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3 years ago

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Serena is making an experiment. For that, she needs 20 grams of a 52% solution of salt. She has two large bottles of salt water:

one with 40% and the other with 70% of salt in them. How much of each must she use to make the solution she needs?

1 answer:

Fudgin [204]3 years ago

5 0

X + y = 20 --> eqn 1
The second equation would be the mass balance shown below:
(x)(0.40) + y(0.70) = 20(0.52) --> eqn 2
Rearranging eqn 1,x = 20 - ySubstitute this to eqn 2:
(20 - y)(0.40) + y(0.70) = 20(0.52)Solve for y,y = 8 grams Finally, solve for x.x = 20 - 8 = 12 grams
Therefore, Serena needs 12 grams of the 40% solution and 8 grams of the 70% solution.

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Answer:

<em>(a) x=2, y=-1</em>

<em>(b) x=2, y=2</em>

<em>(c)</em> $\displaystyle x=\frac{5}{2}, y=\frac{5}{4}$

<em>(d) x=-2, y=-7</em>

Step-by-step explanation:

<u>Cramer's Rule</u>

It's a predetermined sequence of steps to solve a system of equations. It's a preferred technique to be implemented in automatic digital solutions because it's easy to structure and generalize.

It uses the concept of determinants, as explained below. Suppose we have a 2x2 system of equations like:

$\displaystyle \left \{ {{ax+by=p} \atop {cx+dy=q}} \right.$

We call the determinant of the system

$\Delta=\begin{vmatrix}a &b \\c &d \end{vmatrix}$

We also define:

$\Delta_x=\begin{vmatrix}p &b \\q &d \end{vmatrix}$

And

$\Delta_y=\begin{vmatrix}a &p \\c &q \end{vmatrix}$

The solution for x and y is

$\displaystyle x=\frac{\Delta_x}{\Delta}$

$\displaystyle y=\frac{\Delta_y}{\Delta}$

(a) The system to solve is

$\displaystyle \left \{ {{x+y=1} \atop {x-2y=4}} \right.$

Calculating:

$\Delta=\begin{vmatrix}1 &1 \\1 &-2 \end{vmatrix}=-2-1=-3$

$\Delta_x=\begin{vmatrix}1 &1 \\4 &-2 \end{vmatrix}=-2-4=-6$

$\Delta_y=\begin{vmatrix}1 &1 \\1 &4 \end{vmatrix}=4-3=3$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{3}{-3}=-1$

The solution is x=2, y=-1

(b) The system to solve is

$\displaystyle \left \{ {{4x-y=6} \atop {x-y=0}} \right.$

Calculating:

$\Delta=\begin{vmatrix}4 &-1 \\1 &-1 \end{vmatrix}=-4+1=-3$

$\Delta_x=\begin{vmatrix}6 &-1 \\0 &-1 \end{vmatrix}=-6-0=-6$

$\Delta_y=\begin{vmatrix}4 &6 \\1 &0 \end{vmatrix}=0-6=-6$

$\displaystyle x=\frac{\Delta_x}{\Delta}=\frac{-6}{-3}=2$

$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{-6}{-3}=2$

The solution is x=2, y=2

(c) The system to solve is

$\displaystyle \left \{ {{-x+2y=0} \atop {x+2y=5}} \right.$

Calculating:

$\Delta=\begin{vmatrix}-1 &2 \\1 &2 \end{vmatrix}=-2-2=-4$

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$\displaystyle y=\frac{\Delta_y}{\Delta}=\frac{56}{-8}=-7$

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Answer: See explanation

Step-by-step explanation:

Here is the complete question:

A museum requires a minimum number of chaperones proportional to the number of students on a field trip. The museum requires a minimum of 3 chaperones for a field trip with 24 students. Which of the following could be combinations of values for the students and the minimum number of chaperones the museum requires? Choose 2 answers.

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B. Students: 16

Minimum of chaperones: 2

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D. Students: 45

Minimum of chaperones: 5

E. Students: 40

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Since the museum requires a minimum of 3 chaperones for a field trip with 24 students. This means that there will be 24/3 = 8 students per chaperone.

We then divide the number of students given in the question by the number of chaperone to know our answers. This. Will be:

Students: 72

Minimum of chaperones: 9

This will be: 72/9 = 8

Therefore, this is correct.

B. Students: 16

Minimum of chaperones: 2

This will be: 16/2 = 8

This is correct

C. Students: 60

Minimum of chaperones: 6

This will be: = 60/6 = 10.

Therefore, this is wrong

D. Students: 45

Minimum of chaperones: 5

This will be 45/5 = 9

Therefore, this is wrong.

E. Students: 40

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This will be: 40/5 = 8.

Therefore, this is wrong.

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