All categories

3 years ago

Will you guys help me please

Mathematics

2 answers:

pickupchik [31]3 years ago

8 0

1.2, 1.23, 2.31, 3.2

marusya05 [52]3 years ago

7 0

How you do this is you go from left to right with each one, and figure out which number is greater. For this particular problem the answer will be,

1.2 1.23 2.31 3.2

You might be interested in

How many inches are in 2miles?

noname [10]

Answer:

2 miles = 126720 (looked it up)

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

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 (13-2x)(13-2x) ?

Irina-Kira [14]

(13-2x)(13-2x)

(13+-2x)(13+-2x)

(13)(13)+(13)(-2x)+(-2x)(13)+(-2x)(-2x)

169-26x-26x+4x^2

4x^2-52x+169

I hope that's help !

3 0

3 years ago

Read 2 more answers

What is the completely factored form of x3-64?

nataly862011 [7]

a^3 - b^3 = (a - b) (a^2 + ab + b^2)
so
x^3 - 4^3 = (x - 4)(x^2 + 4x + 16)

hope it helps

7 0

3 years ago

Read 2 more answers

Please help ASAP Please help ASAP Please help ASAP Please help ASAP

Alex_Xolod [135]

The volume of the rectangular box is 12 cubic inches

<h3>Volume of a box</h3>

The formula for measuring the volume of a rectangular box is expressed as:

V = lwh

l is the length

w is the width

h is the height

Given the following

l = 2in

w = 3in

h = 2in

Substitute

V = 2* 3 * 2

V = 12 cubic inches

Hence the volume of the rectangular box is 12 cubic inches

Learn more on volume of. rectangular box here: brainly.com/question/8880159

#SPJ1

7 0

2 years ago