Question

If x, y, z, are integers such that 2^x*3^y*7*z=329, Then what is x, y, and z? PLZZZZ HELP THANK YOU

Answer 1

Answer:

$\boxed{\sf \ \ \ 2^0*3^0*7*47=329 \ \ }$

Step-by-step explanation:

hello,

let's try to divide by 7 329 it comes

329 = 47 * 7

and 329 is not divisible by 2 or 3 so

the solution is

x = 0

y = 0

z = 47

$2^0*3^0*7*47=329$

hope this helps