Question

Need help please :) I’m not sure how to solve this one either

Answer 1

The answer is

580|21

Answer 2

Answer: 27 & 13/21

Your teacher may want you to delete the & symbol.

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Explanation:

Let's convert each mixed number to an improper fraction.

We'll start with 6 & 2/3

The formula to use is this

a & b/c = (a*c + b)/c

So,

a & b/c = (a*c + b)/c

6 & 2/3 = (6*3 + 2)/3

6 & 2/3 = (18 + 2)/3

6 & 2/3 = 20/3

The mixed number 6 & 2/3 converts to the improper fraction 20/3

Now do the same for the other mixed number as well.

a & b/c = (a*c + b)/c

4 & 1/7 = (4*7 + 1)/7

4 & 1/7 = (28 + 1)/7

4 & 1/7 = 29/7

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The task is to now multiply these improper fractions

Recall that (a/b)*(c/d) = (a*c)/(b*d)

We multiply across the numerators separately and multiply across the denominators separately.

So,

(20/3)*(29/7) = (20*29)/(3*7) = 580/21

The last step is to convert this to a mixed number.

Use a calculator to get 580/21 = 27.6190476190476 approximately.

The stuff to the left of the decimal is the whole part 27, aka the quotient.

The decimal part is then multiplied with the denominator (21) to get 0.6190476190476*21 = 13 which is the remainder.

Quotient = 27, remainder = 13, so 580/21 = 27 remainder 13 = 27 & 13/21