Question

solve (6^0+27^-1)x3^5

Answer 1

(6^0 + 27 - 1 )* x^35.

hope its help u

Answer 2

Answer:

The value of this exponent is 252

Step-by-step explanation:

Here we have been given to find the value of the expression as given below. Now we know that anything raised to the power of zero is 1. Using this and the properties of exponents we find the value to be;

$(6^{0} + 27^{-1} ) (3^{5} )$

= $(1 + \frac{1}{27} )(3^{5} )$

= $\frac{27+1}{27} (3^{5} )$

= $\frac{28}{27} (3^{5} )$

= $\frac{28}{3^{3} } (3^{5} )$

= 28 × $3^{5-3}$

= 28 × $3^{2}$

= 28 × 9

= 252

Hence the value of the above-given exponential sum is 252