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3 years ago

Which postulates are used to find the perimeter of a shape?

Mathematics

1 answer:

Illusion [34]3 years ago

5 0

Answer:

Hiiiiiiiiiiiii.... lol ooopppp

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Urgent please help!!

wolverine [178]

Answer:

i think it would be B because if AC= 80 and thats like the distane between then abc would be 80 if im thinking right

Step-by-step explanation:

but hey if im wrong ill have my friend punch me in the arm to make it even

4 0

2 years ago

Write an equation of the line that passes through (−5, −2)(−5, −2) and is parallel to y=23x+1y=23x+1.

Klio2033 [76]

Assuming you meant 2/3
y=2/3x+1 and (-5,-2)

paralell means same slope
y=mx+b
m=slope

y=2/3x+1
sllope=2/3

point slope
y-y1=m(x-x1)
for slope=m
a point is (x1,y1)

slope=2/3
point=(-5,-2)

y-(-2)=2/3(x-(-5))
y+2=2/3(x+5)
or
y=2/3x+16/3

7 0

3 years ago

Can anyone help me integrate :

worty [1.4K]

Rewrite the second factor in the numerator as

$2x^2+6x+1=2(x+2)^2-2(x+2)-3$

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

$\displaystyle\int\frac{(\sqrt3\sec t-2)(6\sec^2t-2\sqrt3\sec t-3)}{\sqrt{(\sqrt3\sec t)^2-3}}(\sqrt3\sec t)\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\sec^2t-1}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{\sqrt{\tan^2t}}\,\mathrm dt$
$=\displaystyle\int\frac{(6\sqrt3\sec^3t-18\sec^2t+\sqrt3\sec t+6)\sec t}{|\tan t|}\,\mathrm dt$

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

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

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

$=3\sqrt3\sec^2t+6\sqrt3\ln|\tan t|-18\sec t+18\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|-6\ln|\csc t+\cot t|+C$
$=3\sqrt3\sec^2t-18\sec t+6\sqrt3\ln|\tan t|+12\ln|\csc t+\cot t|-\sqrt3\ln|\csc2t+\cot2t|+C$

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

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

etc.

3 0

3 years ago

PLZZZ HELPPP Hey guys can you help I’m looking for this guy on here named zack we are talking and then my old account got delete

zlopas [31]

Step-by-step explanation: i will try t look for him but i have not seen him at all

8 0

3 years ago

Read 2 more answers

Nora and Owen do business as Profit & Property, a real estate investment partnership. In acting on the firm’s behalf in a de

il63 [147K]

Answer:

breach of the duty of loyalty

Step-by-step explanation:

The deal with Quaint Village Mall was not for her personally but for her firm so for her to make a secret profit and take advantage of such an opportunity is a breach of the duty of loyalty which she is liable of.

If an employee makes a decision for the corporation that profits both him and the corporation. The duty of loyalty is breached when the employee, his or her interest in front of that of the corporation.

4 0

3 years ago