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navik [9.2K]
2 years ago
14

5/9x3/4 in simplest form?

Mathematics
2 answers:
kompoz [17]2 years ago
8 0

Answer:

\frac{5}{12}

Step-by-step explanation:

\frac{5}{9} × \frac{3}{4} = \frac{15}{36}

HCF of 15,36 = 3

\frac{15}{36} ÷ \frac{3}{3} = \frac{5}{12}

balandron [24]2 years ago
6 0
5/9*3/4
=15/36
=divisible by 3
=5/12
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Answer:

The correct option is;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }}

Step-by-step explanation:

The given information is that m(x) = x² - 17·x

The above equation can be written in the form;

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x =  \dfrac{17\pm \sqrt{289+4\cdot y}}{2}= \dfrac{17}{2} \pm \dfrac{1}{2}  \times \sqrt{289+4\cdot y}} = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +\dfrac{4\cdot y}{4} }}

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x = \dfrac{17}{2} \pm \sqrt{\dfrac{289}{4} +y }} = \dfrac{17}{2} \pm \sqrt{y + \dfrac{289}{4} }}

Interchanging x and y in the function of the inverse, m⁻¹(x), we have;

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m'(x) = 2·x - 17 = 0

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m''(x) = 2 > 0, therefore, the point is a minimum point on the graph

Therefore, as x increases past the minimum point of 17/2, m⁻¹(x) increases to give;

The \ domain \ restriction \ x \geq \dfrac{17}{2} \ results \ in \ m^{-1}(x) = \dfrac{17}{2} \pm \sqrt{x + \dfrac{289}{4} }} to increase m⁻¹(x) above the minimum.

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