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- 4 answers
Posted by Aashi Jain 7 years, 4 months ago
- 2 answers
Posted by Mahek Thakur 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
The given points are A(x, -1) and B(5, 3)
Also given, AB = 5
{tex}\Rightarrow{/tex}AB2=25
{tex}\Rightarrow{/tex} (5 - x)2 + (3 + 1)2 =25 [ using distance formula]
{tex}\Rightarrow{/tex} 25 + x2 - 10x + 16 =25
{tex}\Rightarrow{/tex} x2 - 10x + 16 =0
{tex}\Rightarrow{/tex} x2 - 8x - 2x + 16 = 0
{tex}\Rightarrow{/tex} x (x - 8) -2(x - 8) =0
{tex}\Rightarrow{/tex} (x - 8) (x - 2) = 0
{tex}\Rightarrow{/tex} x - 8 = 0 or x - 2 =0
{tex}\Rightarrow{/tex} x = 8 or x = 2
Hence, possible values of x are 8 & 2.
Posted by Abhimanyu Singh 7 years, 4 months ago
- 6 answers
Posted by Bhubaneshwar Vishwakarma 7 years, 4 months ago
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Tanu Choudhary 7 years, 4 months ago
Posted by Aravinth Aravinth 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
Common difference,
{tex}d_1 = \sqrt { 6 } - \sqrt { 3 }{/tex}
{tex}= \sqrt { 3 } ( \sqrt { 2 } - 1 ){/tex}
{tex}d_2= \sqrt { 9 } - \sqrt { 6 }{/tex}
{tex}= \sqrt { 3\times3 } - \sqrt{2\times 3}{/tex}
{tex}= 3 - \sqrt { 6 }{/tex}
{tex}d_3 = \sqrt { 12 } - \sqrt { 9 } {/tex}
{tex}= \sqrt { 4\times3 } - \sqrt { 9 } {/tex}
{tex}= 2 \sqrt { 3 } - 3{/tex}
As common difference does not equal.
Hence, The given series is not in A.P.
Posted by Mirza Adnan Baig Deshmukh 7 years, 4 months ago
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Shivam Namdev 7 years, 4 months ago
Posted by Apoo Sri 7 years, 4 months ago
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Posted by Gopala Krishna 7 years, 4 months ago
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Posted by Elsa B 7 years, 4 months ago
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Bethini S. 7 years, 4 months ago
Posted by Sound Harya 7 years, 4 months ago
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Posted by Mohit Singh 7 years, 4 months ago
- 7 answers
Bethini S. 7 years, 4 months ago
Rajput S 7 years, 4 months ago
Posted by Govind4555 Tiwari 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
Given : In {tex}\triangle A B C{/tex}, DE || BC and intersects AB in D and AC in E.
Prove that : {tex}\frac{AD}{DB} = \frac{AE}{EC}{/tex}
Construction: Join BE, CD and draw EF {tex}\perp{/tex} BA and DG {tex}\perp{/tex} CA.
Now from the given figure we have,
EF is the height of ∆ADE and ∆DBE (Definition of perpendicular)
Area({tex}\triangle{/tex}ADE) ={tex}\frac{AD.EF}{2}{/tex} .....(1)
Area({tex}\triangle{/tex}DBE) = {tex}\frac{DB.EF}{2}{/tex} ....(2)
Divide the two equations we have
{tex}\frac{Area \triangle ADE}{Area \triangle DBE} = \frac{AD}{DB}{/tex} .....(3)
{tex}\frac{Area \triangle ADE}{Area \triangle DEC} = \frac{AE}{EC}{/tex} .....(4)
Therefore, {tex}\triangle \mathrm{DBE} \sim \triangle \mathrm{DEC}{/tex} (Both the ∆s are on the same base and between the same || lines)
Area({tex}\triangle{/tex}DBE) = Area({tex}\triangle{/tex}DEC) (If the two triangles are similar their areas are equal) ....(5)
{tex}\frac{AD}{DB} = \frac{AE}{EC}{/tex} [from equation 3,4 and 5]
Hence proved.
Posted by Pratham Bajpai 7 years, 4 months ago
- 0 answers
Posted by Abhi Kushwaha 7 years, 4 months ago
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Posted by Brijesh Kumar 7 years, 4 months ago
- 2 answers
Posted by Naveen Tiwari 7 years, 4 months ago
- 2 answers
Smriti Yadav 7 years, 4 months ago
Smriti Yadav 7 years, 4 months ago
Posted by Vaibhav Parmar 7 years, 4 months ago
- 4 answers
Adarsh Patil 7 years, 4 months ago
Yash Rajput 7 years, 4 months ago
Yogita Ingle 7 years, 4 months ago
x + y = 14 ........(i)
x - y = 4 .........(ii)
(i) + (ii)
2x = 18
x = 18/2 = 9
Put x = 9 in (i)
9 + y =1 4
y = 14 - 9
y = 5
So, x = 9 and y = 5
Posted by Himanshu Yadav 7 years, 4 months ago
- 1 answers
Posted by Rajprateem Nath 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
Let p be any positive integer and b = 3. Applying division Lemma with p and b =3 ,
we have
p = 3q + r, where 0 {tex}\leq{/tex} r < 3 and q is some integer
So r=0,1,2
If r=0 , p=3q
If r=1, p=3q+1
If r=2, p=3q+2
Therefore any positive integer is of form 3q,3q+1,3q+2 for some integer q.
Posted by Dia Nandy 6 years, 6 months ago
- 1 answers
Sia ? 6 years, 6 months ago
q = a + (p - 1)d ..... (i)
p = a + (q - 1)d ...... (ii)
q - p = (p - 1 - q + 1)d
{tex}\frac{{q - p}}{{p - q}} = d{/tex}
{tex} \Rightarrow d = - 1{/tex}
Put the value of d in eq (i)
q = a + (p - 1) (-1)
{tex} \Rightarrow {/tex} q = a - p + 1
{tex} \Rightarrow {/tex} a = q + p - 1
ap+q = a +(p + q - 1)d
= (q + p - 1) + (p + q - 1) (-1)
= q + p - 1 - p - q + 1
= 0
Posted by Shivani Singh 7 years, 5 months ago
- 1 answers
Posted by Chanda Chuhan 7 years, 5 months ago
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Dheeraj Verma 7 years, 4 months ago
1Thank You