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3 years ago

Understanding Dilations and Similarity continued

Mathematics

1 answer:

Bogdan [553]3 years ago

3 0

Answer

Dilations is a transformation that produces an image that is the same shape as A description of a dilation includes the scale factor (or ratio) and the center of the Most dilations in the coordinate plane use the origin, (0,0), as the center of the. the center of the dilation at point A to the other points B, C and D. The dilation

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4.92 divided by 2.28 give detailed response

e-lub [12.9K]

Answer:

2.1578

Step-by-step explanation:

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3 years ago

Divide the following polynomial using synthetic division, then place the answer in the proper location on the grid. Write answer

Alenkinab [10]

Answer:

The quotient is $x^2-3x+1$

Step-by-step explanation:

The given polynomial division is $(x^3-4x^2+4x-1)\div (x-1)$

To perform the synthetic division we write out the coefficients and arrange them as shown.

1 -4 4 -1

1 -3 1 0

The quotient is $x^2-3x+1$

The descending powers of x means from highest degree to the least

The remainder is 0

4 0

3 years ago

Read 2 more answers

Mangoes cost $4 each and nectarines cost $2 each. Dominic buys m mangoes and n nectarines for $32.

xenn [34]

Answer: 6m+4n=32

Step-by-step explanation:

m=4

n=2

From this, we can deduce that m+n=32.

6*4=24 and 4*2=8. 24+8=32

So, 6m+4n=32. Hope this was what you were looking for!

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3 years ago

Please help I will mark previous

yawa3891 [41]

Answer:

Im sorry! idk the answer but if you have a phone, there is this app called photo math, or ask you mom or dad for help.

Step-by-step explanation:

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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