Question

3^2*2^2-2^2+3^2= ???????????? ​

Answer 1

Answer:

$3^2\cdot \:2^2-2^2+3^2=41$

Step-by-step explanation:

$3^2\cdot \:2^2-2^2+3^2$

$\gray{\mathrm{Follow\:the\:PEMDAS\:order\:of\:operations}}$

$\gray{\mathrm{Calculate\:exponents}\:3^2\::\quad 9}$

$=9\cdot \:2^2-2^2+3^2$

$\gray{\mathrm{Calculate\:exponents}\:2^2\::\quad 4}$

$=9\cdot \:4-2^2+3^2$

$\gray{\mathrm{Calculate\:exponents}\:2^2\::\quad 4}$

$=9\cdot \:4-4+3^2$

$\gray{\mathrm{Calculate\:exponents}\:3^2\::\quad 9}$

$=9\cdot \:4-4+9$

$\gray{\mathrm{Multiply\:and\:divide\:\left(left\:to\:right\right)}\:9\cdot \:4\::\quad 36}$

$=36-4+9$

$\gray{\mathrm{Add\:and\:subtract\:\left(left\:to\:right\right)}\:36-4+9\::}$

$36-4+9$

$\gray{36-4=32}$

$=32+9$

$\gray{32+9=41}$

$=41$