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3 years ago

What is the value of x?

Mathematics

2 answers:

Bingel [31]3 years ago

8 0

X is equall to 9. Very simple, just look carefully.

Liono4ka [1.6K]3 years ago

4 0

X = 9

we know this because the two lines on each side of the triangle indicates that they are congruent and equal the same length.

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2x – y = 6 5x + 10y = –10

Studentka2010 [4]

2x-y=6;5x+10y=-10. -y=-2x+6. -y/-1=-2x+6/-1. y=2x-6. 5x+10y=-10. 5x+10(2x-6)=-10. 25x-60=-10. 25x=50. 25x/25=50/25. y=2x-6. y=(2)(2)-6. y=-2. Therefore, x=2 and y=-2

8 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

What is the equation of a line that passes through the points (3, 6) and (8, 4)?

romanna [79]

9514 1404 393

Answer:

y = -2/5x +36/5 . . . . . slope-intercept form

2x +5y = 36 . . . . . . . . standard form

Step-by-step explanation:

You can use the 2-point form of the equation of a line to find it.

y = (y2 -y1)/(x2 -x1)(x -x1) +y1

y = (4 -6)/(8 -3)(x -3) +6

y = -2/5(x -3) +6

y = -2/5x +36/5 . . . . slope-intercept form

2x +5y = 36 . . . . . . . standard form

5 0

2 years ago

A serviceman with two children has monthly salary of Rs 22800. He has got tax free

Sonbull [250]

Answer:

answer below :) hope this helps!

Step-by-step explanation:

Amount invested in Civil investment fund= 10% of 96000 = $\frac{10}{100}$ x 96000 = 10 x 960= 9600

remaing taxable amount = =96000 -(75000+9600)

= 96000 - 84600

= Rs. 11400

Tax = 15% of 11400

= $\frac{15}{100}$ x 11400 = 15 x 114 = 1710

again, i hope this helps :)

4 0

3 years ago

Are the triangles similar? explain. please help

grin007 [14]

Answer:

The triangles are similar.

Step-by-step explanation:

The reason for this is because they both share two pairs of angles that are the same: 40 degrees and 60 degrees. If these are equal, then that must mean the third angle for both triangles is also the same, making both triangles equal in proportion. Therefore, they must be similar.

5 0

2 years ago

Read 2 more answers