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Ask QuestionPosted by Anmol Yadav 7 years, 4 months ago
- 4 answers
Posted by Rohan Das 7 years, 4 months ago
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Posted by Keshav Goyal 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
GIVEN A right triangle ABC right-angled at A, {tex}A D \perp B C.{/tex}
TO PROVE AD2 = BD {tex}\times{/tex} CD
Since {tex}\Delta{/tex} ABD and {tex}\Delta{/tex}ACD are right triangles.
{tex}\therefore{/tex} AB2 = AD2 + BD2 ......(i)
and, AC2 = AD2 + CD2 .......(ii)
Adding equations (i) and (ii), we get
AB2 + AC2 = 2AD2 + BD2 + CD2
{tex}\Rightarrow{/tex} BC2 = 2 AD2 + BD2 + CD2 [{tex}\because{/tex} {tex}\Delta{/tex}ABC is right-angled at A {tex}\therefore{/tex} AB2 + AC2 = BC2]
{tex}\Rightarrow{/tex} (BD + CD)2 = 2AD2 + BD2 + CD2
{tex}\Rightarrow{/tex} BD2 + CD2 + 2 BD {tex}\times{/tex} CD = 2 AD2 + BD2 + CD2
{tex}\Rightarrow{/tex} 2BD {tex}\times{/tex} CD = 2 AD2
{tex}\Rightarrow{/tex} AD2 = BD {tex}\times{/tex} CD
{tex}\Rightarrow{/tex} Hence, AD2 = BD {tex}\times{/tex} CD.
Posted by Ayushman Singh 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
The given numbers are 320 and 457
Now as 5 and 7 are remainders on division of 320 and 457 by said number
On subtracting the reminders 5 and 7 from 320 and 457 respectively we get:
320 - 5 = 315,
457 - 7 = 450
The prime factorization:of 315 and 405 are
{tex}315 = 3 \times 3 \times 5 \times 7{/tex}
{tex}= 3 ^ { 2 } \times 5 \times 7{/tex}
{tex}450 = 2 \times 3 \times 3 \times 5 \times 5{/tex}
{tex}= 2 \times 3 ^ { 2 } \times 5 ^ { 2 }{/tex}
{tex}\therefore{/tex} H.C.F. of 315 and 450 = {tex}3 ^ { 2 } \times 5 = 9 \times 5 = 45{/tex}
Hence the said number =45
Posted by Smruti Sahu 5 years, 8 months ago
- 1 answers
Sia ? 6 years, 5 months ago
Let the speed of the boat in still water be 'x' km/hr and speed of the stream be 'y' km/
Speed = Distance / Time
{tex}\therefore{/tex} {tex}\frac { 30 } { x - y } + \frac { 28 } { x + y } = 7{/tex}
and {tex}\frac { 21 } { x - y } + \frac { 21 } { x + y } = 5{/tex}
Let {tex}\frac { 1 } { x - y } \text { be } a \text { and } \frac { 1 } { x + y } \text { be } b{/tex}
30a + 28b = 7 ......(i)
21a + 21b = 5 ......(ii)
Multiplying (i) by 3 and (ii) by 4 and then subtracting.
{tex}90a+84b=21{/tex} ..............(iii)
{tex}84a+84b=20 {/tex} ..............(iv)
By solving (iii) and (iv)
{tex}90a-21=84a-20{/tex}
{tex}\Rightarrow{/tex}6a= 1
{tex}\Rightarrow{/tex} {tex}a = \frac { 1 } { 6 }{/tex}
Putting this value of ,a in eqn., (i),
{tex}30 \times \frac { 1 } { 6 } + 28 b = 7{/tex}
{tex}28 b = 7 - 30 \times \frac { 1 } { 6 } = 2{/tex}
{tex}\therefore{/tex}{tex}b = \frac { 1 } { 14 }{/tex}
x + y = 14 ...(iv)
Now, {tex}a = \frac { 1 } { x - y } = \frac { 1 } { 6 }{/tex}
{tex}\Rightarrow{/tex} x - y = 6
{tex}\Rightarrow{/tex}x = y + 6 .....(v)
Putting (iv) in (v)
y + 6 + y = 14
{tex}\Rightarrow{/tex} y = 4
Hence, speed of the boat in still water = 10 km/hr and speed of the stream = 4 km/hr.
Posted by Vanshika Rajput 7 years, 4 months ago
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Posted by Arun Teotia 7 years, 4 months ago
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Posted by Durga Prasad 7 years, 4 months ago
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Posted by Mehman Jaria 7 years, 4 months ago
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Posted by Chanchal Dhiman 7 years, 4 months ago
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Posted by Pulkit Kumar 7 years, 4 months ago
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Posted by Bikram Murmu 6 years, 4 months ago
- 1 answers
Sia ? 6 years, 4 months ago
Let {tex} \alpha{/tex} and {tex} \frac { 1 } { \alpha }{/tex} be the zeros of (a2 + 9)x2 + 13x + 6a.
Then, we have
{tex} \alpha \times \frac { 1 } { \alpha } = \frac { 6 a } { a ^ { 2 } + 9 }{/tex}
⇒ 1 = {tex} \frac { 6 a } { a ^ { 2 } + 9 }{/tex}
⇒ a2 + 9 = 6a
⇒ a2 - 6a + 9 = 0
⇒ a2 - 3a - 3a + 9 = 0
⇒ a(a - 3) - 3(a - 3) = 0
⇒ (a - 3) (a - 3) = 0
⇒ (a - 3)2 = 0
⇒ a - 3 = 0
⇒ a = 3
So, the value of a in given polynomial is 3.
Posted by Abhishek Yadav Kotputli 7 years, 4 months ago
- 1 answers
Harleen Narula 7 years, 4 months ago
Posted by Balakumar Bala 7 years, 4 months ago
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Posted by Abhishek Singh 7 years, 4 months ago
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Posted by Yuvraj Khichi 7 years, 4 months ago
- 5 answers
Posted by Parul Rawat 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
OA = 10 units
{tex}\Rightarrow{/tex} OA = {tex}\sqrt { ( 2 a- 1 + 3 ) ^ { 2 } + ( 7 + 1 ) ^ { 2 } }{/tex}
{tex}\Rightarrow{/tex} 10 = {tex}\sqrt { 4 a ^ { 2 } + 4 + 8 a + 64 }{/tex}
Squaring 100 = 4a2 + 8a + 68
{tex}\Rightarrow{/tex} 4a2 + 8a - 32 = 0
{tex}\Rightarrow{/tex} a2 + 2a - 8 = 0
{tex}\Rightarrow{/tex} a2 + 4a - 2a - 8 = 0
{tex}\Rightarrow{/tex} a(a + 4) - 2(a + 4) = 0
{tex}\Rightarrow{/tex} (a + 4)(a - 2) = 0
{tex}\therefore{/tex} a = - 4, a = 2.
Posted by Bhawana Routela 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let AB be the tree of height 12 metres. Suppose the tree is broken by the wind at point C and the part CB assumes the position CO and meets the ground at O.
Let AC = x. Then, CO = CB = 1 2 - x. It is given that {tex}\angle A O C{/tex} = 60o
In {tex}\triangle O A C,{/tex} we have
{tex}\sin 60 ^ { \circ } = \frac { A C } { O C }{/tex}
{tex}\Rightarrow \quad \frac { \sqrt { 3 } } { 2 } = \frac { x } { 12 - x }{/tex}
{tex}\Rightarrow \quad 12 \sqrt { 3 } - \sqrt { 3 } x = 2 x{/tex}
{tex}\Rightarrow \quad 12 \sqrt { 3 } = x ( 2 + \sqrt { 3 } ){/tex}
{tex}\Rightarrow \quad x = \frac { 12 \sqrt { 3 } } { 2 + \sqrt { 3 } } = \frac { 12 \sqrt { 3 } } { 2 + \sqrt { 3 } } \times \frac { 2 - \sqrt { 3 } } { 2 - \sqrt { 3 } } = 12 \sqrt { 3 } ( 2 - \sqrt { 3 } ){/tex}
{tex}\Rightarrow \quad x = 24 \sqrt { 3 } - 36 = 5.569{/tex} metres
Hence, the tree is broken at a height of 5.569 metres from the ground.
Posted by Apoorv Dixit 7 years, 4 months ago
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Posted by Mohit Kumar 5 years, 8 months ago
- 1 answers
Sia ? 6 years, 5 months ago
Let the speed of the stream = x km/hr
Speed of the boat in upstream = (5 - x) km/hr
Speed of the boat in downstream = (5 + x) km/hr
Distance = 40 km
Time taken in upstream = {tex}{40 \over (5 - x)} hr{/tex}
Time taken in downstream = {tex}{40\over (5 + x)} hr{/tex}
According to the question,
{tex}\frac{40}{{5 - x}} = 3\left( {\frac{40}{{5 + x}}} \right){/tex}
⇒ (5 + x) = 3(5 - x)
⇒ 5 + x = 15 - 3x
⇒ 4x = 10
⇒ x = 2.5
Therefore, speed of the stream = 2.5 km/hr.
Posted by Lovely Boy 7 years, 4 months ago
- 2 answers
Posted by Muky Thakur 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Prism is a solid geometric figure whose two ends are similar, equal, and parallel rectilinear figures, and whose sides are parallelograms.
Posted by Himanshu Mahawar 7 years, 4 months ago
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Posted by Shubham Dayal 6 years, 4 months ago
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Posted by Jaydeepdamor Damor 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Let the number of wickets taken by Zaheer in a cricket match are x , then the number of wickets taken by Harbhajan = (2x - 3)
Given, The product of wickets taken by these two = 20
⇒ (x)(2x -3) = 20
⇒ 2x2 - 3x - 20 = 0
Posted by Samarth M.P 7 years, 4 months ago
- 1 answers
Posted by Vaibhav Lohitashv 6 years, 5 months ago
- 1 answers
Sia ? 6 years, 5 months ago
Here the remainder =6x+2
Therefore, when 6x+2 is subtracted from 8x4 + 14x3 + x2 + 7x + 8, then it will be divisible by 4x2 - 3x + 2.
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Dev Tera Yaar Jaat☢️ 7 years, 4 months ago
2Thank You