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3 years ago

B = 57°, b = 26 cm. Find c in cm. Round to 2 decimal places.

Mathematics

1 answer:

baherus [9]3 years ago

3 0

Answer:

Step-by-step explanation:

Answer:

21.1cm

Step-by-step explanation:

From sine rule which can be expressed as

b/ sinB = c/sin C

B = 57°,

b = 26 cm.

FROM THE COMPLETE QUESTION,

C=43°

Then we can substitute this

26/sin(57) = c/sin(43)

If we cross multiply and make "c" subject of the formula

c=[26 × sin(43)]/ sin57

c= 21.1 cm

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Answer:

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Step-by-step explanation:

The given rational function is

$f(x) = \frac{x - 5}{ - 4x - 2}$

Let

$y = \frac{x - 5}{ - 4x - 2}$

Interchange x and y

$x= \frac{y- 5}{ - 4y - 2}$

Cross multiply

$x( - 4y - 2)= y- 5$

Expand

$- 4xy - 2x= y- 5 \\ - 4xy - y= 2x - 5 \\y( - 4x - 1)= 2x - 5$

Solve for y;

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Step-by-step explanation:

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3 years ago

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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

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

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

$\sec t=\dfrac{x+2}{\sqrt3}$
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$\tan t=\sqrt{\dfrac{x^2+4x+1}3}$
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All other options are false. They should have a less than sign instead of a greater than sign.

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AnnyKZ [126]

Answer:

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Step-by-step explanation:

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B = 57°, b = 26 cm. Find c in cm. Round to 2 decimal places.​

B = 57°, b = 26 cm. Find c in cm. Round to 2 decimal places.