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 2 years ago
use an inequality to solve: 1 side of a rectangle is 5 and the other is x,, find the possible values of x if the area is at least 60...... I found out what x would be it would be 25 but idk how to do the inequality im not to sure anyone
 2 years ago
use an inequality to solve: 1 side of a rectangle is 5 and the other is x,, find the possible values of x if the area is at least 60...... I found out what x would be it would be 25 but idk how to do the inequality im not to sure anyone

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Koran90
 2 years ago
Best ResponseYou've already chosen the best response.0The inequality is simply the x on its own multipled by the indices

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.0first, where'd you get 25 from?

Koran90
 2 years ago
Best ResponseYou've already chosen the best response.0Multiplying an indices gets 25 from the number bank

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.0Koran, you're not being original or funny. I see at least 5 people do what you are doing whenever I come on here

laura80
 2 years ago
Best ResponseYou've already chosen the best response.0well to get the area it is l^2+w^2 so it would be 5+5+x+x=60 so it would be 10+2x=60

Koran90
 2 years ago
Best ResponseYou've already chosen the best response.0Its a hypothetical model to inequalities devised by the indices I am not being funny. I tutor higher mathemtatics

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.05+5+x+x would give you the perimeter, not the area

SUROJ
 2 years ago
Best ResponseYou've already chosen the best response.2area is at least 60 means lenght*bredth >=60 5*x>=60 x>=12

Dockworker
 2 years ago
Best ResponseYou've already chosen the best response.0area=length*width perimeter=2(length+width)

Koran90
 2 years ago
Best ResponseYou've already chosen the best response.0no perimenter*indices*diagnonals(length=width0

SUROJ
 2 years ago
Best ResponseYou've already chosen the best response.2so if you take value of x equal or greater than 12 then area would be at least 60 or more than 60 but not less than 60
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