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3 years ago

1. To paint 30% of a house takes 6 gallons of paint, how many gallons of paint are needed to paint the whole house?

Mathematics

1 answer:

vichka [17]3 years ago

6 0

Answer:

20 gallons of painta

Step-by-step explanation:

30%÷6=5

5% per gallon

100%÷5%=20

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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

How 8/33 into a decimal

Tanzania [10]

To write 8/33 as a decimal you have to divide numerator by the denominator of the fraction.
We divide now 8 by 33 what we write down as 8/33 and we get 0.24242424242424 

And finally we have: 
8/33 as a decimal equals 0.24242424242424

5 0

3 years ago

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What the answer to this all of those

zvonat [6]

So much stuff at once...
I'm only going to give you the information needed to do this.

All of these involve the 4 main exponent laws.

1) Multiplying.
When you multiply two 'numbers' that share the same base, you add exponents and keep the base the same. (ex. 4g²(4g⁴) = 4g²⁺⁴ = 4g⁶)

2) Dividing.
When you divide two 'numbers' that share the same base, you subtract exponents and keep the base the same. (ex. 3a⁷ / 3a⁴ = 3a⁷⁻³ = 3a³)

3) To the power of 1.
Anything to the power of 1 is the same. (ex. 5a¹ = 5a)

4) To the power of 0.
Anything to the power of 0 = 1. (ex. 2d⁰ = 2(1) = 2)

This should help.

8 0

3 years ago

A satellite is orbiting the Earth at the rate of 25 kilometers per minute. How far will it travel in two hours? a. 3000 kilomete

topjm [15]

3000 kilometers
hope this helps :)

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3 years ago

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F(1)=16 f(n)=(n-1)+2n what is f(4)

Wittaler [7]

F(4)=(4-1)+2*4=11...

This seems a bit too easy though, did you mean f(n)=f(n-1)+2n ? In this case :
f(1)=16
f(2)=16+2*2=20
f(3)=20+2*3=26
f(4)=26+2*4=34

5 0

3 years ago

Read 2 more answers