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3 years ago

HELP!!! Name the space figure you can form from the net.

Mathematics

1 answer:

schepotkina [342]3 years ago

3 0

I believe it is a triangular prism

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The range for the set of data represented by the following box-and-whisker plot is 8.

DIA [1.3K]

Answer:

True

Step-by-step explanation:

The range is the difference between the lowest and the highest values of the set:

19 - 11 = 8

The answer is TRUE

3 0

2 years ago

Read 2 more answers

Cameron and Cadena go out to eat for lunch. Part A If their food and beverages cost $22.50 and there is a 9% meals tax, how much

Elan Coil [88]

Answer:

24.53

Step-by-step explanation:

9% of 22.50 = 2.03

22.50 + 2.03 = 24.53

4 0

3 years ago

Read 2 more answers

A raft and a motorboat simultaneously leave from the same point and move down the river. After 2 hours the motorboat turns aroun

lesya692 [45]

Answer:

2 hours

Step-by-step explanation:

Let x mph be the speed of motorboat in still water and y mph be the speed of current.

A raft and a motorboat simultaneously leave from the same point and move down the river, then speeds down the river are:

motorboat = x + y mph

raft = y mph

In two hours they will cover:

motorboat = 2(x+y) miles

raft = 2y miles

After 2 hours the motorboat turns around and starts moving towards the raft (against the current), so its speed is now x - y mph.

Let t be the time they move towards each other. In t hours, they cover

motorboat = t(x - y) miles

raft = ty miles

The distance from the starting point to the turning point is the same as the sum of the distances the raft covers and the motorboat covers after turning around, so

$2(x+y)=2y+t(x-y)+ty \\ \\2x+2y=2y+tx-ty+ty\\ \\2x=tx\\ \\t=2\ hours$

6 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 slip of the line shown below ? A - 11/4 B- 4/11 C- -4/11 D- 11/4

RUDIKE [14]

The picture is blurry

5 0

3 years ago