The equation 5x-10y=30is written in standard form. What is the first step when solving for x? Add 5x to both sides of the equati
2 answers:
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Well, what I can make out is.
"20% of the drug present in the preceding time...", is mostly mumble jumble, in short is just saying, the patient took a pill, then before taking another pill, the previous pill isn't really gone from the body stream, is there still, 20% of it, so by the time the patient drops another pill in the following day as prescribed, she's dropping 500mg of the drug, which is 100% of the drug, but besides that, she already has 20% of the drug from the pill she took the previous time, the day before, so, from yesterday.
a)
how much of the drug is there when she takes another pill?
well, 20% of 500mg is just 100mg 0.2 * 500
so, she's dropping 500mg today, and she has in her body stream 100mg(20%) from the one she took yesterday, so her body will have then 500 + 100 milligrams of the drug.
b)
now, the second equation there, is fine too, is just that it is a difference, but the result is the same 1.2 or 120% of the drug.
in essence the difference equation is just saying, "ok, grab two pills and then subtract 80% of one pill, that's what your body has right after taking the next pill".
"20% of the drug present in the preceding time...", is mostly mumble jumble, in short is just saying, the patient took a pill, then before taking another pill, the previous pill isn't really gone from the body stream, is there still, 20% of it, so by the time the patient drops another pill in the following day as prescribed, she's dropping 500mg of the drug, which is 100% of the drug, but besides that, she already has 20% of the drug from the pill she took the previous time, the day before, so, from yesterday.
a)
how much of the drug is there when she takes another pill?
well, 20% of 500mg is just 100mg 0.2 * 500
so, she's dropping 500mg today, and she has in her body stream 100mg(20%) from the one she took yesterday, so her body will have then 500 + 100 milligrams of the drug.
b)
now, the second equation there, is fine too, is just that it is a difference, but the result is the same 1.2 or 120% of the drug.
in essence the difference equation is just saying, "ok, grab two pills and then subtract 80% of one pill, that's what your body has right after taking the next pill".