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3 years ago

PLEASE HELP!!!!!

Mathematics

2 answers:

aliya0001 [1]3 years ago

7 0

Answer:

1) g(x) is translated 2 units down from f(x)

2) g(x)=1.25⋅10x

I Hope I Helped!

lina2011 [118]3 years ago

3 0

2 units up!!!!!!!!!!!!!

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Find the number c that satisfies the conclusion of the Mean Value Theorem on the given interval. (Enter your answers as a comma-

coldgirl [10]

Answer:

c is all the points in the open interval (0,25)

Step-by-step explanation:

Here given is a function

$f(x) =x$ , which is continuous in the interval [0,25] and differentiable in (0,25)

Mean value theorem says there exists at least one c in the interval (0,25) such that

$f'(c) = \frac{f(25)-f(0)}{25-0}$

We have

$f(25)=25 and f(0) = 0\\f'(c) = 1$

For the given function

$f'(x) =1$

Hence we have c equals all the points in the open interval (0,25)

7 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:

$g: x_1+x_2+x_3=1000$

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

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

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3 years ago

How to write a polynomial in standard form

Setler79 [48]

Step-by-step explanation: Standard form is when we take a polynomial and we write it in order from the greatest degree to the smallest degree.

Let's look at an example which I provided in the image attached.

In this polynomial, I have 2 degrees, 1 degree, and 1 degree above the <em>x</em>.

This is not in the form of least to greatest so I need to write it in descending order. Our constant which in this is 27 will be last in polynomial.

So, you look at the degree of each term and then write each in term in order of degree from greatest to least (descending order).

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3 years ago

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Lapatulllka [165]

Written as a power of ten is 256.×10^0 or 2.56×10^1

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3 years ago

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vekshin1

Answer:

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Step-by-step explanation:

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2 years ago

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