How many 3 digit even number can be formed from the digit 1 2 3 4 5 6 and 6 can't be repeated ​

90 numbers

Step-by-step explanation:

Total choices(don’t care about anything)

6*6*6= 216 numbers

Half of these numbers are odd because the odd numbers are just 1,3,5

6*6*3=108

the three is there because there is 3 odd digits

we can subtract this from 216 along with the numbers that have a repeated 6

case 1: 66_

66 and any other number

remember: we have 6 choices for the last number

number of choices= 1*1*6=6 (the 1’s are there because 6 is the only digit that can come there).

case 2: 6_6

just like last time we have 1*6*1 choices=6

cool?

case 3: _66

again, we have 6*1*1=6 choices

Plan: total numbers minus the odd numbers and 6 repeats

216-108-18=90

answer 90 numbers

Mark me the brainliest only if you see my effort and you think it makes sense

step-by-step explanation:

mean or average is 40.6

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)