Geometry i need help with this question pls help

A rectangle has a length of 7 units and a width of 4 units. One vertex of the rectangle is (0,0).

Ulleksa [173]

Answer:

28

Step-by-step explanation:

6 0

2 years ago

Solve for x. Enter the solutions from least to greatest.

shepuryov [24]

Answer:

Lesser = -7

Greater = 4

7 0

2 years ago

Find the slope of the tangent to y= -7Sin(x)+2cos(x) at x= 3π/4

BaLLatris [955]

Y = -7sin(x) + 2cos(x)
y = -7sin(3π/4) + 2cos(3π/4)
y = -7sin(9.42/4) + 2cos(9.42/4)
y = -7sin(2.355) + 2cos(2.355)
y = -7(0.03698381721) + 2(0.9993158646)
y = -0.2588867205 + 1.998631729
y = 1.739745009

6 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:

<u>Optimizing With Lagrange Multipliers</u>

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

An artist needs to purchase jars of paint. The table shows Which quantity container is the better buy?

frutty [35]

Answer:8-oz jar

First, we have to find out what is the unit rate of the 2-oz jar. $1.50÷2 is $.75.

Second, we have to find the unit rate of the 4-oz jar. $2.92÷4=$73.

Third, we have to find the unit rate of the 8-oz jar. $5.68=$.71.

Fourth, we have to find the unit rate of the 16-oz jar. The division for this one may be tricky. From dividing $11.62 by 16, is stopped at the 3rd number I got from dividing. I got $.726. This is not a value of cents and the value can't go in the thousandths place. So, I rounded .726. I got $.73. So $11.62÷16=$.73.

Lastly, you have to compare the amounts.

The lowest amount or better buy, is $.71 or 8-oz jar.

8 0

2 years ago

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