All Answers 358504 Answers

3x - 8x^2 =  0

3x            =  8x^2

3              =  8x

3/8           =  x

so that x = 3/8

              = 0.375//

Thank you very muhc Chi...and as i hadnt mentioned.....the first one to post on my laughs will get 5 points!and hurray your the winner!

Very cool picture..

hey....thats rude......i did such a good job on that answer and you came up with just " yeah , if you say so" 

next time i'll ignore this newspaper!

$$S=4\pi r^2=36\pi

r^2=9

r=3

V=4\pi r^2/3=36\pi cm^3$$

ummmm Melody...no thanks......i am happy with mine! :/

Thanks YehChi, I shall have to look at it when I am fresher.    

I just made it an entire fraction. 

Thanks Melody.

Definite yes to learning. I have no interest in just getting the answers right unless I understand how I got them and can get them again. It's sort of cheating otherwise.

I'm glad my answer was also correct but I still can't understand how their's was stating the HCF was 1/2 from 1/2 and 1/4 as obviously they both have 1/4 in them but only 1/2 has 1/2 so can't be common to both?

I'm completely home study relying entirely on publicly available resources on the net such as this so I'd really appreciate it if you could explain this to me as I have no tutor to ask.

Thanks

Graham mixes up some mint green paint for his bathroom. He mixes 2 1/5 litres of white paint with 3 1/3 of green paint and 1/2 litre of blue paint. How many litres of mint green paint had Graham made altogether?

$$\small{\text{
$
2\frac{1}{5}+3\frac{1}{3}+\frac{1}{2}
$
}}$\\$
\small{\text{$
=2+3+\frac{1}{5}+\frac{1}{3}+\frac{1}{2}$
}}$\\$
\small{\text{$
=5+\frac{1}{5}+\frac{1}{3}+\frac{1}{2}
$
}}$\\$
\small{\text{$
=5+\frac{1*3*2}{5*3*2}+\frac{1*5*2}{3*5*2}+\frac{1*5*3}{2*5*3}
$
}}$\\$
\small{\text{$
=5+\frac{6}{5*3*2}+\frac{10}{3*5*2}+\frac{15}{2*5*3}
$
}}$\\$
\small{\text{$
=5+\frac{6+10+15}{5*3*2}
$
}}$\\$
\small{\text{$
=5+\frac{30+1}{30}
$
}}$\\$
\small{\text{$
=5+\frac{30}{30}
+\frac{1}{30}
$
}}$\\$
\small{\text{$
=5+1+\frac{1}{30}
$
}}$\\$
\small{\text{$
=6+\frac{1}{30}
$
}}$\\$
\small{\text{$
\textcolor[rgb]{1,0,0}{=6\frac{1}{30}}
$
}}$$

Rosala this is an easier way

5 3/4 + 4 3/4

=5 +3/4 + 4+ 3/4

=5+4 +3/4 + 3/4

=9 +6/4 

=9+4/4+2/4

=9+1+1/2

=10 1/2

Melody,You are too complicated...

$$lny=ln(e+1)^3+\(\frac{1}{2}(x+1)-ln(x+2)-2ln(sin(3x+2))

\(\frac{y'}{y}=0+\(\frac{1}{2(x+1)}-\(\frac{1}{x+2}-\(\frac{2cos(3x+2)3}{sin(3x+2)}

y'=y[\(\frac{1}{2(x+1)}-\(\frac{1}{x+2}-\(\frac{6cos(3x+2)}{sin(3x+2)}]

y'=\(\frac{(e+1)^3\sqrt{x+1}}{(x+2)sin^2(3x+2)}[\(\frac{1}{2(x+1)}-\(\frac{1}{x+2}-6cot(3x+2)]$$

What is 36pi ?

3x-8x^2=0

x(3-8x)=0

x=0   or   3-8x=0

x=0   or   3=8x

x=0   or   x=3/8

You can simplify it but you cannot find x and y.

If you have 2 variables then you need 2 equations.  :)

2x+95/5=85y-854*9

2x+19=85y-7686

2x-85y+7705=0

It is the equation of a line so there are infinite solutions

Yea if you say so Rosala :)

Thanks Heureka,  I didn't know where to start with that second one.

I think you have done the first one different from me too - I shall have to take a look :)

I. $$f(x)=arctan(8x);f'(x)=?$$

$$\tan[\ \textcolor[rgb]{1,0,0}{\tan^{-1}(8x)} \ ] = 8x \quad | \quad \frac{\ d()}{dx} \quad \small{\text{ and }} \quad \boxed{ [ \ tan(x)\ ]' = 1+\tan^2(x) }\\\\(1+\tan^2(\ \textcolor[rgb]{1,0,0}{\tan^{-1}(8x)} \ ) )\times\left[\ \textcolor[rgb]{1,0,0}{\tan^{-1}(8x)} \ \right]' = 8 \\\\\left[ 1+(8x)^2} \right]\times\left[\textcolor[rgb]{1,0,0}{\tan^{-1}(8x)} \ \right]' = 8 \\\\\boxed{ \left[\ \textcolor[rgb]{1,0,0}{\tan^{-1}(8x)} \ \right]' = \frac{8} {1+(8x)^2} } }$$

II. $$f(x)=sin(x)^{cos(x)};f'(x)=?$$

$$y=\sin{(x)}^{ \cos{(x)} } \quad | \quad \ln() \\ \\
\ln{(y)} = \cos{(x)} *\ln{ (\sin{(x)} ) } \quad | \quad \frac{\ d()}{dx} \\ \\
\frac{y'}{y} = [\cos{(x)}]' * \ln{ (\sin{(x)} ) } + \cos{(x)} *[ \ln{ (\sin{(x)} ) } ]' \\\\
y' = y
\left(
[\cos{(x)}]' * \ln{ (\sin{(x)} ) } + \cos{(x)} *[ \ln{ (\sin{(x)} ) } ]'
\right) \\ \\
y' = \sin{(x)}^{ \cos{(x)} }
\left(
-\sin{(x)} * \ln{ (\sin{(x)} ) } + \cos{(x)} * \frac{ \cos{(x)} } {\sin{(x)} }
\right) \\ \\
y' = \sin{(x)}^{ \cos{(x)} }
\left(
-\sin{(x)} * \ln{ (\sin{(x)} ) } + \frac{ \cos^2{(x)} } {\sin{(x)} }
\right) \\ \\$$

You are probably encouraged to do this via formula but I can never remember the formula so I will show you the long way.

Besides I never do anything the short way - ask anyone       LOL

$$\\y=atan(8x)\\\\
8x=tany\\\\
x=\frac{tany}{8}\\\\
\frac{dx}{dy}=\frac{sec^2y}{8}\\\\
\frac{dx}{dy}=\frac{1}{8cos^2y}\\\\
\frac{dy}{dx}=8cos^2y\\\\$$

At this point I used this triangle

$$\\\frac{dy}{dx}=8[cosy]^2\\\\
\frac{dy}{dx}=8[\frac{1}{\sqrt{1+64x^2}}]^2\\\\
\frac{dy}{dx}=\frac{8}{1+64x^2}\\\\$$

See, these are mixed fractions and you need to solve them to improper fractions to get answer .......

so to change them to improper fractions we will multiply the denominator with the whole number then add the numberator to the product!

5 3/4 + 4 3/4

= 4*5+3/4 + 4*3+3/4

= 23/4 + 15/4

now just add them, tell me if you cant do it.......i'll help you!

Melody you should give some points to me too for my so good answer.....you broke the poor kids heart ....how bad......

okay I will give you a score this time to get you started.  

But these are snark questions - you are not meant to take them seriously so I don't usually give points for the answers.    

But just for you - Mind, you have not give me points fo my answer yet either.  

Edit - Now you have - thank you :)

Anyway here are you points 

and Welcome to the forum   

I thought he would give me Score..

I am going to use quotient rule.

$$\\u=(x+1)^{0.5}\\
u'=0.5*(x+1)^{-0.5}\\\\\\
v=(x+2)[sin(3x+2)]^2\\\\
v'=1*[sin(3x+2)]^2\;\;+\;\;2(sin(3x+2))*cos(3x+2)*3\\\\
v'=sin^2(3x+2)\;\;+\;\;6sin(3x+2)cos(3x+2)\\\\\\
\frac{dy}{dx}=(e+1)^3\frac{vu'+uv'}{v^2}\\\\
vu'=(x+2)sin^2(3x+2)*0.5*(x+1)^{-0.5}\\\\
vu'=\frac{0.5(x+2)sin^2(3x+2)}{(x+1)^{0.5}}\\\\\\
uv'=(x+1)^{0.5}*sin^2(3x+2)\;\;+\;\;6sin(3x+2)cos(3x+2)\\\\
uv'=\frac{(x+1)*sin^2(3x+2)\;\;+\;\;6sin(3x+2)cos(3x+2)(x+1)^{0.5}}{(x+1)^{0.5}}$$

$$\\vu'=\frac{0.5(x+2)sin^2(3x+2)}{(x+1)^{0.5}}\\\\\\
uv'=\frac{(x+1)*sin^2(3x+2)\;\;+\;\;6sin(3x+2)cos(3x+2)(x+1)^{0.5}}{(x+1)^{0.5}}\\\\
vu'+uv'=\frac{0.5(x+2)sin^2(3x+2)+(x+1)sin^2(3x+2)\;\;+\;\;6sin(3x+2)cos(3x+2)(x+1)^{0.5}}{(x+1)^{0.5}}$$

$$\\\frac{vu'+uv'}{v^2}=\frac{[0.5(x+2)sin^2(3x+2)]+[(x+1)sin^2(3x+2)]\;\;+\;\;[6sin(3x+2)cos(3x+2)(x+1)^{0.5}]}{(x+2)^2sin^4(3x+2)(x+1)^{0.5}}
\\\\\frac{vu'+uv'}{v^2}=\frac{[0.5(x+2)sin(3x+2)]+[(x+1)sin(3x+2)]\;\;+\;\;[6cos(3x+2)(x+1)^{0.5}]}{(x+2)^2sin^3(3x+2)(x+1)^{0.5}}$$

$$\\\frac{dy}{dx}=(e+1)^3\times \frac{[0.5(x+2)sin(3x+2)]+[(x+1)sin(3x+2)]\;\;+\;\;[6cos(3x+2)(x+1)^{0.5}]}{(x+2)^2sin^3(3x+2)(x+1)^{0.5}}\\\\$$

There is probably only about 1000 mistakes in there.

Let's see what Heureka found   ( I saw Heureka's pop-up)

$$\small{\text{
$
f(x)=\dfrac{ (e+1)^3\sqrt{x+1} } { (x+2)\sin^2{(3x+2)} } \\
\\
$
}}
\samll{\text{
$ \qquad \textcolor[rgb]{1,0,0}{f'(x) = ?} $
}}$\\\\$
\small{\text{
$
f(x)=\dfrac{ (e+1)^3\sqrt{x+1} } { (x+2)\sin^2{(3x+2)} } = (e+1)^3 \left(
\sqrt{x+1} *\frac{1}{x+2} * \frac{1}{\sin{(3x+2)} } * \frac{1}{\sin{(3x+2)} }
\right)
$
}}$\\\\$
\small{\text{
$
f'(x)=(e+1)^3\dfrac{\sqrt{x+1} } { (x+2)\sin^2{(3x+2)} } \left(
\dfrac { ( \sqrt{x+1} )' } { \sqrt{x+1} }
-\dfrac { ( x+2 )' } { x+2 }
-\dfrac { ( \sin{(3x+2)} )' } { \sin{(3x+2)} }
-\dfrac { ( \sin{(3x+2)} )' } { \sin{(3x+2)} }
\right)
$
}}$\\\\$
\small{\text{
$
f'(x)=(e+1)^3\dfrac{\sqrt{x+1} } { (x+2)\sin^2{(3x+2)} } \left(
\dfrac { 1 } { 2\sqrt{x+1}\sqrt{x+1} }
-\dfrac { 1 } { x+2 }
-\dfrac { 3\cos{(3x+2)} } { \sin{(3x+2)} }
-\dfrac { 3\cos{(3x+2)} } { \sin{(3x+2)} }
\right)
$
}}$\\\\$
\small{\text{
$
f'(x)=(e+1)^3\dfrac{\sqrt{x+1} } { (x+2)\sin^2{(3x+2)} } \left(
\dfrac { 1 } { 2\sqrt{x+1}\sqrt{x+1} }
-\dfrac { 1 } { x+2 }
-\dfrac { 2*3\cos{(3x+2)} } { \sin{(3x+2)} }
\right)
$
}}$\\\\$
\small{\text{
$
f'(x)=\dfrac{ (e+1)^3\sqrt{x+1} } { (x+2)\sin^2{(3x+2)} } \left(
\dfrac { 1 } { 2( x+1 ) }
-\dfrac { 1 } { x+2 }
-\dfrac { 6\cos{(3x+2)} } { \sin{(3x+2)} }
\right)
$
}}$\\\\$
\small{\text{
P.S.
$
(uv)' = uv\left( \frac{u'}{u} + \frac{v'}{v} \right)
$
and
$
(\frac{u}{v})' = \frac{u}{v}\left( \frac{u'}{u} - \frac{v'}{v} \right)
$
and
$
(\frac{u}{v*w})' = \frac{u}{v*w}\left( \frac{u'}{u} - \frac{v'}{v} - \frac{w'}{w} \right)
$
and
$
(\frac{u}{v*w*w})' = \frac{u}{v*w*w}\left( \frac{u'}{u} - \frac{v'}{v} - \frac{w'}{w} - \frac{w'}{w} \right)
$
}}$$