3 years ago

Can someone explain to me HOW to solve these problems

Mathematics

1 answer:

oee [108]3 years ago

5 0

Answer:

(2m)^3(m^4)^5= 8m^23

(-3y^5)^3x2y^2= -54y^17

Step-by-step explanation:

1. For solving first part

(2m)^3(m^4)^5

= $(2^3 m^3) (m^{20} )\\Acoording \,to\,exponential\,rules \, exponent 4 \, and\, 5 \,were\, \,multiplied\\As\\\2^3 =8 \\so\\8m^3 \times m^{20} \\\\Acoording \,to\,exponential\,rules \, exponent \,3 \,and\,20\,will\,be\,added\\x^{2} 8m^{23}$

(2m)^3(m^4)^5= 8m^23

2. For solving second part

(-3y^5)^3 x 2y^2

$(-3^3) y^{5\times3} \times 2y^2\\\\As \,3^3= 27 \,and\\\\applying \,exponential\, rule \,the\, exponents\ 5 \, and\, 3 \,will\,be\,multiplied\\the \,equation will become\\-27 \,y15 \times 2y^2\\\\applying \,exponential\, rule \,the\, exponents\, 15 \,and\, 2 \,wil, be\,added \\-54 y^{17}$