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Hatshy [7]
3 years ago
14

Graph the points and state whether they are collinear.

Mathematics
1 answer:
aev [14]3 years ago
5 0
2) (-1,1), (2,-2), (4,-3) 
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У = 2х – 6<br>у = х2 – 4х + 2​
Dafna11 [192]
Your answer would be (x,y)=(2,-2)
4 0
3 years ago
Square ABCD is shown with four congruent images such that ABCD FGHI JKLM NOPQ RSTU.
Anettt [7]

Answer:

  • <em><u>RSTU</u></em>

Explanation:

The question states the <em>congruent</em> images:

  • ABCD ≅ FGHI ≅ JKLM ≅ NOPQ ≅ RSTU

The order of the letters matters.

For instance, ABCD ≅ FGHI means that the vertex A is transformed into vertex F, the vertex B is transformed into the vertex G, the vertex C is transformed into vertex H, and the vertex D is transformed into vertex I.

From the images, the square ABCD is just shifted 9 units down to form the square FGHI. No refelection is needed.

As per the square RSTU only a reflection of square ABCE accross the y-axis makes the vertex A into vertex R, the vertex B into the vertex S, the vertex C into the vertex T, and the vertex D into the vertex U.

3 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
–10p = –11p − 15<br> What is the answer
tensa zangetsu [6.8K]

Answer: p = −15

Step-by-step explanation:

Add 11p to both sides.

−10p+11p=−11p−15+11p

p=−15

4 0
3 years ago
Read 2 more answers
******PLEASE HELP!!******
expeople1 [14]

Answer:

9^{4}

Step-by-step explanation:

9^-5 = 9^4  x  9^-9

4-9 = -5

7 0
3 years ago
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