anonymous
  • anonymous
What is the least common denominator for the fractions 1/6 and 3/4 ? A. 8 B. 12 C. 18 D. 24 i will give medel
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
what fractions
anonymous
  • anonymous
fractions 1/6 and 3/4
anonymous
  • anonymous
yeah

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

mathstudent55
  • mathstudent55
|dw:1449686598904:dw|
anonymous
  • anonymous
so 22.3?
mathstudent55
  • mathstudent55
To find the LCD, first find the prime factorizations of the numbers. The prime factorization of 6 is 2 * 3 The prime factorization of 4 is 2^2
anonymous
  • anonymous
or 2 to the power of 2 .3
mathstudent55
  • mathstudent55
Then the LCD is the product of common and not common factors with the higher exponent. 2 is a common factor, use the larger exponent, 2^2. 3 is not common, use it too. \(LCD = 2^2 \times 3\) WHat is \(2^2 \times 3\)?
anonymous
  • anonymous
12
mathstudent55
  • mathstudent55
2^2 = 2 * 2 = 4, so 2^2 * 3 = 4 * 3 = ?
mathstudent55
  • mathstudent55
Correct. The LCD is 12.
anonymous
  • anonymous
Thanks!
mathstudent55
  • mathstudent55
You're welcome.

Looking for something else?

Not the answer you are looking for? Search for more explanations.