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Sia ? 4 years, 9 months ago
One-to-One Relationship
Such a relationship exists when each record of one table is related to only one record of the other table.
<i>For example,</i> If there are two entities ‘Person’ (Id, Name, Age, Address)and ‘Passport’(Passport_id, Passport_no). So, each person can have only one passport and each passport belongs to only one person.
One-to-Many or Many-to-One Relationship
Such a relationship exists when each record of one table can be related to one or more than one record of the other table. This relationship is the most common relationship found. A one-to-many relationship can also be said as a many-to-one relationship depending upon the way we view it.
<i>For example,</i> If there are two entity type ‘Customer’ and ‘Account’ then each ‘Customer’ can have more than one ‘Account’ but each ‘Account’ is held by only one ‘Customer’. In this example, we can say that each Customer is associated with many Account. So, it is a one-to-many relationship. But, if we see it the other way i.e many Account is associated with one Customer then we can say that it is a many-to-one relationship.
Many-to-Many Relationship
Such a relationship exists when each record of the first table can be related to one or more than one record of the second table and a single record of the second table can be related to one or more than one record of the first table. A many-to-many relationship can be seen as a two one-to-many relationship which is linked by a 'linking table' or 'associate table'. The linking table links two tables by having fields which are the primary key of the other two tables. We can understand this with the following example.
Example: If there are two entity type ‘Customer’ and ‘Product’ then each customer can buy more than one product and a product can be bought by many different customers.
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Firstly, power sharing is good because it helps to reduce the possibility of conflict between social groups. A democratic rule involves sharing power with those affected by its exercise, and who have to live with its effects. People have a right to be consulted on how they are to be governed.
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Posted by C Akhilesh 4 years, 9 months ago
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Sia ? 4 years, 9 months ago
Now, the point is on x-axis, so the value 'y' of the given point will be zero.
Let, the value of the 'x' value of the given point = x
[Assume,x as a variable to do the further mathematical calculations.]
So,the point is = (x,0)
As mentioned in the question,the two points are equidistant from (x,0).
So,
Distance between (x,0) and (2,-5)
= ✓(x-2)²+(0+5)²
Distance between (x,0) and (-2,9).
=✓(x+2)²+(0-9)²
Now,the points are equidistant.
So,
✓(x-2)²+(0+5)² = ✓(x+2)²+(0-9)²
(x-2)²+(0+5)² = (x+2)²+(0-9)²
x²-4x+4+25 = x²+4x+4+81
x²-4x-x²-4x = 4+81-4-25
-8x = 56
x = -7
Posted by C Akhilesh 4 years, 9 months ago
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The traffic light at three different road crossing change after every 48 seconds,72 seconds and 108 seconds, respectively.
So let us take the LCM of the given time that is 48 seconds, 72 seconds, 108 seconds
⇒ 48 = 2 × 2 × 2 × 2 × 3
⇒ 72 = 2 × 2 × 2 × 3 × 3
⇒ 108 = 2 × 2 × 3 × 3 × 3
Hence, LCM of 48, 72 and 108 = (2 × 2 × 2 × 2 × 3 × 3 × 3)
LCM of 48, 72 and 108 = 432
So after 432 seconds, they will change simultaneously
We know that
60 seconds = 1 minute
so on dividing 432 / 60, we get 7 as quotient and 12 as a reminder
Hence, 432 seconds = 7 min 12 seconds
∴ The time = 7 a.m. + 7 minutes 12 seconds
Hence, the lights change simultaneously at 7:07:12 a.m
Posted by C Akhilesh 4 years, 9 months ago
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Sia ? 4 years, 9 months ago
Let O be the mid point,
As = Ap ( tangent to the given circle)
and since L SAp is 90°
As = OP = Ap = 10cm ( radius)
Now, Let t be the point where Ap meets Cq
and we know that Cq = PR = 27 (tangent)
and Ct = 38 ( given in the fig)
» Cq + qt = 38
» 27cm + qt = 38cm
» qt = 38cm-27cm = 11cm
and now tp = tq ( tangent)
tp = 11cm
So to find x , i.e Apt
we have Apt = Ap + pt
= 10cm + 11cm = 21 cm
Posted by C Akhilesh 5 years, 2 months ago
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Bhoomika Verma 5 years, 2 months ago
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Posted by C Akhilesh 4 years, 9 months ago
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ABC is an equilateral triangle , where D point on side BC in such a way that BD = BC/3 . Let E is the point on side BC in such a way that AE⊥BC .
Now, ∆ABE and ∆AEC
∠AEB = ∠ACE = 90°
AE is common side of both triangles ,
AB = AC [ all sides of equilateral triangle are equal ]
From R - H - S congruence rule ,
∆ABE ≡ ∆ACE
∴ BE = EC = BC/2
Now, from Pythagoras theorem,
∆ADE is right angle triangle ∴ AD² = AE² + DE² ------(1)
∆ABE is also a right angle triangle ∴ AB² = BE² + AE² ------(2)
From equation (1) and (2)
AB² - AD² = BE² - DE²
= (BC/2)² - (BE - BD)²
= BC²/4 - {(BC/2) - (BC/3)}²
= BC²/4 - (BC/6)²
= BC²/4 - BC²/36 = 8BC²/36 = 2BC²/9
∵AB = BC = CA
So, AB² = AD² + 2AB²/9
9AB² - 2AB² = 9AD²
Hence, 9AD² = 7AB²
Posted by C Akhilesh 4 years, 9 months ago
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Sia ? 4 years, 9 months ago
Given cosA/(1+sinA) +(1+sinA)/cosA
On taking the LCM we get,
={cos²A +(1+sinA)²}/cosA.(1+sinA)
Combining the like terms and adding and also we know that,use sin²A+cos²A =1
=(1+1+2sinA)/cosA(1+sinA)
=2(1+sinA)/cosA(1+sinA)
=2/cosA
=2secA
Posted by C Akhilesh 5 years, 2 months ago
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Nargish Akhtar 5 years, 2 months ago
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Himanshu Pathak 5 years, 2 months ago
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