Question

3/5(k-2)-1/4(2k-7)<=3. solve and write in interval

Answer 1

3/5(k-2)-1/4(2k-7)<=3

3/5k-1.2-1/2k-1.75<=3

1/10k-2.95<=3
1/10k<=5.95
\[k \le59.5\]

Interval notation
\[(59.5,-\infty)\]

Answer 2

okay thanks

Answer 3

Hope I helped you!

Answer 4

why decimals for opening 5(k-2)

Answer 5

3/5(k-2)

Answer 6

yes

Answer 7

these is how i was doing it
\[[3/(5k-10)]-[1/(4k-28)\le3\]
since
\[a/b+c/d=(ad+bc)/bd\]
\[[3*(8k-28)-(5k-2)]/[5*4(k-2)(2k-7)]\le3\]
\[[19k-74]/20(k-2)(2k-7)\le3\]
\[(19k-74)/60(k-2)(2k-7)\le1\]
critical values \[k=2;(3,5);(3,897)\]
\[k<2\]
\[7/2<k<74/19\]

Answer 8

Jonask that is incorrect