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Ask QuestionPosted by Rajneesh Payal 6 years, 5 months ago
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Posted by Gauri Chhatwani 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
39y3 (50y2 – 98) ÷ 26y2(5y + 7)
{tex} = \frac{{39{y^3}(50{y^2} - 98)}}{{26{y^2}(5y + 7)}}{/tex}
{tex} = \frac{{39{y^3} \times 2 \times (25{y^2} - 49)}}{{26{y^2}(5y + 7)}}{/tex}
{tex} = \frac{{39{y^3} \times 2 \times \{ {{(5y)}^2} - {{(7)}^2}\} }}{{26{y^2}(5y + 7)}}{/tex}
{tex} = \frac{{39{y^3} \times 2 \times (5y + 7)(5y - 7)}}{{26{y^2}(5y - 7)}}{/tex}. . . . [Using Identity III
= 3y (5y – 7)
Posted by Deepshikha Mishra 6 years, 5 months ago
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Deepshikha Mishra 6 years, 5 months ago
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Kashika Khuranaa.. ? 6 years, 5 months ago
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Yogita Ingle 6 years, 5 months ago
1] to construct it first draw a line of 1 unit AB
2] perpendicular to it draw a line of 1 unit BC
3] join AC and it will be root 2
4] then draw line perpendicular to AC as CD
5] join the AD continue this process upto root 13.
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Sia ? 6 years, 5 months ago
Linear equation is an equation between two variables that gives a straight line when plotted on a graph.
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