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3 years ago

Acil yaparmısınız çok acilllll

Mathematics

1 answer:

Firlakuza [10]3 years ago

5 0

Answer:

i don't understand dawg .

Step-by-step explanation:

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You and your friend are painting the walls in your apartment. You estimate that there is 1000 square feet of space to be painted

leva [86]

After 45 minutes you will have painted 4*45=180 square feet.

so the total number of painted area, y, is:

y=180+x*(3+4) after the 45 minutes

where x is the number of minutes AFTER THE 45 minutes:

if y is 1000 then

1000=180+x*7
820=x*7
x=117.14

so 117 minutes, or one hour and 57 minutes.

5 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

identify an equation in point slope form for the line perpendicular to y=3x+5 that passes through (4,-1)

Alborosie

Answer:

y-4 = (-1/3)(x+1)

Step-by-step explanation:

We need to identify the equation in point slope form.

The standard equation of point slope form is:

(y-y₁) = m (x-x₁)

where m is the slope and x₁ and y₁ are the points

We are given point(4,-1) so,

x₁=4 and y₁=-1

And a perpendicular line: y =3x+5

Which is equal to y = mx+b

where m is slope so, slope m = 3

Since the line is perpendicular, so the slope in negative inverse of actual slope that m = -1/m

i.e, m = -1/3

So, the equation in point slope form is:

y-(-1) = (-1/3)(x-4)

=> y+1 = (-1/3)(x-4)

6 0

3 years ago

Read 2 more answers

A bag contains 2 red beads, 1 yellow bead and 3 green beads. A bead is chosen at random.

Katena32 [7]

1/5

Step-by-step explanation:

there is only one yellow bead 2 red and 3 green you add the red and green with a total of 5 getting a ratio of 1/5

5 0

3 years ago

Read 2 more answers

a. Find the value of each variable to the nearest tenth. B. Find the perimeter of the triangle to the nearest tenth. 1.A = 30.0,

eimsori [14]

Answer:

a = 30.0, c = 36.6, perimeter = 87.6 - this is what I got for my answer so I hope this helps

7 0

3 years ago