4 years ago

If ADB = 70°,

1 answer:

7 0

Diameter of the circle is BD as shown, as indicated and passed through center where BD passes, triangle ABD is a right triangle with angle ABD as the right angle. It means ADB and ABD are complementary which means
AB = 140 degree
AD=40 degree

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Expand $\left(x^2 \frac{1}{x}\right)^3$. (Write the terms with higher degree first, so for example an $x^2$ term would come befo

Ilya [14]

Answer:

$x^6+3x^3+3+\dfrac{1}{x^3}$

Step-by-step explanation:

We can use the pascal triangle to find the coefficients of the expasion of the cube. It holds that

$(x^2 + \dfrac{1}{x})^3=(x^2)^3+3(x^2)^2\dfrac{1}{x}+3x^2(\dfrac{1}{x})^2+(\dfrac{1}{x})^3\\\\=x^6+3x^4\dfrac{1}{x}+3x^2\dfrac{1}{x^2}+\dfrac{1}{x^3}\\\\=x^6+3x^3+3+\dfrac{1}{x^3}$