in fig. 8.141, if l || m, then x =
given that,l ‖ m and n cuts themlet,∠1 = 65o∠2 = x∠3 = 40o∠1 = ∠4 = 65o (alternate angle) (i)∠3 + ∠4 + ∠5 = 180o (angle sum property)40o + 65o + ∠5 = 180o∠5 = 75onow,∠2 + ∠5 = 180o (linear pair)x + 75o = 180ox = 105o
given to prove 

then taking l.h.s.

we have 

                    





                                  

 

  

=  r.h.s.

hence proved. if the figure below represents a parabola, identity the physical&nbsp;quantities Let $y=f(x)$ be a function. Suppose that $a$ and $\mathbf{L}$ are numbers such that:


 


* whenever $\boldsymbol{x}$ is close to $\boldsymbol{a}$ but not equal to $\boldsymbol{a}, \boldsymbol{f}(\boldsymbol{x})$ is close to $\mathbf{L} ;$
* as $\boldsymbol{x}$ gets closer and closer to $\boldsymbol{a}$ but not equal to $\boldsymbol{a}, \boldsymbol{f}(\boldsymbol{x})$ gets closer and closer to $\mathbf{L}$; and
* suppose that $f(x)$ can be made as close as we want to $\mathbf{L}$ by making $x$ close to $a$ but not equal to $a$.


Then we say that the limit of $f(x)$ as $x$ approaches a is $L$ and we write $\lim \_{x \rightarrow a} f(x)=\mathbf{L}$ Definition used:  

Congruence.  

Let m be an integer greater than 1. If xand yare integers, then x is congruent to y modulo m if x — y is divisible by m. It can be represented as \(\displaystyle{x}={y}{\left({b}\ mod\ {m}\right)}\). This relation is called as congruence modulo m.  

Calculation:   

Consider the values and congruence classes modulo m,  

\(\displaystyle{\left[{13}\right]}^{{{6}}}\) and m = 15.  

It is evident that [x]=[y] if \(\displaystyle{\left[{x}={y}{\left({b}\ mod\ {m}\right)}\right]}\). Since \(\displaystyle{13}=-{2}{\left({b} \ mod \ {15}\right)}\).  

Then,  

[13]=[-2]  

\(\displaystyle{\left[{13}\right]}^{{{6}}}={\left[-{2}\right]}^{{{6}}}\)  

\(\displaystyle={\left[{\left(-{2}\right)}^{{{6}}}\right]}\)  

=[64]  

Since, by the principal or congruence \(\displaystyle{64}={4}{\left({b}\ mod \ {15}\right)}\). Therefore, \(\displaystyle{\left[{13}\right]}^{{{6}}}={\left[{4}\right]}\) in \(\displaystyle{Z}\_{{{15}}}\).  

Therefore, the value of the expression is [4]. answer: `1.8706 xx 10^(11) " erg", 4477 cal` solution: ` x = {:((-4/5,-1),(2/5,1)):}` using the long division method: >> [x,y] = ismember(a,b);
>> [~,z] = sort(y(x));
>> v = find(x);
>> r = reshape(v(z),2,[]).'
r =
 2 9
 4 8
 5 7 fill in the blank (xiv). a - ii, b - iv, c - i, d - iii