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Ask QuestionPosted by Shubham Yadav 6 years, 11 months ago
- 0 answers
Posted by Lionel Messi⚽️ 6 years, 11 months ago
- 2 answers
Chētnà Pandey✌️ 6 years, 11 months ago
Posted by Ankit Kumar 6 years, 11 months ago
- 1 answers
Gaurav Seth 6 years, 11 months ago
Let us three consecutive integers be, x, x + 1 and x + 2 respectively.
[• Whenever a number is divided by 3 the remainder obtained is either 0 or 1 or 2. ( it's proved )]
Let x = 3p ...................................(i)
where p belongs to an integer and does not equal to zero ( 0 ).
=> x is divisible by 3.
If x = 3p + 1
where p belongs to an integer and does not equal to zero ( 0 ).
then x + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1)
which is divisible by 3.
If x = 3p + 2
where p belongs to an integer and does not equal to zero ( 0 ).
then x + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1)
which is divisible by 3.
So that x, x + 1 and x + 2 is always divisible by 3.
=> x (x + 1) (x + 2) is divisible by 3.
Similarly, whenever a number is divided 2 we will get the remainder is 0 or 1.
Let x = 2q
where q belongs to an integer and does not equal to zero ( 0 ).
=> x is divisible by 2
If x = 2q + 1
where q belongs to an integer and does not equal to zero ( 0 ).
then x + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1)
which is divisible by 2.
So that x, x + 1 and x + 2 is always divisible by 2.
⇒ x (x + 1) (x + 2) is divisible by 2.
Since x (x + 1) (x + 2) is divisible by 2 and 3.
∴ x (x + 1) (x + 2) is divisible by 6.
Posted by Sahil Kharb 6 years, 11 months ago
- 6 answers
Ram Kushwah 6 years, 11 months ago
963=657×1+306
657=306×2+45
306=45×6 + 36
45=36×1+9
36=9×4+0
So HCF=9
Hence 657x +963x(-15)=9
657x-14445=9
657x=14445+9
657x=14454
x=22 ans
Anshika Pal 6 years, 11 months ago
Anshika Pal 6 years, 11 months ago
Raunak Pandey ?? 6 years, 11 months ago
Posted by Cheshtha Sethi 6 years, 11 months ago
- 2 answers
Sinchana Sheshadri 6 years, 11 months ago
Divyanshi Tongariya 6 years, 11 months ago
Posted by Sita. Devi 6 years, 11 months ago
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Posted by Mann Maan 6 years, 11 months ago
- 1 answers
Ram Kushwah 6 years, 11 months ago
x2+k(4x+k-1)+2=0
X2+4kx+k2-k+2=0
If roots are equal then
b2=4ac
Here a=1,b=4k c=k2-k+2
So (4k)^2=4×1×(k^2-k+2)
4k^2=k^2-k+2
3k^2+k-2=0
3k^2+3k-2k-2=0
3k(k+1)-2(k+1)=0
(3k-2)(k+1)=0
k=2/3, .- 1
Posted by Parampreet Kaur 6 years, 11 months ago
- 1 answers
Ram Kushwah 6 years, 11 months ago
Here A=(1+a^2)b^2,B=2abc C=c^2-m^2
If roots are equal then
B^2=4AC
4a^2b^2c^2=4(1+a^2)b^2(c^2-m^2)
a2b2c2=(b2+a2b2)(c2-m2)
=b2c2-b2m2+a2b2c2-m2a2b2
Or b2c2=b2m2+m2a2b2
b2C2=b2m2(1+a2)
c2=m2(1+a2) proved
Sorry I am facing problem in writing powers
Posted by Navneet Verma 6 years, 11 months ago
- 2 answers
Posted by Anjali Sahu 6 years, 11 months ago
- 2 answers
Priyanshu Manas 6 years, 11 months ago
Posted by Jasline S.? 6 years, 11 months ago
- 1 answers
Posted by Adfer Shakeel 6 years, 11 months ago
- 2 answers
Prahlad Pal 6 years, 11 months ago
Posted by Tanishka Singh 6 years, 11 months ago
- 2 answers
Gaurav Verma 6 years, 11 months ago
Posted by Yameen Malik 6 years, 11 months ago
- 2 answers
Posted by Lionel Messi⚽️ 6 years, 11 months ago
- 3 answers
Gaurav Seth 6 years, 11 months ago
Mid term is nth term and (n+1)th term
=>nth term is 2n and (n+1)th term is 2(n+1)
median = (2n+2(n+1))/2=2n+1
Posted by Pankaj Pal 6 years, 11 months ago
- 2 answers
Posted by Tushar Kumar 6 years, 11 months ago
- 1 answers
Posted by Nithyashree S 6 years, 11 months ago
- 0 answers
Posted by Vikas Agrawal 6 years, 11 months ago
- 6 answers
Posted by Aniket Ssnodiya 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
P is midpoint of QR
or, {tex}\frac { a } { 3 } = \frac { - 5 + ( - 1 ) } { 2 }{/tex}
or, {tex}\frac { a } { 3 } = \frac { - 6 } { 2 }{/tex}
or, a = -9
Posted by Jaya. Prabhu 6 years, 11 months ago
- 1 answers
Gaurav Seth 6 years, 11 months ago
The tangents can be constructed in the following manner:
Step 1
Draw a circle of radius 5 cm and with centre as O.
Step 2
Take a point A on the circumference of the circle and join OA. Draw a perpendicular to OA at point A.
Step 3
Draw a radius OB, making an angle of 120° (180° − 60°) with OA.
Step 4
Draw a perpendicular to OB at point B. Let both the perpendiculars intersect at point P. PA and PB are the required tangents at an angle of 60°.
Justification
The construction can be justified by proving that ∠APB = 60°
By our construction
∠OAP = 90°
∠OBP = 90°
And ∠AOB = 120°
We know that the sum of all interior angles of a quadrilateral = 360°
∠OAP + ∠AOB + ∠OBP + ∠APB = 360°
90° + 120° + 90° + ∠APB = 360°
∠APB = 60°
This justifies the construction.
Posted by Jay Patidar 6 years, 11 months ago
- 1 answers
Gaurav Seth 6 years, 11 months ago
TSA = 462 cm2 2 = 2πrh + 2πr2
LSA = 1/3 of TSA= 1/3 462 = 154 = 2πrh
462 = 154 + 2πr2
462 = 154 + 2 x 22/7 x r2
462 - 154 = 2 x 22/ 7 x r2
308 x7 =r2
2 x 22 It's upon 2 into 22
therefore r2 = 49
r = 7
LSA = 2
TSA = 462 cm2 2 = 2πrh + 2πr2
CSA = 1/3 of TSA= 1/3 462 = 154 = 2πrh
462 = 154 + 2πr2
462 = 154 + 2 x 22/7 x r2
462 - 154 = 2 x 22/ 7 x r2
308 x7 =r2
2 x 22 It's upon 2 into 22
therefore r2 = 49
r = 7 cm
LSA = 2πrh
154 = 2 x 22/7 x 7 x h
h = 3.5 cm
volume = πr2h
22/7 x 7 x 7 x 3.5
= 539 cm3
Posted by Ajay Kumar 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
The given system of equations are:
(a - b)x + (a + b)y = 2a2 - 2b2
So, (a - b)x + (a + b)y - 2a2 - 2b2 = 0
(a - b)x + (a + b)y - 2(a2 - b2) = 0 ........(i)
And (a + b)(x + y) = 4ab
So, (a +b)x + (a + b)y - 4ab = 0 ..........(ii)
The given system of equation is in the form of
a1x + b1y - c1 = 0
and a2x + b2y - c2 = 0
Compare (i) and (ii) , we get
a1 = a - b, b1 = a + b, c1 = -2(a2 + b2)
a2 = a + b, b2 = a + b, c2 = -4ab
By cross-multiplication method
{tex}\frac{x}{{2(a + b)({a^2} - {b^2} + 2ab)}}{/tex} {tex} = \frac{{ - y}}{{2(a - b)({a^2} + {b^2})}}{/tex} {tex} = \frac{1}{{ - 2b(a + b)}}{/tex}
Now, {tex}\frac{x}{{2(a + b)({a^2} - {b^2} + 2ab)}} = \frac{1}{{ - 2b(a + b)}}{/tex} {tex}{/tex}
{tex}⇒ x = \frac{{2ab - {a^2} + {b^2}}}{b}{/tex}
And, {tex}\frac{{ - y}}{{2(a - b)({a^2} + {b^2})}} = \frac{1}{{ - 2b(a + b)}} {/tex}
{tex}⇒ y = \frac{{(a - b)({a^2} - {b^2})}}{{b(a + b)}}{/tex}
The solution of the system of equations are {tex}\frac{{2ab - {a^2} + {b^2}}}{b}{/tex} and {tex}\frac{{(a - b)({a^2} - {b^2})}}{{b(a + b)}}{/tex} respectively.
Posted by Saumya Saini 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5
Prime factorisation of 6n = (2 ×3)n
It can be observed that 5 is not in the prime factorisation of 6n.
Hence, for any value of n, 6n will not be divisible by 5.
Therefore, 6n cannot end with the digit 0 for any natural number n.
Posted by Daksh Sharma 6 years, 11 months ago
- 4 answers
Yaar 5 years, 8 months ago
Daksh Sharma 6 years, 11 months ago
Raunak Pandey ?? 6 years, 11 months ago
Posted by Lol No 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
Let a, and A be the first terms and d and D be the common difference of two A.Ps
Then, according to the question,
{tex}\frac { S _ { n } } { S _ { n } ^ { \prime } } = \frac { \frac { n } { 2 } [ 2 a + ( n - 1 ) d ] } { \frac { n } { 2 } [ 2 A + ( n - 1 ) D ] } = \frac { 7 n + 1 } { 4 n + 27 }{/tex}
or, {tex}\frac { 2 a + ( n - 1 ) d } { 2 A + ( n - 1 ) D } = \frac { 7 n + 1 } { 4 n + 27 }{/tex}
or,{tex}\frac { a + \left( \frac { n - 1 } { 2 } \right) d } { A + \left( \frac { n - 1 } { 2 } \right) D } = \frac { 7 n + 1 } { 4 n + 27 }{/tex}
Putting, {tex}\frac { n - 1 } { 2 } = m - 1{/tex}
{tex}n-1 = 2m - 2{/tex}
{tex}n= 2m - 2 + 1{/tex}
or, {tex}n = 2m - 1{/tex}
{tex}\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \frac { 7 ( 2 m - 1 ) + 1 } { 4 ( 2 m - 1 ) + 27 }{/tex}
{tex}\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \frac { 14 m - 7 + 1 } { 8 m - 4 + 27 }{/tex}
{tex}\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \frac { 14 m - 6 } { 8 m + 23 }{/tex}
Hence, {tex}\frac { a _ { m } } { A _ { m } } = \frac { 14 m - 6 } { 8 m + 23 }{/tex}
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Anshika Pal???? 6 years, 11 months ago
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