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marusya05 [52]
3 years ago
7

A chemist has three different acid solutions. The first acid solution contains 25 % acid, the second contains 45 % and the third

contains 90 %. He wants to use all three solutions to obtain a mixture of 50 liters containing 70 % acid, using 2 times as much of the 90 % solution as the 45 % solution. How many liters of each solution should be used?
the chemist should use:​
Mathematics
1 answer:
liq [111]3 years ago
8 0

Answer:

Wyzant

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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1 Expert Answer

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Best

Arthur D. answered • 06/28/17

TUTOR 4.9 (67)

Forty Year Educator: Classroom, Summer School, Substitute, Tutor

ABOUT THIS TUTOR ›

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

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Una caja de 25 newtons se suspende mediante un cable con diámetro de 2cm¿cuál es el esfuerzo aplicado al cable?
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<h3>¿Cómo calcular el esfuerzo aplicado sobre el cable?</h3>

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1 year ago
Hey guys<br>im new here<br>please solve this for me with steps!<br>ill mark as the best answer​
Vinil7 [7]

Answer:

The factors of  2(x+y)^2-9(x+y)-5 is ((x+y)-5)(2x+2y+1)

Step-by-step explanation:

Given polynomial

=>2(x+y)^2-9(x+y)-5

To Find:

The factors of the polynomial =?

Solution:

Lets assume  k = (x+y)

Then 2(x+y)^2-9(x+y)-5 can be written as 2k^2-9k-5

Now by using quadratic formula

k =\frac{-b\pm\sqrt{(b^2-4ac}}{2a}

where

a= 2

b= -9

c= -5

Substituting the values, we get

k =\frac{-b\pm\sqrt{(b^2-4ac)}}{2a}

k =\frac{-(-9) \pm \sqrt{((-9)^2-4(2)(-5)}}{2(2))}

k =\frac{-(-9) \pm \sqrt{(81+40)}}{4}

k =\frac{-(-9) \pm \sqrt{(121)}}{4}

k =\frac{-(-9) \pm 11}}{4}

k= \frac{ 9 \pm 11}{4}

k =  \frac{20}{4}                         k =  \frac{-2}{4}    

k_1 =5                                      k_2 = -\frac{1}{2}

2k^2-9k-5= 2(k-5)(k+\frac{1}{2})

Solving the RHS we get

\frac{2}{2}(k-5)(2k+1)

(k-5)(2k+1)

Substituting k = x+y

((x+y)-5)(2(x+y+1)

((x+y)-5)(2x+2y+1)

5 0
2 years ago
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