9514 1404 393
Answer:
C. 3x∛(y²z)
Step-by-step explanation:
The relevant rules of exponents are ...
![\sqrt[n]{a^m} = a^\frac{m}{n}\\\\(a^b)^c=a^{bc}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5Em%7D%20%3D%20a%5E%5Cfrac%7Bm%7D%7Bn%7D%5C%5C%5C%5C%28a%5Eb%29%5Ec%3Da%5E%7Bbc%7D)
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Your expression can be rewritten and simplified as follows.
![\displaystyle\sqrt[3]{27x^3y^2z}=(3^3x^3y^2z)^\frac{1}{3}=3xy^\frac{2}{3}z^\frac{1}{3}=3x(y^2z)^\frac{1}{3}=\boxed{3x\sqrt[3]{y^2z}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csqrt%5B3%5D%7B27x%5E3y%5E2z%7D%3D%283%5E3x%5E3y%5E2z%29%5E%5Cfrac%7B1%7D%7B3%7D%3D3xy%5E%5Cfrac%7B2%7D%7B3%7Dz%5E%5Cfrac%7B1%7D%7B3%7D%3D3x%28y%5E2z%29%5E%5Cfrac%7B1%7D%7B3%7D%3D%5Cboxed%7B3x%5Csqrt%5B3%5D%7By%5E2z%7D%7D)
Answer:
The given two triangles are not congruent.
Step-by-step explanation:
See the diagram of the two triangles attached to this question.
The third angle of the second triangle is [180° - (31° + 124°)] = 180° - 155° = 25°.{Since the other two angles of the triangle are 31° and 124°}
Now, for the first triangle, the values of the adjacent angles of the side with length 6 are 45° and 31°, but for the second triangle, the values of the adjacent angles of the side with length 6 are 25° and 31°.
Therefore, two triangles do not satisfy the Angle=Side-Angle i.e. ASA criteria to become congruent.
So, the given two triangles are not congruent. (Answer)
-4, -5/6, 3/7, 0.7 is least to greatest
