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think about it. 80 percent is 4 times 20 percent.
4*1=4 (part 2)
Write an equation in ONE variable that can be used to find the total number of liters of orange juice and water in the can
plz help @Plasmataco
what the heck does \(y=4x\) mean in this context?
i dont know?
lets go slow, read the question carefully, name a variable, and see if we can come up with an equation that solves what we are looking for
oh. Well, picture it in your head. If 80% is OJ, how much percent is water?
the question asks, "Write an equation in one variable that can be used to find the total number of liters of orange juice and water in the can"
so isn't there 4 times more OJ than water?
we actually know the liters of water in the can, it is 1 as you are told that so what we need to find is the number of liters of orange juice in the can
so if there is one liter of water, there should be 4 time more in OJ
now as @Plasmataco said, after he wrote that nonsense about \(y=4x\) which has nothing to do with this, there is in fact 4 times more oj than water, so since there is one liter of water there must be 4 liters of ok, the total being 5 liters lets still come up with an equation that we can solve
and woah! don't attack me. I wuz tryin halp.
if we put say \(x\) as the number of liters of OJ, then we know two things first we know \(x+1\) is the total number of liters of mix, and we know that \(80\%\) of it is OK, so we know \[0.8\times (x+1)=x\]
if you solve that equation in ONE variable (which we have defined as \(x=\) the number of liters of OJ we will get \(x=4\)
whatevs, we're both right.
so confused omg
just... 4 liters of orange juice and 0.8*(x+1)=x
ok? so .8 times (x plus one) equals x?
or... if you want to take it really simple, X=4*1. You should probably use the other one. XD
^^ @pooja195 but the question asks for an equation that gives me the total liters, oj+water in the can, so would x=.8x(t+1) be right, if t=oj and x=total amount?
pleease? @Hero ? can u suggest any1 to help?
no... well, i just dont know if he got the message of the question :( read my last comment... @Hero
and not anymore