question
. with the help of venn-diagram.find n(aub) = n(a) + n(b)where a=(4,8)b=(3,6,9)what do you notice??
. with the help of venn-diagram.find n(aub) = n(a) + n(b)where a=(4,8)b=(3,6,9)what do you notice??
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

``` in the given expressionwe know that(a + b)2 = a2 + 2ab + b2(a-b)2 = a2-2ab + b2hence4[a2 + 2ab + b2] – 9[a2-2ab + b2]4a2 + 8ab + 4b2 - 9a2 + 18ab - 9b226ab – 5a2 - 5b225ab + ab – 5a2 – 5b2[25ab – 5a2] + [ab – 5b2]5a[5b – a] – b[5b – a][5a – b][5b – a]hence the factors 4 (a + b)2 – 9 (a – b)2 are [5a – b] and [5b – a] a - ii, b - iv, c - i, d - iii a=c and b=d
b=d and `anec`
`ab=1 " and "cdne1`
`cd=1" and "ab=ne1`
answer: a
solution: using `a.m.gtg.m,` we have   
 `a+(1)/(b)gt2sqrt((a)/(b))b+(1)/( c )gt2sqrt((b)/( c )),c+(1)/(d)gt2sqrt(( c )/(d))" and "d+(1)/(a)gt2sqrt((d)/(a))`   
 `implies(a+(1)/(b))(b+(1)/( c ))(c+(1)/(d))(d+(1)/(a))gt16`   
 but,`(a+(1)/(b))(b+(1)/( c ))(c+(1)/(d))(d+(1)/(a))=4xx1xx4xx1=16`   
 `:." "a=(1)/(b),b=(1)/( c ),c=(1)/(d)" and "d=(1)/(a)`   
 `implies" "a=(1)/(b)=2,b=(1)/( c )=(1)/(2),c=(1)/(d)=2" and "d=(1)/(a)=(1)/(2)`   
 `implies" "a=2,b=(1)/(2),c=2" and "d=(1)/(2)`   
 `implies" "a=c" and "b=d` a, b and c >> [x,y] = ismember(a,b);
>> [~,z] = sort(y(x));
>> v = find(x);
>> r = reshape(v(z),2,[]).'
r =
 2 9
 4 8
 5 7 (i) co-ordinates of point p are `((1xx17+2xx(-1))/(1+2),(1xx10+2xx1)/(1+2))` `=((17-8)/3,(10+2)/3)` `=(9/3,12/3)` `=(3,4)` (ii) `op=sqrt((0-3)^2+(0-4)^2)` `op=sqrt(9+16)` `op=sqrt25` `op=5` units (iii) let ab be divided by the point p(0, y) lying on y-axis in the ratio `therefore (0,y)=((kxx17+1xx(-4))/(k+1),(kxx10+1xx1)/(k+1))` `=>(0,y)=((17k-4)/(k+1),(10k+1)/(k+1))` `=>0=(17k-4)/(k+1)` `=>17k-4=0` `=>k=4/17` thus, the ratio in which the y-axis divide the line ab is 4: 17. in this we are asked to divide it in 2:3 ratioso, 3+2 = 55, number congruent of line segment is formed.i. firstly, we have to draw a line segment ab of length 7 cm. draw a line with an acute angle from point aii. point the compass on a, and cut the new line with some radiusplace the compass along the line and make 5 arcs. since we need 5 congruent parts, and name the last one as “g”iii. take length of bg as radius draw an arc from a just below it.again, take the length of ag as radius and draw an arc which cuts the previous arc from b, name the intersection point as hconnect b and hiv. taking the same length that we used in line ag, draw arcs along bh and name the points.v. connect the points along ag and bh. name the intercepts on the line ab as m, n, o & p.vi. thus we can divide a line segment as ao:ob = 6:4= 3:2 f(x) = u/v....i.e. u = 5x² + 4x + 3 and v = 2x² - 2x - 14

now, f '(x) = (u'v - uv')/v²

so, f '(x) = [(10x + 4)(2x² - 2x - 14) - (5x² + 4x + 3)(4x - 2)]/(2x² - 2x - 14)²   

a bit of tidying up with the denominator gives:

f '(x) = -2[9x² + 76x + 25]/(2x² - 2x - 14)²

:)> several ways to do this, here's one:

to go from (-6,4) to (-2,5) you need to go 4 in the x direction and 1 in the y direction. if you move again 4 in the x direction and 1 in the y direction you end up at (2, 6). that's p2