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2 years ago

Please help me so i dont fail

Mathematics

2 answers:

Nina [5.8K]2 years ago

7 0

I think it would the last one G

Advocard [28]2 years ago

4 0

It is the second one

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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

2 years ago

Need help ASAP with my math

yaroslaw [1]

U need a calculator for that lol

7 0

2 years ago

Read 2 more answers

Hello I have uploaded the question via image

maksim [4K]

Labour. They are man-made and used in agriculture.

3 0

2 years ago

A hog producer is feeding soybean meal and corn to his pigs. He needs to know how many pounds he needs to feed of each in order

Mazyrski [523]

Answer:

Feed 1, Soybean meal required = $\dfrac{2000}{3} \text{ lbs.}$

Feed 2, Corn meal required = $\dfrac{4000}{3} \text{ lbs.}$

Step-by-step explanation:

Total feed is 1 ton i.e. 2000 lbs.

Let x be the amount of Feed 1 required.

Feed 1 has $45\%$ of protein.

$\text{Protein in Feed 1 = }x \times \dfrac{45}{100} ..... (1)$

Then, amount of Feed 2 required = $(2000 - x) \text{ lbs.}$

Feed 2 has $15\%$ of protein.

$\text{Protein in Feed 2 = }(2000 - x) \times \dfrac{15}{100} ..... (2)$

As per question, total protein required is $25\%$ of 2000 lbs .

Adding (1) and (2) and putting it equal to total protein required.

$\Rightarrow x \times \dfrac{45}{100} + (2000 - x) \times \dfrac{15}{100} = 2000 \times \dfrac{25}{100}\\\Rightarrow 45x + 30000 - 15x = 50000\\\Rightarrow 30x = 20000\\\Rightarrow x = \dfrac{2000}{3}$

Feed 1 required = $\dfrac{2000}{3}\text{ lbs .}$

Feed 2 required = $2000 - \dfrac{2000}{3}\\$

$\Rightarrow \dfrac{4000}{3}\text{ lbs.}$

7 0

3 years ago

Write a problem that by comparing feet and inches using a model.include a solution. Explain why are you changing from a larger u

ziro4ka [17]

Change 4 feet to inches.

You need to know that for every feet, there are 12 inches. So, if you have 4 feet, you just multiply 4 by 12 and you will get 48 inches. Note that the feet unit is larger than inches.

8 0

2 years ago