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2 years ago

The measures of the angles of a triangle are 25 degrees, 87 degrees, and 68 degrees. Is the triangle acute, right, or obtuse?

Mathematics

1 answer:

Goryan [66]2 years ago

5 0

Answer:

their all less than 90 degress so its acute

Step-by-step explanation:

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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

What is 36:39 in simplest form

Vanyuwa [196]

36:39 in simplest form is 12:13

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2 years ago

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Twenty middle school students are outside a movie theater waiting for it to open. There are four more girls than

Lapatulllka [165]

Answer:

0.4 , 0.1

Step-by-step explanation:

Let number of boys be = b , number of girls = b + 4 . As total students = 20 b + b + 4 = 20 → 2b + 4 = 20 → 2b = 20 - 4 = 16 → b = 16 / 2 = 8 (boys) , so girls = 12

Prob (a random first child is a boy) = boys / total students = 8 c 1 / 20 c1, or = 8 / 20 = 0.4

Prob ( a 'Tina' out of two Tina is the last) = 2 c 1 / 20 c 1 , or = 2 / 20 = 0.1

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2 years ago

There are 8employees on The Game Shop's sales team. Last month, they sold a total of g games. One of the sales team members, Chr

Anuta_ua [19.1K]

Answer:

(g/8) - 17

Step-by-step explanation:

Given that:

Number of employees = 8

Employees sold a total of g games

Chris sale = 17 fewer than team Average

Number of games Chris sold :

Team average :

Total sales made / number of employees

Team average = g/ 8

Number of games sold by Chris:

(g/8) - 17

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2 years ago

Divide to find the quotient. Please help ! Just tell me what write in boxes

vichka [17]

Answer: 8 r 2

Step-by-step explanation:

The closest thing that goes into 26 is 24, which goes into 3 8 times. Subtract 24 from 26, and you are left with 2. The 2 you are left with will be your remainder.

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3 years ago

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