anonymous
  • anonymous
If A:B=2:3 and B:C=4:5 and 5A:3C is --------- ?
Mathematics
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anonymous
  • anonymous
If A:B=2:3 and B:C=4:5 and 5A:3C is --------- ?
Mathematics
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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.

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anonymous
  • anonymous
A:B is 2:3 which means A/B=2/3. Also, B:C=4:5 which means B/C=4/5. You need a combination with A and C in your answer. You just have to solve for A and C in each equation and not worry about the fact there's a B still present. In these types of questions, it will usually disappear, as I'm about to show you: If A/B=2/3, then A = 2B/3. Also, C = 5B/4. Therefore, 5A = 10B/3 and 3C = 15B/4. Dividing, you have: \[\frac{5A}{3C}={\frac{\frac{10B}{3}}{\frac{15B}{4}}}=\frac{10B}{3} \times \frac{4}{15B}=\frac{2}{3}\]Note how the Bs cancelled. Good luck with the rest - I need sleep :)
anonymous
  • anonymous
Oh - to be technically finished, you should write: 5A:3C = 2:3

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