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harkovskaia [24]
3 years ago
12

How do you know that 21 over 30 is greater than 2 over 3

Mathematics
1 answer:
Bond [772]3 years ago
7 0
The problem deals with fractions comparison, lets do it:
21/30 > 2/3
we begin solving:
21 > (2/3)*30
21 > 2*10
<span>21 > 20
</span>therefore the proposed inequality is true, <span>21/30 > 2/3

You can solve as well getting same denominator for both fractions and comparing directly, in this case we need to get 2/3 to be divided by 30:
2/3 = (10/10)(2/3) = 20/30
So we have:
</span><span>21/30 > 2/3
</span>which is equal to:
<span>21/30 > 20/30
</span>and we compare directly because both fractions are divided by the same number, and we can see that the inequality is true.
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X/2 - y/3 = 3/2<br>X/3 + y/2 = 16/3
Ivenika [448]
X/2-y/3=3/2
(6×x/2)-(6×y/3)=6×3/2
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3 years ago
6x^2-4xy-15xy+10y^2 factor by grouping
PtichkaEL [24]
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Best Answer
We will need to split the middle term and use the grouping method. To do this multiply the coefficient of the first term (6) against the coefficient of the last term (10):

6 * 10 = 60
Factors of 60 = +-(1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60)

From the list of factors find two numbers that when added together give -19 and when multiplied together give 60. -15 and -4 added together give -19 and multiplied together give 60 so split the middle term by rewriting these values back into the expression:

6x^2 - 15xy - 4xy + 10y^2

Now use the grouping method, take out the highest common factor between the two sets of terms:

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Group the outside terms:

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Answers
6 = -3*-2
10 = 5*2
5*3+2*2=19
thus
(2y-3x)(5y-2x)
8 0
3 years ago
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