What’s the question LOLOL.
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ALGEBRA WORD PROBLEM
Dorian S. asked • 06/28/17
THREE LINEAR EQUATIONS WITH THREE VARIABLES
A chemist has three different acid solutions. The first acid solution contains 15% acid, the second contains 25%,
and the third contains 70%. He wants to use all three solutions to obtain a mixture of 40 liters containing 45%
acid, using 2 times as much of the 70% solution as the 25% solution. How many liters of each solution should be used?
How to set up as three linear equations to find the answer?
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Arthur D. answered • 06/28/17
TUTOR 4.9 (67)
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set up 2 equations
the first equation...
15%*x+25%*y+70%*2y=45%*40
0.15x+0.25y+1.4y=0.45*40
0.15x+1.65y=18
multiply all terms by 100
15x+165y=1800
divide all terms by 15
x+11y=120
the second equation...
x+y+2y=40
x+3y=40
x=40-3y
substitute into the first equation x+11y=120
40-3y+11y=120
40+8y=120
8y=120-40
8y=80
y=80/8
y=10 liters
x+3*10=40
x+30=40
x=40-30
x=10 liters
10 liters of 15%, 10 liters of 25%, and 20 liters of 70%
check:
10*15%=1.5
10*25%=2.5
20*70%=14
1.5+2.5+14=18
18/40=9/20=45/100=45%
45%*40=18
hope it helps
please mark brainliest
A: it would be 2 fives outta 8 then he picked one more bill out so chances are 4 fives outta 16. making it 1/4 of a probability.
V(x) = (2x - 3)/(5x + 4) The domain is all Real numbers except x = -4/5, because if x = -4/5 the denominator would be zero and you cannot divide by zero.{x | x ∈ R, x ≠ -4/5} w(x) = (5x + 4)/(2x - 3)similarly, x ≠ 3/2so, {x| x ∈ R, x ≠ 3/2}
Answer:
CB And A F
Step-by-step explanation:
