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3 years ago

Simplify 2x(5x + 3) + 7(5x + 3)

1 answer:

6 0

Use distributive property to solve.
2x(5x+3) + 7(5x+3)
1. distribute

2x•5x = 10x^2
2x•3= 6x

10x^2+6x

7•5x = 35x
7•3= 21

35x+21

2. combine like terms

10x^2+6x+35x+21

10x^+ 41x +21 [final answer]

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$\pi \div \frac{1}{4} = 4\pi$

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3 years ago

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Determine the equation of a circle whose diameter has the endpoints (-1, 2) and (7.

Viefleur [7K]

Answer:

(x - 3)^2 + (y + 1) = 25

Step-by-step explanation:

First find the midpoint of the diameter, because that represents the center of the circle.

-1 + 7

The x-coordinate of the midpoint is xm = ---------- = 3

2 - 4

and the y-coordinate is ym = ---------- = -1

And so the center of this circle is at (3, -1).

Use the Pythagorean Theorem to determine the square of the radius:

square of radius = 4^2 + (-3)^2 = 16 + 9 = 25

And so the equation of this circle is (x - 3)^2 + (y + 1) = 25

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3 years ago

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Which value below will make 3-6x=21

Lorico [155]

We are given the following equation and want to find which value of x will make it true:

$3-6x=21$

Subtract 3 from both sides

$-6x=18$

Divide both sides by -6

$x=-3$

Thus, the value that will make 3-6x=21 is x=-3. Let me know if you need any clarifications, thanks!

~ Padoru

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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

5 0

4 years ago

An evergreen nursery usually sells a certain shrub after 7 years of growth and shaping. The growth rate during those 7 years is

Ne4ueva [31]

Answer:

( 0.6 t^2 + 3t + 11 ) cm

Step-by-step explanation:

dh/dt = 1.2t + 3

at t = 0, h = 11 cm

(a)

dh / dt = 1.2 t + 3

dh = (1.2 t + 3) dt

integrate on both sides

h = 0.6 t^2 + 3t + c .... (1)

where c is the integrating constant

put t = 0

11 = c

Put in equation (1) , we get

h = ( 0.6 t^2 + 3t + 11 ) cm

Thus, teh height of tree after t years is given by

( 0.6 t^2 + 3t + 11 ) cm.

3 0

3 years ago