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3 years ago

Which equation represents a line which is perpendicular to the line 2x + y = -5?

Mathematics

1 answer:

Gnesinka [82]3 years ago

7 0

Step-by-step explanation:

Given expression;

2x + y = -5;

The given equation must be written in a form where the slope is the negative inverse of the given one.

A line perpendicular to another will have a negative inverse slope.

2x + y = -5

Equation of a straight line is generally written as;

y = mx + c

y and x are the coordinates

m is the slope

c is y - intercept

2x + y = -5

y = -2x -5

The slope of the perpendicular = $\frac{1}{2}$

So,

the new line;

y = $\frac{1}{2}$ x -5

Any line with slope of $\frac{1}{2}$ will be perpendicular

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Find the local maximum and minimum values and saddle point(s) of the function.

iogann1982 [59]

$f(x,y)=9e^y(y^2-x^2)$
$\nabla f=\left\langle-18xe^y,9ye^y(2+y)\right\rangle$

Critical points occur where the gradient is zero. This is guaranteed whenever $x=0$ and either $y=0$ or $y=-2$ .

The Hessian matrix for this function looks like

$H(x,y)=\begin{bmatrix}f_{xx}&f_{xy}\\f_{yx}&f_{yy}\end{bmatrix}=\begin{bmatrix}-18e^y&-18xe^y\\-18xe^y&9e^y(2-x^2+4y+y^2)\end{bmatrix}$

and has determinant

$|H(x,y)|=-162e^{2y}(2+x^2+4y+y^2)$

Maxima occur whenever the determinant is positive and $f_{xx}$ . Minima occur whenever both the determinant and $f_{xx}$ are positive. Saddle points occur whenever the determinant is negative.

At $(0,0)$ , you have a saddle point since the determinant reduces to -324, so $(0,0,0)$ is the saddle point.

At $(0,-2)$ , the determinant is $\dfrac{324}{e^4}>0$ and $f_{xx}(0,-2)=-\dfrac{18}{e^2}$ , so $\left(0,-2,\dfrac{36}{e^2}\right)$ is a local maximum.

No other critical points remain, so you're done.

4 0

3 years ago

HELP IM STUCK PLEASE

morpeh [17]

Answer:

C and F

Step-by-step explanation:

3 0

3 years ago

More and more and more

Lemur [1.5K]

Since a and 5 aren't like terms, you would write the expression as $\frac{a+5}{b}$