Will someone check my awnsers on my geometry b exam for me

Answer all for points

IceJOKER [234]

Answer:

1. If we drink 8 glasses, we are drinking 2 L of water, so we are exceeding 1 L of water per day.

2. We need 100 sections.

I used metric conversion to convert km to meters.

3. 230 mL of lemonade per person

I used metric conversion to convert L to ML

Step-by-step explanation:

1. 250 mL = 1 glass

We are drinking 8 glasses

8 glasses * 250 ml/ 1 glass = 2000 mL

1 L = 1000 mL

2000 mL * 1 L/1000 mL =2 L

If we drink 8 glasses, we are drinking 2 L of water, so we are exceeding 1 L of water per day.

I used metric convertsion to convert mL to liters.

Guardrails sections to be 5m long

Road length that needs a guardrail is .5 km. We need to convert this to meters 1000 m = 1 km

.5km * 1000m / 1km = 500 meters

We have 500 meters that needs guardrails.

500 meters * 1 section/ 5 meters = 100 sections

We need 100 sections.

I used metric conversion to convert km to meters.

3. 6.9 L of lemonade to serve 30 people. We want to serve ml

Convert L to mL

1L = 1000 ML

6.9 L * 1000 ml/1L = 6900 mL

We need to divide 6900 Ml by 30 people

6900mL/30 people = 230 mL of lemonade per person

I used metric conversion to convert L to ML

5 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

Please please please help!!!

dalvyx [7]

9514 1404 393

Answer:

2

Step-by-step explanation:

Fill in the argument value and look up the function value on the graph.

$(f\circ f)(-2) = f(f(-2)) = f(2) = \boxed{2}$

7 0

2 years ago

2+2=? A- 4 B- no it's A C- nop look at B D- come on really? Go to College

anygoal [31]

Answer:

E- 48-x+y=1.2353783763

6 0

2 years ago

Read 2 more answers

Point Q is plotted on the coordinate grid. Point P is at (10, −20). Point R is vertically above point Q. It is at the same dista

bixtya [17]

Answer:

Point R is at (−20, 10), a distance of 30 units from point Q

Step-by-step explanation:

Q has coordinates (-20,-20).

P has coordinates (10,-20)

Since point R is vertically above point Q, it will have the same x-coordinate as Q.

Let R have coordinates (-20,y).

It was given that;

$|RQ|=|PQ|$

$\Rightarrow |y--20|=|10--20|$

$\Rightarrow y+20=10+20$

$\Rightarrow y=10+20-20$

$\Rightarrow y=10$ .

The coordinates of R are (-20,10).

The dstance from Q is 30 units.

3 0

3 years ago

Read 2 more answers