cartesian plane showing work?

Points: (s, 2s) and (3s, 8s) Slope: (8s-2s)/(3s-s) = 6s/2s = 3 Perpendicular slope: -1/3 Midpoint: (s+3s)/2 and (2s+8s)/2 = (2s, 5s) Equation: y-5s = -1/3(x-2s) => 3y-15s = -1(x-2s) => 3y = -x+17x Perpendicular bisector equation in its general form: x+3y-17s = 0


What is the perpendicular bisector equation of a line joined by the points of s 2s and 3s 8s showing key stages of work?

It is found as follows:- Points: (s, 2s) and (3s, 8s) Slope: (2s-8s)/(s-3s) = -6s/-2s = 3 Perpendicular slope: -1/3 Midpoint: (s+3s)/2 and (2s+8s)/2 = (2s, 5s) Equation: y-5s = -1/3(x-2s) Multiply all terms by 3: 3y-15s = -1(x-2s) => 3y = -x+17s In its general form: x+3y-17s = 0


How many 8s go into 24?

3


How do you determine the equation for the perpendicular bisector of the straight line joining the points s 2s and 3s 8s?

A = (s, 2s), B = (3s, 8s) The midpoint of AB is C = [(s + 3s)/2, (2s + 8s)/2] = [4s/2, 10s/2] = (2s, 5s) Gradient of AB = (8s - 2s)/(3s - s) = 6s/2s = 3 Gradient of perpendicular to AB = -1/(slope AB) = -1/3 Now, line through C = (2s, 5s) with gradient -1/3 is y - 5s = -1/3*(x - 2s) = 1/3*(2s - x) or 3y - 15s = 2s - x or x + 3y = 17s


What is 2.6666666666667 as a fraction?

It's 26,666,666,666,667/10,000,000,000,000 .(If you had said that the 6s go on forever and never end, then it would have been 22/3 or 8/3 .)


How many 8s are you 24?

3


What is the short form for 6 out of 8?

Fraction- 6/8 or 3/4 You simplify 6/8s to 3/4s by dividing the numerator and denominator by a common number, in this case the number 2.


What is the improper answer to 3 and 3 8s?

27 / 8


Why do you have 3 leads in one junction box?

Junction boxes can have as many leads in them as required by the circuits that need splicing. The only limitations to the amount of wires in the box is the internal cubic inches. Two standard size junctions boxes are , the four inch round (15 cubic in.) deep allows 14 #14s, 12 #12s, 9 #10s, 7 #8s and4 #6s. The four inch square box (30 cubic in.) deep allows 20 #14s, 17 #12s, 13 #10s, 10 #8s and 6 #6s.