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3 years ago

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 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:

RSTU

Explanation:

The question states the congruent 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 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