How do you evaluate #5r-8.35s# when #r=12# and #s=4#?

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Answer 1 · 2018-07-06T15:47:13

#color(crimson)(26.6#

#r = 12, s = 4#

#5 r - 8.35 s = (5*12) - (8.35 * 4)#

#=> 60 - 33.4 = 26.6#

Answer 2 · 2018-07-06T16:10:02

#26.6#

#"substitute the given values for r and s into the expression"#

#(5xxcolor(red)(12))-(8.35xxcolor(blue)(4))#

#=60-33.4#

#=26.6#

Answer 3 · 2018-07-06T17:10:29

#26.6#

We can just plug the values for the variables in our expression to get:

#5color(blue)((12))-8.35color(red)((4))#

#=>60-33.4#

#=26.6#

Hope this helps!