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alex41 [277]
3 years ago
5

Solve the inequality. Check your solutions.

Mathematics
1 answer:
Komok [63]3 years ago
8 0
The inequality is 

7- \frac{2}{b}\ \textless \ \frac{5}{b}

write 7 as 7b/b to have all the expressions in common denominator:

\frac{7b}{b} - \frac{2}{b}\ \textless \ \frac{5}{b}

\frac{7b-2}{b} \ \textless \ \frac{5}{b}

\frac{7b-2}{b}- \frac{5}{b}\ \textless \ 0

\frac{7b-2-5}{b}\ \textless \ 0

\frac{7b-7}{b}\ \textless \ 0

\frac{7(b-1)}{b}\ \textless \ 0

here b=1 is a root and b=0 is not in the domain of the expression, but it still has an effect in the sign of the expression.

the sign table of \frac{7(b-1)}{b} is :

+++++++[0] --------[1] +++++

this means that for values of b to the left of 0 and to the right of 1, the expression is positive, and for values of b in (0, 1), the expression is negative.

that is \frac{7(b-1)}{b}\ \textless \ 0 for b∈(0, 1)

Answer: (0, 1)
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3 years ago
In a purse there are 30 coins, twenty one-rupee and remaining 50-paise coins. Eleven coins are picked simultaneously at random a
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The solution to the answer is as follows:


<span>0votes</span><span>answered <span>Aug 8, 2015 </span><span><span>by </span>Shashi Kumar Evangelist <span>(3,082<span> points)</span></span></span></span><span>total coins 30
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7 0
4 years ago
Find the greatest number that divides451, 328 and 673 leaving 1, 4 and 1,respectively as remainders.
Vesna [10]

You're looking for the largest number <em>x</em> such that

<em>x</em> ≡ 1 (mod 451)

<em>x</em> ≡ 4 (mod 328)

<em>x</em> ≡ 1 (mod 673)

Recall that

<em>x</em> ≡ <em>a</em> (mod <em>m</em>)

<em>x</em> ≡ <em>b</em> (mod <em>n</em>)

is solvable only when <em>a</em> ≡ <em>b</em> (mod gcd(<em>m</em>, <em>n</em>)). But this is not the case here; with <em>m</em> = 451 and <em>n</em> = 328, we have gcd(<em>m</em>, <em>n</em>) = 41, and clearly

1 ≡ 4 (mod 41)

is not true.

So there is no such number.

4 0
3 years ago
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