How many three digit locker combinations are possible if the last number must be an even number?

for the 1st 2 you have a variety of 10 numbers; 1,2,3,4,5,6,7,8,9,0

and its 2 so 20+?

there are only 4 even numbers possible;2,4,6,8

so 20+4=24 combinations

y=−9/2x+10

step-by-step explanation:

let's solve for y.

0−9x−2y=−20

step 1: add 9x to both sides.

−9x−2y+9x=−20+9x

−2y=9x−20

step 2: divide both sides by -2.

−2y

−2

=

9x−20

−2

y=

−9

2

x+10

answer: the first graph

step-by-step explanation: