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2 years ago

I need help with math

Mathematics

1 answer:

Keith_Richards [23]2 years ago

4 0

$\frac{1}{5} + \frac{3}{4} \times \frac{3}{5} = \\$

$\frac{1}{5} + \frac{9}{20} = \\$

$\frac{4}{20} + \frac{9}{20} = \\$

$\frac{13}{20} \\$

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What is the vertex point of y= - 2+3? Enter your answer as an ordered pair, e.g. (1,2).

Anettt [7]

Answer:

y=1, so (0, 1)

Step-by-step explanation:

Just subtract 3 from -2. Since there is no x in the problem, it means that this line is horizontal, meaning that the x can change but the y has to stay at 1.

The line is horizontal so there is no vertex point

6 0

2 years ago

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.

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2 years ago

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cluponka [151]

Answer:

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Step-by-step explanation:

10!•4!

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6!•5!

Rewriting as

10 * 9*8*7*6! * 4!

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6!•5*4!

Canceling like terms

10 * 9*8*7

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