find the intercepts of given polynomial equation 2×⁴–11׳+11ײ–11×9=0?
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Answer:
The actual amount if 40% of the amount is 60 rupees is 150.
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Answer:
Given : x=2+\sqrt{3}x=2+3
To Find : find the value of x+\frac{1}{x}x+x1
Solution:
x+\frac{1}{x}x+x1
Since x=2+\sqrt{3}x=2+3
To find \frac{1}{x}x1
x=2+\sqrt{3}x=2+3
\frac{1}{x} =\frac{1}{2 +\sqrt{3} } \times \frac{2 - \sqrt{3} }{2 - \sqrt{3}}x1=2+31×2− 32−3
\frac{1}{x} =\frac{2 -\sqrt{3} }{(2) {}^{2} - ( \sqrt{3}) {}^{2} }x1=(2)2−(3)22−3
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