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tangare [24]
3 years ago
6

If A ⊂ B, then A ∩ B = ∅.

Mathematics
1 answer:
S_A_V [24]3 years ago
3 0

Answer:i don't know the answer

Step-by-step explanation:

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Find, correct to four decimal places, the length of the curve of intersection of the cylinder 4x2 1 y2 − 4 and the plane x 1 y 1
charle [14.2K]

<u>Answer-</u> Length of the curve of intersection is 13.5191 sq.units

<u>Solution-</u>

As the equation of the cylinder is in rectangular for, so we have to convert it into parametric form with

x = cos t, y = 2 sin t   (∵ 4x² + y² = 4 ⇒ 4cos²t + 4sin²t = 4, then it will satisfy the equation)

Then, substituting these values in the plane equation to get the z parameter,

cos t + 2sin t + z = 2

⇒ z = 2 - cos t - 2sin t

∴ \frac{dx}{dt} = -\sin t

  \frac{dy}{dt} = 2 \cos t

  \frac{dz}{dt} = \sin t-2cos t

As it is a full revolution around the original cylinder is from 0 to 2π, so we have to integrate from 0 to 2π

∴ Arc length

= \int_{0}^{2\pi}\sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}+(\frac{dz}{dt})^{2}

=\int_{0}^{2\pi}\sqrt{(-\sin t)^{2}+(2\cos t)^{2}+(\sin t-2\cos t)^{2}

=\int_{0}^{2\pi}\sqrt{(2\sin t)^{2}+(8\cos t)^{2}-(4\sin t\cos t)

Now evaluating the integral using calculator,

=\int_{0}^{2\pi}\sqrt{(2\sin t)^{2}+(8\cos t)^{2}-(4\sin t\cos t) = 13.5191




8 0
3 years ago
How do you simplify 12+3(8+x)
nadya68 [22]
<span>12+3(8+x)
= 12 + 24 + 3x
= 3x + 36</span>
3 0
3 years ago
Read 2 more answers
£750 is divided between Bridget, Caroline &amp; Sarah so that Bridget gets twice as much as Caroline, and Caroline gets three ti
Sladkaya [172]

Answer: £225

Step-by-step explanation:

Let Sarah's amount be represented by x.

Since Caroline gets three times as much as Sarah, Caroline will get: 3x

Bridget gets twice as much as Caroline, therefore Bridget will get: 2 × 3x = 6x

Sarah = x

Caroline = 3x

Bridget = 6x

Total = x + 3x + 6x = 10x

Caroline's faction is 3/10. We then multiply the fraction by £750. This will be:

= 3/10 × £750

= 0.3 × £750

= £225

Caroline will get £225

4 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
Find the value of c. 16 = 4 - 3c
dybincka [34]

Answer:

c = -4

Step-by-step explanation:

16 = 4 - 3c

subtract 4 from each side

16-4 = 4-4 - 3c

12 = -3c

Divide each side by -3

12/-3 = -3c/-3

-4 =c

6 0
3 years ago
Read 2 more answers
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