Last one for tonight. This really confused me. Please help.
1 answer:
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"Compare ratios" covers a lot of territory. If all you want to do is find which one is larger, you can subtract or divide.
(a/b) - (c/d) > 0 . . . . means a/b is larger
(a/b) / (c/d) > 1 . . . . means a/b is larger
These operations with fractions are done using the methods of arithmetic with fractions that you have been taught.
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"Simplify" when applied to ratios usually means common factors are removed from each of the terms.
For example, the ratio 2:12 is a ratio of even numbers, so we know both numbers have a factor of 2. (2 is a factor common to both numbers.) When we divide them both by 2, we have the reduced ratio 1:6. That is, 2:12 simplifies to become 1:6.
(It is helpful to have a good working knowledge of multiplication tables when you approach problems in simplifying ratios.)
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In "quick and dirty" terms, you can do the subtraction ...
That is, you really only need to find out if (ad) > (bc) to determine if (a/b) > (c/d). A similar result is obtained when you consider division ...
This will be greater than 1 if (ad) > (bc), signifying that (a/b) > (c/d).
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Here's an example with numbers.
... Compare 6/7 to 43/50.
... The relationship of interest will be revealed by the products 6·50 = 300 and 7·43 = 301.
... Since 300 < 301, these tell us 6/7 < 43/50.
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We say the above methods are "quick and dirty" because the results may need to be simplified when you're done.
For example, subtracting 1/6 from 1/2 gives
... 1/2 - 1/6 = (1·6 - 2·1)/(2·6) = 4/12
Here, numerator and denominator have a common factor of 4. Removing that gives 4/12 = 1/3.
Even if you do this by the method of common denominators, you still need to reduce the result.
... 1/2 - 1/6 = 3/6 - 1/6 = (3-1)/6 = 2/6
This must be reduced to 1/3 by removing a common factor of 2 from numerator and denominator.
We know that
In any quadrilateral where the vertices are on the circumference of a circle, the opposite angles add up to
degrees.
So,
in this problem
therefore
the answer is
In any quadrilateral where the vertices are on the circumference of a circle, the opposite angles add up to
So,
in this problem
therefore
the answer is