Answer:
![\sqrt[4] {x^3}](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%20%7Bx%5E3%7D)
Step-by-step explanation:
At this point, we can transform the square root into a fourth root by squaring the argument, and bring into the other root:
![\sqrt x \cdot \sqrt[4] x =\sqrt [4] {x^2} \cdot \sqrt[4] x = \sqrt[4]{x^2\cdot x} = \sqrt[4] {x^3}](https://tex.z-dn.net/?f=%5Csqrt%20x%20%5Ccdot%20%5Csqrt%5B4%5D%20x%20%3D%5Csqrt%20%5B4%5D%20%7Bx%5E2%7D%20%5Ccdot%20%5Csqrt%5B4%5D%20x%20%3D%20%5Csqrt%5B4%5D%7Bx%5E2%5Ccdot%20x%7D%20%3D%20%5Csqrt%5B4%5D%20%7Bx%5E3%7D)
Alternatively, if you're allowed to use rational exponents, we can convert everything:
![\sqrt x \cdot \sqrt[4] x = x^{\frac12} \cdot x^\frac14 = x^{\frac12 +\frac14}= x^{\frac24 +\frac14}= x^\frac34 = \sqrt[4] {x^3}](https://tex.z-dn.net/?f=%5Csqrt%20x%20%5Ccdot%20%5Csqrt%5B4%5D%20x%20%3D%20x%5E%7B%5Cfrac12%7D%20%5Ccdot%20x%5E%5Cfrac14%20%3D%20x%5E%7B%5Cfrac12%20%2B%5Cfrac14%7D%3D%20x%5E%7B%5Cfrac24%20%2B%5Cfrac14%7D%3D%20x%5E%5Cfrac34%20%3D%20%5Csqrt%5B4%5D%20%7Bx%5E3%7D)
Answer:
Perron–Frobenius theorem for irreducible matrices. Let A be an irreducible non-negative n × n matrix with period h and spectral radius ρ(A) = r. Then the following statements hold. The number r is a positive real number and it is an eigenvalue of the matrix A, called the Perron–Frobenius eigenvalue.
Answer:
Ratio of area of circles: = πr2 :πc2 = πr2 :π r2 4 = 4 :1. 1. The small equilateral triangle is rotated through 60° about O, the center of the circle. The arm of the rotation is the radius of the circle
Step-by-step explanation:
36/20
36:20
36 to 20
hope this helps
