Ask question

Ask question

All categories

3 years ago

11

The graph of y = 4x - 11 is translated up 8 units. Which equation represents the translated graph? f y = 4x - 19 g y = 4x - 3 h

y = 12x - 11 j y = 12x - 3

2 answers:

d1i1m1o1n [39]3 years ago

7 0

$\underbrace{y=f(x)\xrightarrow{T_{\vec{a}=[0;\ 8]}} y=f(x)+8}_{tranlation\ up\ 8\ units}\\\\y=4x-11\xrightarrow{T_{\vec{a}=[0;\ 8]}} y=4x-11+8=4x-3\\\\Answer:\fbox{g}$

lakkis [162]3 years ago

3 0

The correct answer is y=4x-3, because:
y=4x-11 u=[0,8]
y=4x-11+u[0,8]
y=4x-11+8
y=4x-3
g(y)=4x-3

You might be interested in

HELP 20 POINTS PLZ RAHHHH

schepotkina [342]

Answer:

answer is option D : 42m+12

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:

$g: x_1+x_2+x_3=1000$

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

How many times does seven go into three hundred ninety four

Nookie1986 [14]

Answer: 56 times.

Step-by-step explanation:

394/7 = 56.3

56 * 7 = 392

392 + 7 = 399

399 is more than 394, therefore the answer is 56.

6 0

3 years ago

Read 2 more answers

The GCF of an odd number and an even number will always be even

Gelneren [198K]

This is false for example, the GCF of 13 and 26 is 13, an odd number.

8 0

3 years ago

21 is what percent of 40

-Dominant- [34]

To solve for this, set up a proportion.
21/40 = x/100.
Cross multiply the fraction's numerators with the denominators;
21 x 100, 40 and x.
2100/40x.
Divide both numbers by 40 to get x.
2100/40 = 52.5
21 is 52.5% of 40.
I hope this helps!

5 0

3 years ago