Answer:
Step-by-step explanation:
The index of a radical is the denominator of a fractional exponent, and vice versa. If you think about the rules of exponents, you know this must be so.
For example, consider the cube root:
![\sqrt[3]{x}\cdot \sqrt[3]{x}\cdot \sqrt[3]{x}=(\sqrt[3]{x})^3=x\\\\(x^{\frac{1}{3}})^3=x^{\frac{3}{3}}=x^1=x](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D%5Ccdot%20%5Csqrt%5B3%5D%7Bx%7D%5Ccdot%20%5Csqrt%5B3%5D%7Bx%7D%3D%28%5Csqrt%5B3%5D%7Bx%7D%29%5E3%3Dx%5C%5C%5C%5C%28x%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%29%5E3%3Dx%5E%7B%5Cfrac%7B3%7D%7B3%7D%7D%3Dx%5E1%3Dx)
That is ...
![\sqrt[3]{x}=x^{\frac{1}{3}} \quad\text{radical index = fraction denominator}](https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7Bx%7D%3Dx%5E%7B%5Cfrac%7B1%7D%7B3%7D%7D%20%5Cquad%5Ctext%7Bradical%20index%20%3D%20fraction%20denominator%7D)
Given: ∠ DEF
To construct: ∠TSZ ≅ ∠DEF
Construction: Consider the attachment
Step-01: Draw a line XY and choose a point S on it as a vertex of the required angle. Further marks point T such that DE = ST
Step-02: Take an arc AB from point E in ∠DEF of any length and draw at point S which cuts at point P on XY line.
Step-03: Take another arc of length AB from point B in ∠DEF and draw from point P which cuts to the previous arc at Q.
Step-04: Now, join the point SQ and extend up to Z such that EF = SZ
Hence, ∠ TSZ will be the required congruent constructed angle to∠DEF
Answer:
CODEINE
Step-by-step explanation: