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jeka94
3 years ago
10

Describe fully the single transformation that maps triangle a onto triangle b

Mathematics
1 answer:
Katena32 [7]3 years ago
4 0
Reflection across y=-2 followed by a translation 2 units up
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Klio2033 [76]
By looking at the answer a is eliminated by default since it is not between 6 and 7 which also eliminates e. now you’re left with b and c and you have to remember the definition of an irrational number (in short, it’s a number that never ends) which would leave you with c
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3 years ago
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If a graph of y=-4x +2 were changed to a graph of y=-4x+5, how would the y- intercept change?
Dmitry [639]

Answer:

So if y=-4x+2 was changed to y=-4x+5, then the y-intercept would increase by 3.

The y-intercept was (0,2) then it becomes (0,5) in the new line.

Step-by-step explanation:

The slope-intercept form a line is y=mx+b where m is the slope and the y-intercept is b.

Both of these equations given are in this form.

y=-4x+2 when compared to y=mx+b you see that m=-4 and b=2.

Since b=2 then the y-intercept is 2.

y=-4x+5 when compared to y=mx+b you see that m=-4 and b=5.

Since b=5 then the y-intercept is 5.

So if y=-4x+2 was changed to y=-4x+5, then the y-intercept would increase by 3.

8 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
4 years ago
Anyone mind helping me out?
Airida [17]
<h2>6 is the answer.</h2>

just simple

3 0
3 years ago
Complete the following ratio table
Nady [450]
Could help if you showed the rest of the table
6 0
3 years ago
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