+0  
 
+5
856
15
avatar+1314 

what is the methods used to solve both equations:

StartFraction d y Over d t EndFraction equals 0.8left-parenthesis y minus 400right-parenthesis,  y equals 80 when  t equals 0, in the form y(t)=

and

(please solve the below question) 

StartFraction d w Over d theta EndFraction equals theta w squared s i n theta squared comma w left-parenthesis 0 right-parenthesis equals 12, in the form w(theta)= 

 

thanks.

 Sep 11, 2014

Best Answer 

 #7
avatar+109719 
+13

 

$$\\w=\frac{-1}{-1/2cos(\theta^2)+(7/16)}\\\\
=-1\div \left(\frac{-1}{2cos(\theta^2)}+\frac{7}{16}\right)\\\\
=-1\div \left(\frac{-8}{16cos(\theta^2)}+\frac{7cos(\theta^2)}{16cos(\theta^2)}\right)\\\\
=-1\div \left(\frac{7cos(\theta^2)-8}{16cos(\theta^2)}\right)\\\\
=-1\times \left(\frac{16cos(\theta^2)}{7cos(\theta^2)-8}\right)\\\\
=\frac{16cos(\theta^2)}{8-7cos(\theta^2)}\\\\$$

 

Is that simple enough?     

 Sep 11, 2014
 #1
avatar+109719 
0

Your pictures are not displaying for me.  

 Sep 11, 2014
 #2
avatar+30270 
0

I can't see the images either.

 Sep 11, 2014
 #3
avatar+1314 
0

Can you simplfy the following please.
$$w=\frac{-1}{-1/2cos(theta)^2+(7/16)}$$

.
 Sep 11, 2014
 #4
avatar+109719 
0

$$w=\frac{-1}{-1/2cos(theta)^2+(7/16)}$$

firstly, do you mean

 

 $$\\(cos\theta)^2\;\;or\;\;cos(\theta^2)\\\\
note:\;\;cos^2\theta \;\;means\;\; (cos\theta)^2$$

.
 Sep 11, 2014
 #5
avatar+1314 
0

sintheta ^2

 Sep 11, 2014
 #6
avatar+1314 
0

h

 Sep 11, 2014
 #7
avatar+109719 
+13
Best Answer

 

$$\\w=\frac{-1}{-1/2cos(\theta^2)+(7/16)}\\\\
=-1\div \left(\frac{-1}{2cos(\theta^2)}+\frac{7}{16}\right)\\\\
=-1\div \left(\frac{-8}{16cos(\theta^2)}+\frac{7cos(\theta^2)}{16cos(\theta^2)}\right)\\\\
=-1\div \left(\frac{7cos(\theta^2)-8}{16cos(\theta^2)}\right)\\\\
=-1\times \left(\frac{16cos(\theta^2)}{7cos(\theta^2)-8}\right)\\\\
=\frac{16cos(\theta^2)}{8-7cos(\theta^2)}\\\\$$

 

Is that simple enough?     

Melody Sep 11, 2014
 #8
avatar+25225 
+10

Can you simplfy the following please

$$w=\frac{-1}{-1/2cos(theta)^2+(7/16)}
=
\dfrac{-1}{-\dfrac{1}{2} \cos{
( \theta^2 )}
+\dfrac{7}{16}
}
=
\dfrac{-2}
{
\dfrac{7}{8}
-
\cos{
( \theta^2 )}
}
=
\dfrac{2}
{
\cos{
( \theta^2 )}
-
\dfrac{7}{8}
}$$

.
 Sep 11, 2014
 #9
avatar+109719 
+5

Heureka and I have interpreter it a little differently.

Heureka's answer is better.     

 Sep 11, 2014
 #10
avatar
0

OK

I tink I it! What about you Stu?

 Sep 11, 2014
 #11
avatar+1314 
+5

Will let you know if the online test accepts either answer  j have a feeling it will and that I am really weak working with fractions like this. Also, there is a problem with mobile view/normal view on this site showing a large portion of the pagebin white and unable to zoom in to the section that is condesed to the lhs.

 Sep 11, 2014
 #12
avatar+1314 
+5

Hey, neither answer works, nor the step before the solution with the theta ^2 in side or outside the brackets.

 Sep 22, 2014
 #13
avatar
0

Try monkeying around with it some more, Stu. You might get it. If not, there's still room in the bartending classes.

I'll help after I finish playing this game of chess. For my fee, fix me a drink. A banana daiquiri. May as well get started on your mixology skills.

 Sep 24, 2014
 #14
avatar
0

Thanks anon,

Sorry Stu but this is really hilarious. 

You really should feel honoured.  No stranger has EVER paid me this much attention.   I haven't laughed so hard since anon's last post.

ROF LOL          

 Sep 25, 2014
 #15
avatar+1314 
+5

solved. thanks for the assit.

 Sep 29, 2014

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