apply the loop rule to loops abgefa and cbgedc in figure | studysoup
?problem 23cq


apply the loop rule to loops abgefa and cbgedc in figure 21.47.
?solution 23cq


a) equation for loop abgefa is (35.5)i1 + (6..25 = 21 v
b) equation for loop cbgedc is ( 9.25i2 - ( 6.25i3 = +15 v
dear student

kindly refer to the link for the similar query :


https://www meritnation
com/ask-answer/question/q-23-and-24-q23-give-the-formula-of-each-of-the-follow/the-p-block-elements/11940305

regards The given case is SAS (two sides and included angle) so use the Law of Cosines:
  

\(\displaystyle{b}^{{2}}={a}^{{2}}+{c}^{{2}}-{2}{a}{c}{\cos{{B}}}\)
  

\(\displaystyle{b}^{{2}}={5}^{{2}}+{3}^{{2}}-{2}{\left({5}\right)}{\left({3}\right)}{\cos{{92}}}°\)
  

\(\displaystyle{b}^{{2}}={34}-{30}{\cos{{92}}}°\)
  

\(\displaystyle{b}=\sqrt{{{34}-{30}{\cos{{92}}}°}}\)
  

b ~ 5.92 The next shapes in the ‘growing’ pattern are as follows:


![](https://mathsgee.com/?qa=blob&qa_blobid=5951663371716016349) let a, b, c be the sides of any triangle abc. then by applying the sine rule, we get⇒c = k sin csimilarly, b = k sin band a = k sin ahere we will consider rhs, so we getrhs = (b2 – c2) sin asubstituting corresponding values in the above equation, we get⇒ = [( k sin b)2 - ( k sin c)2] sin a⇒ = k2(sin2 b – sin2 c )sin a……….(ii)but,substituting the above values in equation (ii), we get⇒ = k2(sin(b + c) sin(b - c)) sin abut a + b + c = π ⇒ b + c = π –a, so the above equation becomes,⇒ = k2(sin(π –a) sin(b - c))sin abut sin (π - θ) = sin θ⇒ = k2(sin(a) sin(b - c))sin arearranging the above equation we get⇒ = (k sin(a))( sin(b - c))(k sin a)from sine rule, a = k sin a, so the above equation becomes,⇒ = a2 sin(b - c) = rhshence proved Here's the diagram you requested. All the best!  

Greg  

![I need a serpentine belt routing diagram for a - c528ac5.gif](http://i.fixya.net/uploads/images/c528ac5.gif) solution: `abs(veca)=abs(vecb)=abs(vecc)=lamda`   

`veca*vecb=vecb*vecc*vecc*veca=0`   

`abs(veca+vecb+vecc)^2=(veca+vecb+vecc)*(veca+vecb+vecc)`   

`=abs(veca)^2+0+0+abs(vecb)^2+abs(vecc)^2`   

`=lambda^2+lambda^2+lambda^2`   

`=3lambda^2`   

`abs(veca+vecb+vecc)=sqrt3lambda`   

`costheta\_1=(veca*(veca+vecb+vecc))/(abs(veca)abs(veca+vecb+vecc))=lambda^2/(lambdasqrt3lambda)=1/sqrt3=theta\_1=cos^(-1)(1/sqrt3)`   

similarly,   

`theta\_1=theta\_2=theta\_3=cos^(-1)(1/sqrt3)` given: ab=acto prove: ad>acproof:in ∆abc,∠acd = ∠b + ∠bac= ∠acb + ∠bac …as ∠c = ∠b as ab = ac= ∠cad + ∠cda +∠bac …as ∠acb = ∠cad + ∠cda∴∠acd > ∠cdaso the side opposite to ∠acd is the longest∴ad > ac yes for current region you can use a range 
  
something like this 
  

  

  

```
With Worksheets("sheet1")
.range("A1:e10").copy
```

  

  
the dot before "range" is important.
  

  
it will copy only from A1 to E10. modify to suit you given : five defective bulbs are accidently mixed with 20 good ones.to find : probability distribution from this lotformula used :five defective bulbs are accidently mixed with 20 good ones.total number of bulbs = 25x denote the number of defective bulbs drawnx can draw the value 0 , 1 , 2 , 3 , 4.since the number of bulbs drawn is 4, n = 4p(0) = p(getting a no defective bulb) = = = p(1) = p(getting 1 defective bulb and 3 good ones) = = p(1) = = p(2) = p(getting 2 defective bulbs and 2 good one) = p(2) = = = p(3) = p(getting 3 defective bulbs and 1 good one) = p(3) = = p(4) = p(getting all defective bulbs) = = = p(4) = the probability distribution table is as follows, to find: limitsnote: first check the form of imit. used this method if the limit is satisfying any one from 7 indeterminate form.in this case, indeterminate form is formula used: = 1so, by using the above formula, we have = = therefore, =