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Ask QuestionPosted by Aryan Sindhi 6 years, 10 months ago
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Posted by Swathi Poojary 6 years, 10 months ago
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Sivalinga Raja 6 years, 10 months ago
Posted by Parth Sharma 6 years, 10 months ago
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Geetanand Yadav 6 years, 10 months ago
Posted by Chirag Surya 6 years, 10 months ago
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Ram Kushwah 6 years, 10 months ago
ax=3x+2
a1=a=3(1)+2=5
a2=a+d=3(2)+2=8
5+d=8-, d=3
S25=25/2(10+24x3)
=25/2(10+72)=25/2 x 82=25x41=1025
Honey ? 6 years, 10 months ago
Posted by Ujjwal Gautam 6 years, 10 months ago
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Shashikala Shashisudhi 6 years, 10 months ago
Posted by Akhil Gavara 6 years, 10 months ago
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Ram Kushwah 6 years, 10 months ago
kx+3y-k+2=0
12x+ky-k=0
Eqns are having no solution if
{tex}\begin{array}{l}\frac{\mathrm k}{12}=\frac3{\mathrm k}\neq\frac{-\mathrm k+2}{-\mathrm k}\\\mathrm{or}\;\mathrm k^2=12\times3=36\\\mathrm k=\pm6\end{array}{/tex}
Tera Baap 6 years, 10 months ago
Posted by Tera Baap 6 years, 10 months ago
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Ram Kushwah 6 years, 10 months ago
Mr. X this not matths
may this is your new invention in maths then why not publish in news papers.
Posted by Dhiraj Singh 6 years, 10 months ago
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Posted by Tera Baap 6 years, 10 months ago
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Posted by Dr.Soniya ? Kaushik 6 years, 10 months ago
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Puja Sahoo? 6 years, 10 months ago
Posted by Mukesh Gupta 6 years, 10 months ago
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Posted by Jayashree Das 6 years, 10 months ago
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Sia ? 6 years, 6 months ago
Check exam pattern here :https://mycbseguide.com/cbse-syllabus.html
Posted by Neh Mahesh 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
To prove :
{tex}3 \left( A B ^ { 2 } + B C ^ { 2 } + C A ^ { 2 } \right) = 4 \left( A D ^ { 2 } + B E ^ { 2 } + C F ^ { 2 } \right){/tex}
In triangle sum of squares of any two sides is equal to twice the square of half of the third side, together with twice the square of median bisecting it.
If AD is the median
{tex}A B ^ { 2 } + A C ^ { 2 } = 2 \left\{ A D ^ { 2 } + \frac { B C ^ { 2 } } { 4 } \right\}{/tex}
or, {tex}2 \left( A B ^ { 2 } + A C ^ { 2 } \right) = 4 A D ^ { 2 } + B C ^ { 2 }{/tex} ...(i)
Similarly by taking BE & CF as medians,
{tex}2 \left( A B ^ { 2 } + B C ^ { 2 } \right) = 4 B E ^ { 2 } + A C ^ { 2 }{/tex} ...(ii)
& {tex}2 \left( A C ^ { 2 } + B C ^ { 2 } \right) = 4 C F ^ { 2 } + A B ^ { 2 }{/tex} ...(iii)
By adding, (i), (ii) and (iii), we get
{tex}3 \left( A B ^ { 2 } + B C ^ { 2 } + A C ^ { 2 } \right) = 4 \left( A D ^ { 2 } + B E ^ { 2 } + C F ^ { 2 } \right){/tex}
Hence proved.
Posted by Astha A 6 years, 10 months ago
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Honey ? 6 years, 10 months ago
Posted by Devyanshi Negi 6 years, 10 months ago
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Posted by Sarah Subiah 6 years, 10 months ago
- 2 answers
Posted by Riya Dfgh 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
it is clearly shown that ABC is a triangle, in which AD is a median on BC.
construction :- draw a line AM perpendicular to BC.
we have to prove : AB² + AC² = 2(AD² + BD²)
proof :- case 1 :- when , it means AD is perpendicular on BC and both angles are right angle e.g., 90°
then, from ∆ADB,
according to Pythagoras theorem,
AB² = AD² + BD² ..... (1)
from ∆ADC ,
according to Pythagoras theorem,
AC² = AD² + DC² ...... (2)
AD is median.
so, BD = DC .......(3)
from equations (1) , (2) and (3),
AB² + AC² = AD² + AD² + BD² + BD²
AB² + AC² = 2(AD² + BD²) [hence proved ]
case 2 :- when
Let us consider that, ADB is an obtuse angle.
from ∆ABM,
from Pythagoras theorem,
AB² = AM² + BM²
AB² = AM² + (BD + DM)²
AB² = AM² + BD² + DM² + 2BD.DM ......(1)
from ∆ACM,
according to Pythagoras theorem,
AC² = AM² + CM²
AC² = AM² + (DC - DM)²
AC² = AM² + DC² + DM² - 2DC.DM ......(2)
from equations (1) and (2),
AB² + AC² = 2AM² + BD² + DC² + 2DM² + 2BD.DM - 2DC.DM
AB² + AC² = 2(AM² + DM²) + BD² + DC² + 2(BD.DM - DC.DM) ...........(3)
a/c to question, AD is median on BC.
so, BD = DC .....(4)
and from ADM,
according to Pythagoras theorem,
AD² = AM² + DM² ........(5)
putting equation (4) and equation (5) in equation (3),
AB² + AC² = 2AD² + 2BD² + 2(BD.DM - BD.DM)
AB² + AC² = 2(AD² + BD²) [hence proved].
Posted by Sameer Khan 6 years, 10 months ago
- 1 answers
Khushboo Giri 6 years, 10 months ago
Posted by Devashish Kumar 6 years, 10 months ago
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Gaurav Seth 6 years, 10 months ago
Q.
In the figure, ∠BED = ∠BDE and E is the mid-point of BC. Prove that AF / CF = AD / BE
Construction : Draw CG || FD
By adding 1 on both sides,
Posted by Abhi Raj 6 years, 10 months ago
- 1 answers
Gaurav Seth 6 years, 10 months ago
No.of bulbs which are defective = 12
Total no.of bulbs = 200
Probability of not getting defective bulb = (200-12)/200 = 188/200 = 94/100 = 0.94
Posted by Honey Tyagi 6 years, 10 months ago
- 2 answers
Gaurav Seth 6 years, 10 months ago
Let the speed of train be x km /h
Distance = 180 km
So, time = 180 / x
When speed is 9 km/h more, time taken = 180 / x+9
According to the given information:
180 / x - 180 / x+9 = 1
180 (x+9-x) / x(x+9) = 1
180 * 9 = x(x+9)
1620 = x2 + 9x
x2 + 9x - 1620 = 0
x2 + 45x - 36x - 1620 = 0
x(x+45) - 36(x+45) = 0
(x-36)(x+45) = 0
x = 36 or -45
But x being speed cannot be negative.
So, x = 36
Hence, the speed of the train is 36km/h.
Posted by Harshit Singh 6 years, 10 months ago
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Khushboo Giri 6 years, 10 months ago
Posted by #Itz_Desi_Gujjar_Ravi???? Birthday On 21St Feb???? 6 years, 10 months ago
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Posted by Muhammed Shahabas 6 years, 10 months ago
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Muhammed Shahabas 6 years, 10 months ago
Posted by Rohit Kamal 6 years, 10 months ago
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Yogita Ingle 6 years, 10 months ago
If 'n' is an odd positive integer, show that (n2-1) Is divisible by 8
Any odd positive integer is in the form of 4p + 1 or 4p+ 3 for some integer p.
Let n = 4p+ 1,
(n2 – 1) = (4p + 1)2 – 1 = 16p2 + 8p + 1 = 16p2 + 8p = 8p (2p + 1)
⇒ (n2 – 1) is divisible by 8.
(n2 – 1) = (4p + 3)2 – 1 = 16p2 + 24p + 9 – 1 = 16p2 + 24p + 8 = 8(2p2 + 3p + 1)
⇒ n2– 1 is divisible by 8.
Therefore, n2– 1 is divisible by 8 if n is an odd positive integer.
Posted by Dipika Rai 6 years, 10 months ago
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Aditya Choudhary 6 years, 10 months ago
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Sumitra Thapa 6 years, 10 months ago
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Sumedh Dabir 6 years, 10 months ago
3Thank You