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2 years ago

What's the inequality?

Mathematics

2 answers:

Ad libitum [116K]2 years ago

7 0

Answer:

I believe the answer is D coorrect me if im wrong

Step-by-step explanation:

if im right brainliest?

or 5 stars and heart

lesya692 [45]2 years ago

7 0

I’m pretty sure that the answer is D

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Help pls!! Find the area

Dima020 [189]

Answer:

17.5 for perpendicular triangle

Step-by-step explanation:

A= 1/2 ( b*h )

A= 1/2 ( 7*5)

A= 1/2 ( 35)

A= 17.5

8 0

1 year 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

2 years ago

a line with slope 3 passes through point (0,10). What is the y-coordinate of the point on the line with x-coordinate 2

skad [1K]

The y-coordinate is 16

<h3><u>Solution:</u></h3>

Given that a line with slope 3 passes through point (0, 10)

To find the y-coordinate of the point on the line with x-coordinate 2

Which means the point is (2, y)

Let us find the required y co-ordinate using slope formula

<em><u>The slope of line is given as:</u></em>

For a line containing points $(x_1 , y_1)$ and $(x_2 , y_2)$ is given as:

$\text {slope}=m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$

$\text {Here } x_{1}=0 ; y_{1}=10 ; x_{2}=2 ; y_{2}=y$

Given that slope "m" = 3

Substituting the values we get,

$\begin{array}{l}{3=\frac{y-10}{2-0}} \\\\ {3=\frac{y-10}{2}} \\\\ {6=y-10} \\\\ {y=10+6=16}\end{array}$

Thus the y-coordinate is 16

8 0

2 years ago

Text me imsooo bored and lonely

netineya [11]

Answer:

I got u homie wassup with you

8 0

2 years ago

Read 2 more answers

PLEASE HELP ME <br> I WILL MARK BRAINLIST!!!!

elixir [45]

Answer: A

Step-by-step explanation: The slope of perpendicular lines is opposite reciprocal. In this case the slope would be 4/3. Plug in for x and y, you get -7 for b.

y = mx +b
9 = 4/3 (12) + b
9 = 16 + b
-7 = b

Rewrite the equation --> y = 4/3x - 7

5 0

2 years ago