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2 years ago

I posted one question at parts pls help me answer it!!!!!

Mathematics

2 answers:

Fantom [35]2 years ago

5 0

Answer:

i'll try my best to help !!

DerKrebs [107]2 years ago

3 0

I’ll look for it and try to answer it :)

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31 and 81 are in the bottom corners of the triangle X is at the top . What is X?

kati45 [8]

Answer:

The top angle of the triangle is, X = 68°.

Step-by-step explanation:

Let there is a triangle named PQR with bottom corner angles ∠ P and ∠ Q and the ∠ R is the angle at the top.

Now, given that ∠ P = 31° and ∠ Q = 81° and we have to find out ∠ R i.e. X.

In Δ PQR, we know, ∠ P + ∠ Q + ∠ R = 180°

⇒ 31° + 81° + X = 180°

⇒ X = 68°

Therefore, the top angle of the triangle is 68°. (Answer)

7 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

75383 divided by 14 with a remainder

Lisa [10]

Answer:

5384.5

Step-by-step explanation:

Ez please brainllest

7 0

2 years ago

Read 2 more answers

Five times the sum of a number and 6 equals 3

Sever21 [200]

Answer:

5x+6=3

Step-by-step explanation:

"A number" is a variable.

"Sum" is addition.

"Equals 3" is =3.

"and six" relates to "sum".

So when you put these together it would be 5x+6=3

3 0

3 years ago

Read 2 more answers

2. A group of activists wants to raise money for a good cause, so they decide to sponsor a bicycle race. It receives $60 per cyc

Scorpion4ik [409]

Answer

Step-by-step explanation:

The number of cyclists the group needs to raise at least $57,000 must be at least 1000 cyclists
Let x represent the number of cyclists.
Since, the activists receive $60 per cyclist entry and $15,000 in donations, hence:
Revenue = 60x + 15000
Also, the cost of the race is $18 per entry, hence:
Cost = 18x
They need to raise at least $57,000, hence:
Profit ≥ 57000
Revenue - Cost ≥ 57000
(60x + 15000) - 18x ≥ 57000
42x ≥ 42000
x ≥ 1000
Therefore the number of cyclists the group needs to raise at least $57,000 must be at least 1000 cyclists

5 0

2 years ago