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lyudmila [28]
3 years ago
15

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:

5 0
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

<u>1 |     1    -3   1</u>

   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!

5 0
3 years ago
Please help I will mark previous
yawa3891 [41]

Answer:

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Step-by-step explanation:

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
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