Answer:
2nd option
Step-by-step explanation:
makes most sense
Answer:
Step-by-step explanation:
Simplify expression with rational exponents can look like a huge thing when you first see them with those fractions sitting up there in the exponent but let's remember our properties for dealing with exponents. We can apply those with fractions as well.
Examples
(a) ![(p^4)^{\dfrac{3}{2}}](https://tex.z-dn.net/?f=%28p%5E4%29%5E%7B%5Cdfrac%7B3%7D%7B2%7D%7D)
From above, we have a power to a power, so, we can think of multiplying the exponents.
i.e.
![(p^{^ {\dfrac{4}{1}}})^{\dfrac{3}{2}}](https://tex.z-dn.net/?f=%28p%5E%7B%5E%20%7B%5Cdfrac%7B4%7D%7B1%7D%7D%7D%29%5E%7B%5Cdfrac%7B3%7D%7B2%7D%7D)
![(p^{^ {\dfrac{12}{2}}})](https://tex.z-dn.net/?f=%28p%5E%7B%5E%20%7B%5Cdfrac%7B12%7D%7B2%7D%7D%7D%29)
Let's recall that when we are dealing with exponents that are fractions, we can simplify them just like normal fractions.
SO;
![(p^{^ {\dfrac{12}{2}}})](https://tex.z-dn.net/?f=%28p%5E%7B%5E%20%7B%5Cdfrac%7B12%7D%7B2%7D%7D%7D%29)
![= (p^{ 6})](https://tex.z-dn.net/?f=%3D%20%28p%5E%7B%206%7D%29)
Let's take a look at another example
![\Bigg (27x^{^\Big{6}} \Bigg) ^{{\dfrac{5}{3}}}](https://tex.z-dn.net/?f=%5CBigg%20%2827x%5E%7B%5E%5CBig%7B6%7D%7D%20%5CBigg%29%20%5E%7B%7B%5Cdfrac%7B5%7D%7B3%7D%7D%7D)
Here, we apply the
to both 27 and ![x^6](https://tex.z-dn.net/?f=x%5E6)
![= \Bigg (27^{{\dfrac{5}{3}}} \times x^\Big{\dfrac{6}{1}\times {{\dfrac{5}{3}}} }\Bigg)](https://tex.z-dn.net/?f=%3D%20%5CBigg%20%2827%5E%7B%7B%5Cdfrac%7B5%7D%7B3%7D%7D%7D%20%5Ctimes%20x%5E%5CBig%7B%5Cdfrac%7B6%7D%7B1%7D%5Ctimes%20%7B%7B%5Cdfrac%7B5%7D%7B3%7D%7D%7D%20%7D%5CBigg%29)
![= \Bigg (27^{{\dfrac{5}{3}}} \times x^\Big{\dfrac{2}{1}\times {{\dfrac{5}{1}}} }\Bigg)](https://tex.z-dn.net/?f=%3D%20%5CBigg%20%2827%5E%7B%7B%5Cdfrac%7B5%7D%7B3%7D%7D%7D%20%5Ctimes%20x%5E%5CBig%7B%5Cdfrac%7B2%7D%7B1%7D%5Ctimes%20%7B%7B%5Cdfrac%7B5%7D%7B1%7D%7D%7D%20%7D%5CBigg%29)
Let us recall that in the rational exponent, the denominator is the root and the numerator is the exponent of such a particular number.
∴
![= \Bigg (\sqrt[3]{27}^{5} \times x^{10} }\Bigg)](https://tex.z-dn.net/?f=%3D%20%5CBigg%20%28%5Csqrt%5B3%5D%7B27%7D%5E%7B5%7D%20%5Ctimes%20x%5E%7B10%7D%20%7D%5CBigg%29)
![= \Bigg (3^{5} \times x^{10} }\Bigg)](https://tex.z-dn.net/?f=%3D%20%5CBigg%20%283%5E%7B5%7D%20%5Ctimes%20x%5E%7B10%7D%20%7D%5CBigg%29)
![= 249x^{10}](https://tex.z-dn.net/?f=%3D%20249x%5E%7B10%7D)
Answer:
1 Expert Answer
X + Y + Z = 264. Since Y is 1 bigger than X, Y = X + 1. Since Z is 1 bigger than Y, Z = Y + 1. But Y = X + 1, so Z = (X + 1) + 1 = X + 2.
Combine like terms
Then you’ll get -7a-2b+8c-14 which is the answer.
Answer:
no
Step-by-step explanation:
If it is not a perfect square, then the original number cannot be a perfect square. 125 is not a perfect square [nor is 53 or 35], so 1953125 cannot be a perfect square.