question
write a pair of intergers whose sum gives
write a pair of intergers whose sum gives
2, 0 ( any two integers with one of them as 0). multiply 2 in both lhs and rhs,integrating on both the sides, hint: first of all, multiply and divide 9 in both the series after taking common 5 and 0.6 from them respectively. then make a g.p of the form \[10,{{10}^{2}},{{10}^{3}}....\] by writing 9 = 10 – 1, 99 = 100 – 1, and so on. finally, use the formula for the sum of n terms of g.p that is \[\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\].complete step-by-step solution -in this question, we have to find the sum of the n terms of the series.(i) 5 + 55 + 555 + ….(ii) .6 + .66 + .666 + ….let us consider the first seriess = 5 + 55 + 555 + ….. n termsby taking out 5 common from the above series, we get, s = 5 (1 + 11 + 111 + 1111 …. n terms)by multiplying 9 on both the sides of the above equation, we get, 9s = 5 [9 + 99 + 999 …… n times]\[\dfrac{9s}{5}=9+99+999.....\text{ n times}\]we can also write the above equation as, \[\dfrac{9s}{5}=\left( 10-1 \right)+\left( 100-1 \right)+\left( 1000-1 \right)....\text{ n times}\]by simplifying the above equation and taking the term 1 separately, we get, \[\dfrac{9s}{5}=\left( {{10}^{1}}+{{10}^{2}}+{{10}^{3}}.....\text{ n times} \right)-\left( 1+1+1......\text{ n times} \right)\]we know that 1 added n times is equal to n and the nth term of 1, 2, 3…. n. so, we get, \[\dfrac{9s}{5}=\left( {{10}^{1}}+{{10}^{2}}+{{10}^{3}}{{....10}^{n}} \right)-\left( n \right)\]we know that the series of the form \[a,ar,a{{r}^{2}},a{{r}^{3}}.....a{{r}^{n}}\] is in g.p and sum of the n terms of this g.p is given by \[\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\]in the above equation, we can see that \[10,{{10}^{2}},{{10}^{3}}....\] are in g.p with the first term and common ratio as 10. so, we get, \[\dfrac{9s}{5}=\dfrac{10\left( 1-{{\left( 10 \right)}^{n}} \right)}{1-10}-n\]\[\dfrac{9s}{5}=\dfrac{10\left( 1-{{\left( 10 \right)}^{n}} \right)}{-9}-n\]\[\dfrac{9s}{5}=\dfrac{10}{9}\left( {{10}^{n}}-1 \right)-n\]\[\dfrac{9s}{5}=\dfrac{10\left( {{10}^{n}}-1 \right)-9n}{9}\]by multiplying \[\dfrac{5}{9}\] on both the sides of the above equation, we get, \[s=\dfrac{5}{9}\left[ \dfrac{10\left( {{10}^{n}}-1 \right)-9n}{9} \right]\]\[s=\dfrac{50\left( {{10}^{n}}-1 \right)}{81}-\dfrac{5}{9}n\]hence, the sum of the sequence 5 + 55 + 555….. up to n terms is equal to \[\dfrac{50\left( {{10}^{n}}-1 \right)}{81}-\dfrac{5n}{9}\]let us consider the second set.s = 0.6 + 0.66 + 0.666 + ….. n times.by taking out 0.6 common, we can write the sequence as, s = 6 [0.1 + 0.11 + 0.111 ….. n terms]by multiplying and dividing rhs by 9, we get,\[s=\dfrac{6}{9}\left[ 0.9+0.99+0.999.....n\text{ terms} \right]\]\[s=\dfrac{2}{3}\left[ \dfrac{9}{10}+\dfrac{99}{100}+\dfrac{999}{1000}+\dfrac{9999}{10000}.....n\text{ terms} \right]\]we can write the above equation as, \[s=\dfrac{2}{3}\left[ \dfrac{\left( 10-1 \right)}{10}+\dfrac{\left( 100-1 \right)}{100}+\dfrac{\left( 1000-1 \right)}{1000}+\dfrac{\left( 10000-1 \right)}{10000}.....n\text{ terms} \right]\]\[s=\dfrac{2}{3}\left[ \left( 1-\dfrac{1}{10} \right)+\left( 1-\dfrac{1}{100} \right)+\left( 1-\dfrac{1}{1000} \right)+\left( 1-\dfrac{1}{10000} \right).....n\text{ terms} \right]\]\[s=\dfrac{2}{3}\left[ \left( 1+1+1......n\text{ terms} \right)-\left( \dfrac{1}{10}+\dfrac{1}{100}+\dfrac{1}{1000}.+...n\text{ terms} \right) \right]\]here, we can see that \[\dfrac{1}{10}+\dfrac{1}{{{\left( 10 \right)}^{2}}}+\dfrac{1}{{{\left( 10 \right)}^{3}}}+.....n\text{ terms}\] is in gp with \[a=\dfrac{1}{10}\] and \[r=\dfrac{1}{10}\].we know that the sum of n terms of the g.p is \[\dfrac{a\left( 1-{{r}^{n}} \right)}{\left( 1-r \right)}\]. so, we get, \[s=\dfrac{2}{3}\left[ \left( n \right)-\dfrac{\dfrac{1}{10}\left( 1-{{\left( \dfrac{1}{10} \right)}^{n}} \right)}{1-\dfrac{1}{10}} \right]\]\[s=\dfrac{2}{3}\left[ n-\dfrac{\dfrac{1}{10}\left( 1-\dfrac{1}{{{10}^{n}}} \right)}{\dfrac{9}{10}} \right]\]\[s=\dfrac{2}{3}\left[ n-\dfrac{1}{9}\left( 1-\dfrac{1}{{{10}^{n}}} \right) \right]\]we know that \[\dfrac{1}{{{a}^{n}}}={{a}^{-n}}\]. by using this, we get, \[s=\dfrac{2}{3}\left( n-\dfrac{1}{9}\left( 1-{{10}^{-n}} \right) \right)\]so, \[s=\dfrac{2}{3}n-\dfrac{2}{27}\left[ 1-{{\left( 10 \right)}^{-n}} \right]\]hence, we get the sum of the sequence 0.6 + 0.66 + 0.666 + ….. n terms as \[\dfrac{2}{3}n-\dfrac{2}{27}\left[ 1-{{\left( 10 \right)}^{-n}} \right]\].note: in this question, students must note that whenever we are in need to find the sum of series of the type a + aa + aaa + aaaa ….., we always have to make the series of 9 + 99 + 999 + ….. type by multiplying and dividing the whole series by 9 so that we can get a g.p of the type \[10,{{10}^{2}},{{10}^{3}}\]…. also in this question, some students make this mistake of considering 5 + 55 + 555 + …. or 0.6 + 0.66 + 0.666 + …. in g.p which is wrong because we can see that there is no common ratio in these sequences. let i = `(1)/(sinx*(3 + 2cosx))*dx` = `int sinx/(sin^2x*(3 + 2cosx))*dx` = `int sinx/((1 - cos^2x)(3 + 2cosx))*dx` = `int sinx/((1 - cosx)(1 + cosx)(3 + 2cosx))*dx` put cos x = t∴ – sinx.dx = dt∴ sinx.dx = – dt ∴ i = `int (1)/((1 - t)(1 + t)(3 + 2t))*(-dt)` = `int (-1)/((1 - t)(1 + t)(3 + 2t))*dt` let `(-1)/((1 - t)(1 + t)(3 + 2t)) = "a"/(1 - t) + "b"/(1 + t) + "c"/(3 + 2t)` ∴ – 1 = a(1 + t)(3 + 2t) + b(1 - t)(3 + 2t) + c(1 - t)(1 + t)put 1 – t = 0, i.e. t = 1, we get– 1 = a(2)(5) + b(0)(5) + c(0)(2) ∴ – 1 = 10a ∴ a = `(-1)/(10)`put 1 + t = 0, i.e. t = – 1, we get– 1 = a(0)(1) + b(2)(1) + c(2)(0) ∴ – 1 = 2b ∴ b = `-(1)/(2)` put 3 + 2t = 0, i.e. t = `-(3)/(2)`, we get – 1 = `"a"(-1/2)(0) + "b"(5/2)(0) + "c"(5/2)(-1/2)` ∴ – 1 = `-(5)/(4)"c"` ∴ c = `(4)/(5)` ∴ `(-1)/((1 - t)(1 + t)(3 + 2t)) = (((-1)/(10)))/(1 - t) + ((-1/2))/(1 + t) + ((4/5))/(3 + 2t)` ∴ i = `int [(((-1)/10))/(1 - t) + ((-1/2))/(1 + t) + ((4/5))/(3 + 2t)]*dt` = `-(1)/(10) int 1/(1 - t)*dt - (1)/(2) int 1/(1 + t)*dt + (4)/(5) int 1/(3 + 2t)*dt` = `-(1)/(10) (log|1 - t|)/(-1) - (1)/(2) log | 1 + t| + 4/5 (log|3 + 2t|)/(2) + c` = `(1)/(10)log|1 - cosx| - (1)/(2)log|1 + cosx| + (2)/(5)log|3 + 2cos| + c`. answer: k=7 Assuming you're joining on columns with no duplicates, which is a very common case:

* An inner join of A and B gives the result of A intersect B, i.e. the inner part of a [Venn diagram](http://en.wikipedia.org/wiki/Venn_diagram) intersection.
* An outer join of A and B gives the results of A union B, i.e. the outer parts of a Venn diagram union.

**Examples**

Suppose you have two tables, with a single column each, and data as follows:


```
A    B
-    -
1    3
2    4
3    5
4    6

```
Note that (1,2) are unique to A, (3,4) are common, and (5,6) are unique to B.

**Inner join**

An inner join using either of the equivalent queries gives the intersection of the two tables, i.e. the two rows they have in common.


```
select * from a INNER JOIN b on a.a = b.b;
select a.*, b.*  from a,b where a.a = b.b;

a | b
--+--
3 | 3
4 | 4

```
**Left outer join**

A left outer join will give all rows in A, plus any common rows in B.


```
select * from a LEFT OUTER JOIN b on a.a = b.b;
select a.*, b.*  from a,b where a.a = b.b(+);

a |  b
--+
1 | null
2 | null
3 |    3
4 |    4

```
**Right outer join**

A right outer join will give all rows in B, plus any common rows in A.


```
select * from a RIGHT OUTER JOIN b on a.a = b.b;
select a.*, b.*  from a,b where a.a(+) = b.b;

a    |  b
+
3    |  3
4    |  4
null |  5
null |  6

```
**Full outer join**

A full outer join will give you the union of A and B, i.e. all the rows in A and all the rows in B. If something in A doesn't have a corresponding datum in B, then the B portion is null, and vice versa.


```
select * from a FULL OUTER JOIN b on a.a = b.b;

 a   |  b
+
   1 | null
   2 | null
   3 |    3
   4 |    4
null |    6
null |    5

``` The problem is that your pair's closure must return two different types of objects: int if you ask for "con", and the next closure on the line if you ask for "crd".


Here's one possible implementation:


Use a speaker so you can come back.



```
public static Func<string, dynamic> Pair(int x, Func<string, dynamic> y)
{
    Func<string, dynamic> pair =
           (a) =>
           {
               if (a == "con") return x;
               if (a == "crd") return y;
               throw new Exception();
           };

    return pair;
}

public static void Print(Func<string, dynamic> pair)
{
    while (true)
    {
        var next = pair("crd");
        Console.WriteLine((next == null ? "-":"+") +"> " + pair("con"));
        if (next != null)
        {
            Console.WriteLine("|");
            pair = (x) => next(x);
            continue;
        }
        break;
    }
}

        

        
        
        
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```


  




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Test:



```
var pair = Pair(1,
            Pair(2,
                  Pair(3,
                        Pair(4, null))));

Print(pair);

        

        
        
        
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```

output:



```
+> 1
|
+> 2
|
+> 3
|
-> 4

        

        
        
        
 (adsbygoogle = window.adsbygoogle || []).push({});
 
        

    
``` 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 = let, logx = z⇒ so,where c is the integrating constant. 1/
a/ the x-intersect between 2 graphs , x^2+4x-5=0>x = -5 or 1
v = pi { integral of [(4x)^2 - (5-x^2)^2] dx , from x=-5 to x=1 }