Respuesta :
a) 1% lidocaine -x g, 0.01x pure lidocaine
5% lidocaine -y g, 0.05y pure lidocaine
(1)0.01x+0.05y=350*0.02
0.01x g we get from1% lidocoine and 0.05y g we get from 5% lidocoine
(2) x+y=350
x g is mass 1% lidocoine, we combine it with y g 5%,
so we get x+y, and we need 350g,
so x+y should be equal 350, x+y=350
we get system of 2 equations that we need to solve to find x and y
(1)0.01x+0.05y=350*0.02
(2) x+y=350
we can solve it by substitution, find x from (2) equation
x=350-y, substitute it into (1)
instead of x we put (350-y)
0.01(350-y)+0.05y=7
3.5-0.01y+0.05y=7
0.04y =3.5
y=87.5g 5% lidocaine
b)x=350-87.5=262.5 g 1% lidocaine
c) 350g =350000 mg
5% lidocaine -y g, 0.05y pure lidocaine
(1)0.01x+0.05y=350*0.02
0.01x g we get from1% lidocoine and 0.05y g we get from 5% lidocoine
(2) x+y=350
x g is mass 1% lidocoine, we combine it with y g 5%,
so we get x+y, and we need 350g,
so x+y should be equal 350, x+y=350
we get system of 2 equations that we need to solve to find x and y
(1)0.01x+0.05y=350*0.02
(2) x+y=350
we can solve it by substitution, find x from (2) equation
x=350-y, substitute it into (1)
instead of x we put (350-y)
0.01(350-y)+0.05y=7
3.5-0.01y+0.05y=7
0.04y =3.5
y=87.5g 5% lidocaine
b)x=350-87.5=262.5 g 1% lidocaine
c) 350g =350000 mg
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Define x and y:
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Let x be the amount of 1% lidocaine needed.
Let y be the amount of 5% lidocaine needed.
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Construct Equations:
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x + y = 350
0.01x + 0.05y = 0.02 x 350
0.01x + 0.05y = 7
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Solve x and y:
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x + y = 350 ------------- (Equation 1)
0.01x + 0.05y = 7 ------------- (Equation 2)
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From equation 1:
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x + y = 350
x = 350 - y
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Substitute x = 350 - y into equation 2:
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0.01(350 - y) + 0.05y = 7
3.5 - 0.01y + 0.05y = 7
3.5 + 0.04y = 7
0.04y = 7 - 3.5
0.04y = 3.5
y = 87.5
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Substitute y = 87.5 into equation 1:
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x + y = 350
x + 87.5 = 350
x = 350 - 87.5
x = 262.5
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Find x and y:
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x = 262.6g (5% lidocaine cream)
y = 87.5g (1% lidocaine cream)
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Convert 350g to milligram:
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350g = 350 x 1000 mg
350g = 350 000mg
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Answer:
(a) Amount of 5% lidocaine needed = 87.5g
(b) Amount of 1% lidocaine needed = 262.g
(c) 350g = 350 000mg
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Define x and y:
------------------------------------------------------------------
Let x be the amount of 1% lidocaine needed.
Let y be the amount of 5% lidocaine needed.
------------------------------------------------------------------
Construct Equations:
------------------------------------------------------------------
x + y = 350
0.01x + 0.05y = 0.02 x 350
0.01x + 0.05y = 7
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Solve x and y:
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x + y = 350 ------------- (Equation 1)
0.01x + 0.05y = 7 ------------- (Equation 2)
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From equation 1:
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x + y = 350
x = 350 - y
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Substitute x = 350 - y into equation 2:
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0.01(350 - y) + 0.05y = 7
3.5 - 0.01y + 0.05y = 7
3.5 + 0.04y = 7
0.04y = 7 - 3.5
0.04y = 3.5
y = 87.5
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Substitute y = 87.5 into equation 1:
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x + y = 350
x + 87.5 = 350
x = 350 - 87.5
x = 262.5
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Find x and y:
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x = 262.6g (5% lidocaine cream)
y = 87.5g (1% lidocaine cream)
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Convert 350g to milligram:
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350g = 350 x 1000 mg
350g = 350 000mg
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Answer:
(a) Amount of 5% lidocaine needed = 87.5g
(b) Amount of 1% lidocaine needed = 262.g
(c) 350g = 350 000mg
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