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3 years ago

2*10^3 in scientific notation

Mathematics

1 answer:

Natali5045456 [20]3 years ago

7 0

2000, you just need to put three zeros at the end of the number.

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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

3 years ago

Choose the slope Y intercept that corresponds with the graph

Jet001 [13]

Answer:

Step-by-step explanation:

First, find the y-intercept by seeing where the line goes through the y-axis

This is at (0, -2) so the y-intercept is -2.

Then, use rise over run to find the slope.

The slope is -3

8 0

3 years ago

Read 2 more answers

Mr. Toshiro manages a company that supplies a variety of domestic and imported nuts to supermarkets. He received an order for 12

alexandr1967 [171]

Answer:

n = 120i + 310j + 60k

p = 29i + 18j + 21 k

Total cost = $10,320

Step-by-step explanation:

Let n and p represent the vectors for number of products and prices respectively.

Also, the coordinates i,j and k represent cashews, walnut and Brazil nut respectively.

The vector form of the total number of bags ordered and the cost are;

n = 120i + 310j + 60k

p = 29i + 18j + 21 k

We can obtain the total cost by obtaining the dot product of the two vectors.

Total cost = n.p = (120i + 310j + 60k).(29i + 18j + 21 k)

C = 120×29 + 310×18 + 60×21

C = $10,320

7 0

3 years ago

Please help fast ASAP

ololo11 [35]

A is the answer because is quadrilateral is a four sided shape that can be congruent and a parallelogram is a four-sided plane rectilinear figure with opposite sides parallel.

6 0

3 years ago

F(x)=3x^2-4 and g(x)=x+2 find f-g (x)

ratelena [41]

Answer:

Step-by-step explanation:

Here we subtract g(x)=x+2 from f(x)=3x^2-4

Properly labeled, we get:

(f - g)(x) = 3x^2-4 - (x+2) (Use of parentheses is essential here.)

(f - g)(x) = 3x^2 - x - 2

3 0

3 years ago