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Ask QuestionPosted by Manisha Dash 5 years, 1 month ago
- 3 answers
Yogita Ingle 5 years, 1 month ago
By Euclid's division algorithm
117 = 65x1 + 52.
65 = 52x1 + 13
52 = 13x4 + 0
Therefore 13 is the HCF (65, 117).
Posted by Yash . 5 years, 1 month ago
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Bishwajeet Kumar 5 years, 1 month ago
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Vaishu ? 5 years, 1 month ago
Bishwajeet Kumar 5 years, 1 month ago
Posted by Account Deleted 5 years, 1 month ago
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Posted by Bhakti Gupta 5 years, 1 month ago
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Gaurav Seth 5 years, 1 month ago
To draw y = x, we will plot the points (1,1) and (0,0).
To draw y = 2x, we will plot the points (1,2) and (0,0).
To draw x + y = 6, we will plot the points (0,6) and (6,0).
Coordinates of A (0,0), B (2,4), D (3,3).
Posted by Archana Nair 5 years, 1 month ago
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Bishwajeet Kumar 5 years, 1 month ago
Posted by Aarzoo Chaudhary 5 years, 1 month ago
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Gaurav Seth 5 years ago
O be the center of circles.
AB be the chord to the larger circle and tangent to smaller circle at P.
OP be radius of smaller circle i.e., OP = 15 cm
OA & OB be radius of larger circle i.e., OA = OB = 17 cm
now,
OP ⊥ AB ( radius and tangent of a circle are ⊥ to each other)
∴ In ΔOAP by PGT,
=
+
=
+
289 = 225 +
289 - 225 =
64 =
= AP
AP = 8 cm.
Now,
In ΔOAP & ΔOPB
OA = OB = 17 cm (Radii of same circle)
OP = OP (Common Side)
∠OPA = ∠OPB = 90° ( OP⊥AB)
∴ ΔOAP ≡ ΔOPB (By R.H.S axiom)
∴ AP = PB by C.P.C.T
Now,
AB = AP + PB
AB = 2 x AP
AB = 2 x 8 cm
AB = 16 cm
∴ Length of Chord is 16 cm.
Posted by Vaibhav Soni 5 years, 1 month ago
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Vaishu ? 5 years, 1 month ago
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Vaishu ? 5 years, 1 month ago
Posted by Raj Dewangan 5 years, 1 month ago
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Prachi Jain 5 years, 1 month ago
Honey Jain 5 years, 1 month ago
Yogita Ingle 5 years, 1 month ago
Suppose one number is X,
And another be Y.
By first condition,
X+Y=16………….(1)
By second condition,
X-Y =18………….(2)
Adding both the equation,
X+Y=16
+X -Y=18
…………….
2x =34 (y,-y get cancelled)
X=17.
Substitute x=17 in (1),
17+y=16
Y=16–17
Y= -1
(x=17,y=-1).
Posted by Simran Rehal 5 years, 1 month ago
- 1 answers
Yogita Ingle 5 years, 1 month ago
It is a non leap year. That means the year consist of 365 days.
Now
The birthday of first friend will be on any day of the total 365 days.
So, P(1) =
Again
The birthday of the 2nd friend will be on any day of the total of 364 days as his birthday date will be on the different day than the first friend.
So, P(2) =
So, P =
P =
Posted by Sunita Sharma 5 years, 1 month ago
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Divija Bansal 5 years, 1 month ago
Posted by Saloni Gupta 5 years, 1 month ago
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Posted by K.H.Disha Kandan 5 years, 1 month ago
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Posted by K.H.Disha Kandan 5 years, 1 month ago
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Yogita Ingle 5 years, 1 month ago
Mr. Hardy quipped that he came in a taxi with the number '1729' which seemed a fairly ordinary number. Ramanujan said that it was not. 1729, the Hardy-Ramanujan Number, is the smallest number which can be expressed as the sum of two different cubes in two different ways.
1729 is the sum of the cubes of 10 and 9 - cube of 10 is 1000 and cube of 9 is 729; adding the two numbers results in 1729.
1729 is also the sum of the cubes of 12 and 1- cube of 12 is 1728 and cube of 1 is 1; adding the two results in 1729.
K.H.Disha Kandan 5 years, 1 month ago
Posted by Sandhya Chahar 5 years, 1 month ago
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Ganesh G 5 years, 1 month ago
Gaurav Seth 5 years, 1 month ago
In the given Fig, altitudes AD and CE of 
intersects each other at the point P. Show that:
(i) ∆AEP ~ ∆CDP
(ii) ∆ABD ~ ∆CBE
(iii) ∆AEP ~ ∆ADB
(iv) ∆PDC ~ ∆BEC.
AD and CE are altitudes, which intersect each other at P.
(i) In ∆AEP and ∆CDP
∠AEP = ∠CDP = 90° [given]
and ∠APE = ∠CPD
[vertically opposite angles]
Therefore, by using AA similar condition
∆AEP ~ ∆CDP.
(ii) In ∆ABD and ∆CBE
∠ADB = ∠CEB = 90° [given]
and ∠B = ∠B [common]
Therefore, by using AA similar condition
∆ABD ~ ∆CBE.
(iii) In ∆AEP and ∆ADB
∠AEP = ∠ADB = 90° [given]
and ∠PAE = ∠DAB [common]
Therefore, by using AA similar condition
∆AEP ~ ∆ADB
(iv) In ∆PDC and ∆BEC
∠PDC = ∠CEB = 90° [given]
∠PCD = ∠ECB [common]
Therefore, by using AA similar condition
∆PDC ~ ∆BEC.
Posted by Priya ✧*｡٩(๑˙╰╯˙๑)و✧*｡ 5 years, 1 month ago
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Sawan Goja 5 years, 1 month ago
Posted by Priya ✧*｡٩(๑˙╰╯˙๑)و✧*｡ 5 years, 1 month ago
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Posted by Sunena Dutta 5 years, 1 month ago
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Priya ✧*｡٩(๑˙╰╯˙๑)و✧*｡ 5 years, 1 month ago
Posted by Prachi Jain 5 years, 1 month ago
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Yogita Ingle 5 years, 1 month ago
In △BPA, we have
DC∣∣AP [Given]
Therefore, by basic proportionality theorem, we have
BC/ CP = BD/DA ........ (I)
In △BCA, we have
DE∣∣AC
Therefore, by basic proportionality theorem, we have
BE/EC = BD/DA........ (iI)
From (i) and (ii), we get
BC/CP = BE/ EC or BE/EC = BC/CP [Hence proved]
Posted by Anushka Shivhare 5 years, 1 month ago
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Posted by Anchal Prakash 5 years, 1 month ago
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Prachi Jain 5 years, 1 month ago
Yogita Ingle 5 years, 1 month ago
- sine: Sine of an angle is defined as the ratio of the side opposite(perpendicular side) to that angle to the hypotenuse.
- cosine: Cosine of an angle is defined as the ratio of the side adjacent to that angle to the hypotenuse.
- tangent: Tangent of an angle is defined as the ratio of the side opposite to that angle to the side adjacent to that angle.
- cosecant: Cosecant is a multiplicative inverse of sine.
- secant: Secant is a multiplicative inverse of cosine.
- cotangent: Cotangent is the multiplicative inverse of the tangent.
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Prachi Jain 5 years, 1 month ago
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