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3 years ago

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Which expression represents the height of the pyramid? StartFraction 3 V Over y squared EndFraction units (3 V minus y squared)

units (V minus 3 y squared) units StartFraction V Over 3 y squared EndFraction units

2 answers:

maxonik [38]3 years ago

6 0

Answer:

Height = 3v/y² units

StartFraction 3 V Over y squared EndFraction units

Step-by-step explanation:

The volume of a solid right pyramid with a square base is v units3 and the length of the base edge is y units. which expression represents the height of the pyramid? units (3v – y2) units (v – 3y2) units units

Volume of a solid right pyramid = 1/3 × area of the base × height

Volume of a solid right pyramid = v units³

Area of the base = y² unit²

Volume of a solid right pyramid = 1/3 × area of the base × height

v = 1/3 × y² × height

Height = v ÷ 1/3 × y²

= v × 3/1y²

= (v × 3) / y²

= 3v / y²

Height = 3v/y² units

StartFraction 3 V Over y squared EndFraction units

alexgriva [62]3 years ago

6 0

Answer:

B

Step-by-step explanation:

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Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

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Then in the entire integrand, set $x+2=\sqrt3\sec t$ , so that $\mathrm dx=\sqrt3\sec t\tan t\,\mathrm dt$ . The integral is then equivalent to

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
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$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

Note that by letting $x+2=\sqrt3\sec t$ , we are enforcing an invertible substitution which would make it so that $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ requires $0\le t$ or $\dfrac\pi2$ . However, $\tan t$ is positive over this first interval and negative over the second, so we can't ignore the absolute value.

So let's just assume the integral is being taken over a domain on which $\tan t>0$ so that $|\tan t|=\tan t$ . This allows us to write

$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\tan t}\,\mathrm dt$
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We can show pretty easily that

$\displaystyle\int\csc t\,\mathrm dt=-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec t\csc t\,\mathrm dt=-\ln|\csc2t+\cot2t|+C$
$\displaystyle\int\sec^2t\csc t\,\mathrm dt=\sec t-\ln|\csc t+\cot t|+C$
$\displaystyle\int\sec^3t\csc t\,\mathrm dt=\frac12\sec^2t+\ln|\tan t|+C$

which means the integral above becomes

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
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Back-substituting to get this in terms of $x$ is a bit of a nightmare, but you'll find that, since $t=\mathrm{arcsec}\dfrac{x+2}{\sqrt3}$ , we get

$\sec t=\dfrac{x+2}{\sqrt3}$
$\sec^2t=\dfrac{(x+2)^2}3$
$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
$\cot t=\sqrt{\dfrac3{x^2+4x+1}}$
$\csc t=\dfrac{x+2}{\sqrt{x^2+4x+1}}$
$\csc2t=\dfrac{(x+2)^2}{2\sqrt3\sqrt{x^2+4x+1}}$

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Read 2 more answers

Of all rectangles with area 324324, which one has the minimum perimeter? Let P and w be the perimeter and width, respectively

ra1l [238]

Answer:

For w = 18 units perimeter is minimum

P = 2(18 + w)

Step-by-step explanation:

Given;

Area of the rectangle = 324 units²

P is the perimeter

w is the width

Let L be the length of the rectangle

therefore,

P = 2(L + w) ............(1)

also,

Lw = 324

or

L = $\frac{324}{w}$ ..........(2)

substituting 2 in 1

P = $2(\frac{324}{w} + w)$

now,

for minimizing the perimeter

$\frac{dp}{dw}=\frac{d(2(\frac{324}{w} + w))}{dw}$ = 0

or

$2((-1)\frac{324}{w^2}+1)$ = 0

or

$(-1)\frac{324}{w^2}+1$ = 0

or

$(-1)\frac{324}{w^2}$ = -1

or

w² = 324

or

w = 18 units

For w = 18 units perimeter is minimum

therefore,

from 2

L = $\frac{324}{18}$

or

L = 18 units

objective function for P is:

P = 2(18 + w)

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