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2 years ago

7

If the length of a box is 17 cm breadth is 6 cm and volume is 1020 cube find the height of the box

1 answer:

alexandr1967 [171]2 years ago

4 0

Answer:

10 cm

Step-by-step explanation:

17×6×h=1020

102h=1020

h=1020/102=10

h=10 cm

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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.

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2 years ago

-2(x+3y)=18 solve for y!

jeka94

$-2(x+3y)=18\ \ \ \ /:(-2)\\\\x+3y=9\\\\3y=9-x\ \ \ \ /:3\\\\y=\frac{9-x}{3}\\\\y=\frac{9}{3}-\frac{x}{3}\\\\y=3-\frac{1}{3}x\\\\y=-\frac{1}{3}x+3$

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3 years ago

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2 years ago

Jason and Alison are hiking in the woods when they spot a rare owl in a tree. Jason stops and measures an angle of elevation of

Shalnov [3]

Answer:

$Owl= 67.14ft$

Step-by-step explanation:

<em>See comment for complete question.</em>

The given information is represented in the attached figure.

First convert 22°8'6'' and 30° 40’ 30” to degrees

$22^{\circ} 8'6'' = 22 + \frac{8}{60} + \frac{6}{3600}$

$22^{\circ} 8'6'' = 22.135^{\circ}$

$30^{\circ} 40'30'' = 30 + \frac{40}{60} + \frac{30}{3600}$

$30^{\circ} 40'30'' = 30.675^{\circ}$

Considering Jason's position:

$tan(22.135^{\circ}) = \frac{H}{x + 48}$

Where x = distance between the tree and Alison

Make H the subject

$H = (x + 48)tan(22.135^{\circ})$

Considering Alison's position

$tan(30.675^{\circ}) = \frac{H}{x}$

Make H the subject

$H = xtan(30.675^{\circ})$

$H = H$

$(x + 48)tan(22.135^{\circ}) = xtan(30.675^{\circ})$

$(x + 48) *0.4068 = x* 0.5932$

Open bracket

$x *0.4068 + 48 *0.4068 = 0.5932x$

$0.4068x + 19.5264 = 0.5932x$

Collect Like Terms

$-0.5932x+0.4068x = -19.5264$

$-0.1864x = -19.5264$

$0.1864x = 19.5264$

Make x the subject

$x = \frac{19.5264}{0.1864}$

$x = 104.76$

Substitute 104.76 for x in $H = xtan(30.675^{\circ})$

$H = 104.76 * tan(30.675^{\circ})$

$H = 104.76 * 0.5932$

$H = 62.14$

The above represents the height of the tree.

The height of the owl is:

$Owl= H + 5$

$Owl= 62.14 + 5$

$Owl= 67.14ft$

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2 years ago