All categories

3 years ago

What is the equation of the line that passes through (-3, -1) and has a slope of 2/5 ? Put your answer in slope-intercept form.

Mathematics

2 answers:

Scrat [10]3 years ago

8 0

Answer:

a. y = 2/5x + 1/5

Step-by-step explanation:

In point-slope form, the equation of a line with slope m through point (h, k) can be written

y = m(x -h) +k

Then a line with slope 2/5 through the point (-3, -1) will have point-slope equation ...

y = (2/5)(x +3) -1

This can be simplified to the desired form:

y = 2/5x +6/5 -1 . . . . . . eliminate parentheses; next, collect terms

y = 2/5x + 1/5

Nesterboy [21]3 years ago

6 0

I don’t know what the answer is I need some help

You might be interested in

Suppose you roll a die. What is the probability that you do not roll a 4?

7nadin3 [17]

Answer:

5/6

Step-by-step explanation:

There are six sides on a die, and they are asking the probability of not rolling a 4. So it would be 5 out of 6

Have a good day :)

6 0

3 years ago

Which expression has a value of 16 when x=2

ELEN [110]

Answer:

Step-by-step explanation:

point sorry

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

5 0

4 years ago

Can I have some help with this problem

Rufina [12.5K]

Corresponding measures have the same ratio.
.. (x -1)/4 = (2x +1)/10
.. 5(x -1) = 2(2x +1) . . . . . . multiply by 20
.. 5x -5 = 4x +2 . . . . . . . . eliminate parentheses
.. x = 7 . . . . . . . . . . . . . . . add 5 -4x

x = 7

7 0

3 years ago

Mileage tests are conducted for a particular model of automobile. If a 98% confidence interval with a margin of error of 1 mile

Llana [10]

Answer: 37

Step-by-step explanation:

Given : Significance level : $\alpha:1-0.98=0.02$