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luda_lava [24]
3 years ago
11

From a point on a circle, two perpendicular chords are drawn. One is 6 cm from the center and the other one is 3 cm from the cen

ter. Find the lengths of these chords.
Mathematics
2 answers:
sergeinik [125]3 years ago
8 0
The chord 3cm away is 12cm and chord 6cm away is 6cm. 
The two perpendicular chords will create a rectangle with the center. each side will be half of the parallel sides length.
Ira Lisetskai [31]3 years ago
4 0
The chord 3cm away is 12cm and chord 6cm away is 6cm. 
The two perpendicular chords will create a rectangle with the center. each side will be half of the parallel sides length. 
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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
HELPP ASaPP I’ll mark you as brainlister
nika2105 [10]

Answer:

BC ≈ 8.9 units

Step-by-step explanation:

Using the cosine ratio in the right triangle

cos36° = \frac{adjacent}{hypotenuse} = \frac{BC}{AB} = \frac{BC}{11} ( multiply both sides by 11 )

11 × cos36° = BC , then

BC ≈ 8.9 ( to the nearest tenth )

5 0
2 years ago
Read 2 more answers
How to find what numbers a number is divisible by
nevsk [136]
If the number ends with an even number or a number that is always divisible by 2. Or every number that ends with 5 or zero is always divisible by 5 too. Bdause for example in a fraction we do 10/5 and it will be equal to 2/1 or 2. Hope it helped you!
7 0
3 years ago
In a sheet metal operation, three identical notches and four identical bends are required. If the operations can be done in any
goldfiish [28.3K]

Answer:

<h2>There are 5040 different possible sequences.</h2>

Step-by-step explanation:

Three identical notches and four identical bends are required in the sheet metal operation.

In total 7 things are required in the metal operation.

We can think it as we need to put the 3 notches and 4 bends in 7 places.

First, lets put the 3 notches in 3 places.

In order to do so, we need to choose 3 places from the 7 places.

We can choose 3 places in \frac{7!}{3!\times4!} = \frac{5\times6\times7}{6} = 35 ways.

The 3 notches can be arrange in 3! = 6 ways.

The 4 bends can arrange in 4! = 24 ways.

Thus, in total 35\times24\times6 = 5040 different possible sequences.

3 0
3 years ago
PLEASE SOMEONE HELP
Yanka [14]
Infinity Many Solutions


- Harvard university professor
5 0
3 years ago
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