How do I do this math problem??

Question

h = 9 3/11 - 6 2/3 + 2 1/11

Answer 1

Hi,

Re-write numbers as improper fractions. Then multiply them to get a common denominator. then add their numerators. Reduce if possible. Convert back into a mixed number if desired.

9 3/11 - 6 2/3 + 2 1/11 As improper fractions you'd have:

102/11 - 20/3 + 23/11 Now multiply by (3/3) and (11/11)

(3/3)(102/11) -(11/11)(20/3) + (3/3)(23/11) =

306/33 - 220/33 + 69/33 =

155/33 = 4 23/33

I hope that helps!

Answer 2

Those are mix fractions:

for example: 9 3/11 is the same as: 102/11. Why? well, first multiply the denominator for the number in front of the fraction, then add up the numerator:

9 3/11 = [(9*11) + 3]/11

Same thing with the others. Then add up all fractions and you'll get the result

Answer 3

first, change the complex fractions into real fractions by multiplying the bottom number of the fraction by the leading number and then adding the top number...put the result over the bottom to get the fraction...
that was probably pretty confusing so,
9 3/11 first 9 times 11= 99...then, plus 3 =102...the fraction is 102/11.

so, all in all you get:

102/11 - 20/3 + 23/11

Then get a common denominator: in this case it's 33. So, you need to multiply everything that's over 11 by 3 and everything over 3 by 11.

That gives 306/33 - 220/33 + 69/33

Then, just add to get 155/33

Answer 4

h = 9 3/11 - 6 2/3 + 2 1/11

h = 102/11 - 20/3 + 23/11

h = 306/33- 220/33 + 69/33

h = 86/33 + 69/33

h = 155/33

- - - - - - - -s-

Answer 5

turn them into pure fractions
h = 102/11 - 20/3 + 23/11
put 3*11 on the bottom of them all
h = 306/33 - 220/33 + 69/33
h = 155/33 = 4 23/33

Answer 6

93/11+21/11-62/3=
114/11-62/3=

342-682/33=
-340/33

Answer 7

just give all 3 a common denominator, which is 33, and it will be as easy as ABC.

Answer 8

=((9*11+3)/11)-((6*3+2)/3)+((2*11+1)/11)
=(102/11)-(20/3)+(23/11)
=((102*3)-(20*11)+(23*3))/33
=(306-220+69)/33
=4.6969..

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Answer 9

12.8181818181818181818181...............