Solve.

Beakers X and Y contain some water. If 50 ml of water is transferred from Beaker X

eduard

Answer:

140 ml

Step-by-step explanation:

Let x be the amount of water in Beaker X and y be the amount of water in Beaker Y.

Then we get following equations:

First part

x - 50 = 3/7(y + 50)
7x -350 = 3y + 150
7x = 3y + 500

Second part

x + 100 = 4(y - 100)
x + 100 = 4y - 400
x = 4y - 500

Substitute x in first equation:

7(4y - 500) = 3y + 500
28y - 3500 = 3y + 500
28y - 3y = 500 + 3500
25y = 4000
y = 4000/25
y = 160 ml

Then finding x:

x = 4*160 - 500
x = 640 - 500
x = 140 ml

Initial amount of water in Beaker X is 140 ml, in Beaker Y is 160 ml

4 0

3 years ago

Area model multiplication: 2x2 digits what to do forget how to do it

valkas [14]

Answer:

Step-by-step explanation:

Yes because

4 0

3 years ago

Lagrange multipliers have a definite meaning in load balancing for electric network problems. Consider the generators that can o

Ivahew [28]

Answer:

The load balance $(x_1,x_2,x_3)=(545.5,272.7,181.8)$ Mw minimizes the total cost

Step-by-step explanation:

Optimizing With Lagrange Multipliers

When a multivariable function f is to be maximized or minimized, the Lagrange multipliers method is a pretty common and easy tool to apply when the restrictions are in the form of equalities.

Consider three generators that can output xi megawatts, with i ranging from 1 to 3. The set of unknown variables is x1, x2, x3.

The cost of each generator is given by the formula

$\displaystyle C_i=3x_i+\frac{i}{40}x_i^2$

It means the cost for each generator is expanded as

$\displaystyle C_1=3x_1+\frac{1}{40}x_1^2$

$\displaystyle C_2=3x_2+\frac{2}{40}x_2^2$

$\displaystyle C_3=3x_3+\frac{3}{40}x_3^2$

The total cost of production is

$\displaystyle C(x_1,x_2,x_3)=3x_1+\frac{1}{40}x_1^2+3x_2+\frac{2}{40}x_2^2+3x_3+\frac{3}{40}x_3^2$

Simplifying and rearranging, we have the objective function to minimize:

$\displaystyle C(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)$

The restriction can be modeled as a function g(x)=0:

$g: x_1+x_2+x_3=1000$

Or

$g(x_1,x_2,x_3)= x_1+x_2+x_3-1000$

We now construct the auxiliary function

$f(x_1,x_2,x_3)=C(x_1,x_2,x_3)-\lambda g(x_1,x_2,x_3)$

$\displaystyle f(x_1,x_2,x_3)=3(x_1+x_2+x_3)+\frac{1}{40}(x_1^2+2x_2^2+3x_3^2)-\lambda (x_1+x_2+x_3-1000)$

We find all the partial derivatives of f and equate them to 0

$\displaystyle f_{x1}=3+\frac{2}{40}x_1-\lambda=0$

$\displaystyle f_{x2}=3+\frac{4}{40}x_2-\lambda=0$

$\displaystyle f_{x3}=3+\frac{6}{40}x_3-\lambda=0$

$f_\lambda=x_1+x_2+x_3-1000=0$

Solving for \lambda in the three first equations, we have

$\displaystyle \lambda=3+\frac{2}{40}x_1$

$\displaystyle \lambda=3+\frac{4}{40}x_2$

$\displaystyle \lambda=3+\frac{6}{40}x_3$

Equating them, we find:

$x_1=3x_3$

$\displaystyle x_2=\frac{3}{2}x_3$

Replacing into the restriction (or the fourth derivative)

$x_1+x_2+x_3-1000=0$

$\displaystyle 3x_3+\frac{3}{2}x_3+x_3-1000=0$

$\displaystyle \frac{11}{2}x_3=1000$

$x_3=181.8\ MW$

And also

$x_1=545.5\ MW$

$x_2=272.7\ MW$

The load balance $(x_1,x_2,x_3)=(545.5,272.7,181.8)$ Mw minimizes the total cost

5 0

3 years ago

There are 800 students in a high school. Two-fifths of the students are freshman, and 3/4 of the upperclassmen (sophomores thru

Vlada [557]

Answer:

1/4 upperclassmen are not involved in enrichment.

Step-by-step explanation:

"3/4 of the upperclassmen (sophomores thru seniors) are involved in an enrichment." That leaves 1/4 upperclassmen students left. Verification: 3/4 + 1/4 = 4/4. 4/4 is also 1 whole.

6 0

2 years ago

Determine if line AB is tangent to the circle. Type

adoni [48]

Answer:

If AB is a tangent to the circle, the triangle ABO is right angled, as the angle where a tangent meets the circumference is always 90 degrees.

We also know that Pythogoras' theorem only holds for right angled triangles.

The hypotenuse is 12 + 8 as 12 is the radius so is 20.

16^2+12^2 = 256 + 144 = 400 = 20^2 so AB must be tangent.

6 0

3 years ago