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Sia ? 6 years, 6 months ago
We can write 10 as
{tex}10 = 9 + 1 = 3^2 + 1^2{/tex}
Draw OA = 3 units, on the number line
Draw BA = 1 unit, perpendicular to OA.
Join OB
Figure:
Now, by Pythagoras theorem,
{tex}OB^2 = AB^2 + OA^2\\ OB^2 = 1^2 + 3^2 = 10{/tex}
{tex}\Rightarrow \quad O B=\sqrt{10}{/tex}
Taking O as centre and OB as a radius, draw an arc which intersects the number line at point C.
Clearly, OC corresponds to {tex}\sqrt{10}{/tex} on the number line.
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Sia ? 6 years, 6 months ago
(103)3
= (100 + 3)3
= 1003 + 33 + 3(100)(3)(100 + 3)
= 1000000 + 9 + 900(100 + 3)
= 1000000 + 9 + 9000 + 2700
= 1011709
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{tex}x ^ { 4 } - 1 = \left( x ^ { 2 } + 1 \right) ( x + 1 ) ( x - 1 ){/tex}
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Muskan Yadav 6 years, 6 months ago
2Thank You