What is the term to this, Choices are given

Mathematics

1 answer:

kakasveta [241]3 years ago

6 0

Greater than the third side of the triangle

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Help!!!!

Luda [366]

Answer:

$\large\boxed{\dfrac{5}{10}}$

Step-by-step explanation:

$\sqrt[3]{\dfrac{125}{1000}}\qquad\text{use}\ \sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}}\\\\=\dfrac{\sqrt[3]{125}}{\sqrt[3]{1000}}=\dfrac{5}{10}\\\\\\\sqrt[3]{125}=5\ because\ 5^3=125\\\\\sqrt[3]{1000}=10\ because\ 10^3=1000$

8 0

3 years ago

Solve :1/1-x^a-b +1/1-x^b-a

Olegator [25]

Answer:

Step-by-step explanation:

$\frac{1}{1-x^{a-b}}+\frac{1}{1-x^{b-a}}\\\\\frac{1}{1-x^{b-a}}=\frac{1}{1-x^{-(a-b)}}=\frac{1}{1-\frac{1}{x^{a-b}}}\\=\frac{1}{\frac{x^(a-b)-1}{x^(a-b)}}=\frac{x^{a-b}}{x^{a-b}-1}\\\\=\frac{x^{a-b}}{-(1-x^{a-b})}=-\frac{x^{a-b}}{1-x^{a-b}}\\\\\frac{1}{1-x^{a-b}}+\frac{1}{1-x^{b-a}}=\frac{1}{1-x^{a-b}}-\frac{x^{a-b}}{1-x^{a-b}}\\\\=\frac{1-x^{a-b}}{1-x^{a-b}}=1$