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Ask QuestionPosted by Devanshu Kumawat 6 years, 9 months ago
- 3 answers
Kusagra Verma 6 years, 9 months ago
Posted by Prem Parkash 6 years, 9 months ago
- 2 answers
Posted by Harish Chand Prasad 6 years, 9 months ago
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Atharva Dixit 6 years, 9 months ago
Posted by Kishan Sengar 6 years, 9 months ago
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Ram Kushwah 6 years, 9 months ago
let no. are x and x+5
so 1/x-1/(x+5=1/10
10(x+5)-10x=x(x+5)
10x+50-10x=x2+5x
x2+5x-50=0
x2+10x-5x-50=0
x(x+10)-5(x+10)=0
(x-5)(x+10)=0
x=5 or -10
so number are 5 and 10 or -10 and -.5
Posted by Shiipra Jain 6 years, 9 months ago
- 4 answers
Posted by Ishit__ __? 6 years, 9 months ago
- 2 answers
Gaurav Seth 6 years, 9 months ago
The solution is:
Let the girl be at point D on the ground from the lamppost after 4 sec.
∴ AD = 1.2 m/sec × 4 sec = 4.8 m = 480 cm
Suppose the length of the shadow of the girl be x cm when she at position D.
∴ CD = x cm
In ∆CDE and ∆CAB
∠CDE = ∠CAB (90°)
∠DCE = ∠ACB (Common)
∴ ∆CDE ∼ ∆CAB (AA similarity)
∴ Length of her shadow after 4 second is 160 cm
Posted by Diya Namdev 6 years, 9 months ago
- 1 answers
Vishakha Singh? 6 years, 9 months ago
Posted by Vikas Sharma 6 years, 9 months ago
- 1 answers
Gaurav Seth 6 years, 9 months ago
Given p2x2 + ( p2 - q2)x - q2 = 0.
⇒ p2x2 + xp2 - xq2 - q2 = 0.
⇒ xp2(x + 1) - q2 (x + 1) = 0.
⇒ (xp2- q2)(x + 1) = 0
⇒ x = q2 / p2 or -1.
Posted by Rishu Raj 6 years, 9 months ago
- 2 answers
Gaurav Seth 6 years, 9 months ago
Step-by-step explanation:
3y+5-(3y-1) = 5y+1-(3y+5)
6 = 2y-4
Y= 5
So the three terms are 14,20,26
Posted by Owais Ansari 6 years, 9 months ago
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Posted by Sahiba Khan 6 years, 9 months ago
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#Miss_Raagini ???? 6 years, 9 months ago
Posted by Ayush Nema 6 years, 9 months ago
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Posted by Garv Taneja 6 years, 9 months ago
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Priya Darshni 6 years, 9 months ago
Posted by Sanyam Dhembla 6 years, 4 months ago
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Sia ? 6 years, 4 months ago
Let AB and CD be two poles of heights a metres and b metres respectively such that the poles are p metres apart i.e. AC= p metres. Suppose the lines AD and BC meet at O such that OL = h metres. Let CL= x and LA = y. Then, x + y = p.
In {tex}\Delta{/tex} ABC and {tex}\Delta{/tex} LOC, we have
{tex}\angle{/tex} CAB = {tex}\angle{/tex}CLO [Each equal to 90°]
{tex}\angle{/tex} C = {tex}\angle{/tex} C [Common]
{tex}\therefore \quad \Delta C A B \sim \Delta C L O{/tex} [By AA-criterion of similarity]
{tex}\Rightarrow \quad \frac { C A } { C L } = \frac { A B } { L O }{/tex}
{tex}\Rightarrow \quad \frac { p } { x } = \frac { a } { h }{/tex}
{tex}\Rightarrow \quad x = \frac { p h } { a }{/tex} ...(i)
In {tex}\Delta{/tex} ALO and {tex}\Delta{/tex} ACD, we have
{tex}\angle{/tex} ALO = {tex}\angle{/tex}ACD [Each equal to 90°]
{tex}\angle{/tex} A= {tex}\angle{/tex} A [Common]
{tex}\therefore{/tex} {tex}\Delta{/tex}ALO {tex}\sim{/tex} {tex}\Delta{/tex}ACD [By AA-criterion of similarity]
{tex}\Rightarrow \quad \frac { A L } { A C } = \frac { O L } { D C }{/tex}
{tex}\Rightarrow \quad \frac { y } { p } = \frac { h } { b }{/tex}
{tex}\Rightarrow \quad y = \frac { p h } { b }{/tex} [{tex}\because{/tex} AC = x+y+=p]...(ii)
From (i) and (ii), we have
{tex}x + y = \frac { p h } { a } + \frac { p h } { b }{/tex}
{tex}\Rightarrow \quad p = p h \left( \frac { 1 } { a } + \frac { 1 } { b } \right){/tex} [ {tex}\because{/tex} x+y=p]
{tex}\Rightarrow \quad 1 = h \left( \frac { a + b } { a b } \right) \Rightarrow h = \frac { a b } { a + b }{/tex} metres
Hence, the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is {tex}\frac { a b } { a + b }{/tex} metres.
Posted by Mansi Dwivedi 6 years, 9 months ago
- 2 answers
Posted by Preethi? L 6 years, 9 months ago
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Purshottam Gehlot 6 years, 9 months ago
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Posted by Divanshu Rawat 6 years, 9 months ago
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Divanshu Rawat 6 years, 9 months ago
Vishakha Singh? 6 years, 9 months ago
Posted by Zarfa Umer 6 years, 9 months ago
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#Miss_Raagini ???? 6 years, 9 months ago
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Samir Sinha 6 years, 9 months ago
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