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mariarad [96]
3 years ago
10

What is 0.11 as a fraction in simplest form

Mathematics
1 answer:
bazaltina [42]3 years ago
3 0
11 hundredths , or 11/100
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Answer is A. Subtract 18
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Study the geometric sequence {-4, -84, -1764, -37044, ...}.
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Answer:

              \bold{a_n=(-4)21^{n-1}}

Step-by-step explanation:

a_1=-4\,,\ a_2=-84\,,\ a_3=-1764\,,\ a_4=-37044\quad\implies\ \ r=21

{-84:(-4)=21 and -1764:(-84)=21 and -37044:(-1764)=21}

geometric sequence formula:   a_n=a_1r^{n-1}

a_1=-4\quad and \ \ r=21

a_n=(-4)21^{n-1}

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If a/b < c/d with b > 0, d > 0, prove that a+c / b+d lies between the two fractions a/b and c/d .​
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Divide through everything by <em>b</em> :

\dfrac{a+c}{b+d} = \dfrac{\dfrac ab + \dfrac cb}{1 + \dfrac db}

Since <em>a/b</em> < <em>c/d</em>, it follows that

\dfrac{a+c}{b+d} < \dfrac{\dfrac cd+\dfrac cb}{1 + \dfrac db}

Multiply through everything on the right side by <em>b/d</em> to get

\dfrac{a+c}{b+d} < \dfrac{\dfrac{bc}{d^2}+\dfrac cd}{\dfrac bd+1} = \dfrac{\dfrac cd\left(\dfrac bd+1\right)}{\dfrac bd+1} = \dfrac cd

and so (<em>a</em> + <em>c</em>)/(<em>b</em> + <em>d</em>) < <em>c/d</em>.

For the other side, you can do something similar and divide through everything by <em>d</em> :

\dfrac{a+c}{b+d} = \dfrac{\dfrac ad + \dfrac cd}{\dfrac bd + 1}

and <em>a/b</em> < <em>c/d</em> tells us that

\dfrac{a+c}{b+d} > \dfrac{\dfrac ad + \dfrac ab}{\dfrac bd + 1}

Then

\dfrac{a+c}{b+d} > \dfrac{\dfrac ab + \dfrac{ad}{b^2}}{1 + \dfrac db} = \dfrac{\dfrac ab\left(1+\dfrac db\right)}{1 + \dfrac db} = \dfrac ab

and so (<em>a</em> + <em>c</em>)/(<em>b</em> + <em>d</em>) > <em>a/b</em>.

Then together we get the desired inequality.

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3 years ago
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