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Ask QuestionPosted by Kuldeep Singh 5 years, 3 months ago
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Kap Sharma 5 years, 3 months ago
Posted by Abhay Patel 5 years, 3 months ago
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Posted by Badboi Vicky 5 years, 3 months ago
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Sia ? 5 years, 3 months ago
Given : ABC is a triangle in which altitude BE and CF to side AC and AB are equal.
To Prove : {tex}\triangle \mathrm { ABE } \cong \triangle \mathrm { ACF }{/tex}
- AB = AC i.e. {tex}\triangle{/tex}ABC is an isosceles triangle.
Proof : BE = CF ...... [Given]
∠BAE = ∠CAF ...... [Common]
∠AFB = ∠AFC ....... [Each 90o]
{tex}\therefore{/tex} {tex}\triangle \mathrm { ABE } \cong \triangle \mathrm { ACF }{/tex} ........ [By AAS property] - {tex}\triangle A B E \cong \triangle A C F{/tex} ....... [As proved]
{tex}\therefore{/tex} AB = AC . . .[c.p.c.t.]
{tex}\therefore{/tex} {tex}\triangle{/tex}ABC is an isosceles triangle.
Posted by Karan Chauhan 5 years, 3 months ago
- 1 answers
Pratham Jaim 5 years, 3 months ago
Posted by Baba Yadav 5 years, 3 months ago
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Posted by Ashi Thakur 5 years, 3 months ago
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Sia ? 5 years, 3 months ago
Given: AD and BC are equal perpendiculars to a line segment AB.
To Prove: CD bisects AB
Proof : In OAD and OBC
{tex}\angle{/tex}AOD = {tex}\angle{/tex}BOC ...[Vertically opposite angles]
{tex}\angle{/tex}OAD = {tex}\angle{/tex}OBC ...[each 90°]
AD = BC ...[Given]
{tex}\therefore{/tex} DOAD {tex}\cong{/tex} DOBC ...[By AAS property]
{tex}\therefore{/tex} OA = OB ...[c.p.c.t.]
{tex}\therefore{/tex} CD bisects AB
Posted by Sumit Bansal 5 years, 3 months ago
- 1 answers
Posted by Manpreet Kaur 5 years, 3 months ago
- 4 answers
Posted by Arpita Tiwari 5 years, 3 months ago
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Khushi Sharma 5 years, 3 months ago
Khushi Sharma 5 years, 3 months ago
Posted by Stella Antony 5 years, 3 months ago
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Sia ? 5 years, 3 months ago
Let p(x) = x3 + 3x2 + 3x + 1
5 + 2x = 0
{tex}\Rightarrow{/tex} 2x = -5
{tex}\Rightarrow{/tex} x = -{tex}\frac{5}{2}{/tex}
= (-{tex}\frac{5}{2}{/tex})3 + 3(-{tex}\frac{5}{2}{/tex})2 + 3(-{tex}\frac{5}{2}{/tex}) + 1
= -{tex}\frac{125}{8}{/tex} + {tex}\frac{75}{4}{/tex} - {tex}\frac{15}{2}{/tex} + 1 = -{tex}\frac{27}{8}{/tex}
Posted by Ankit Patel 5 years, 3 months ago
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Posted by Yuvraj Kasana 5 years, 3 months ago
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Yogita Ingle 5 years, 3 months ago
Given x3+ 1/x3 =110
Recall that (x+1/x)3 = x3+ 1/x3 + 3(x+1/x)
Let (x+1/x)=a
⇒ a3 = 110 + 3a
⇒a3 − 3a −110=0
Put a=5
⇒ 53 − 3(5) −110
=125 −15 −110=0
Hence (a − 5) is a factor.
On dividing (a3 − 3a −110) with (a − 5) we get the quotient as (a2+5a+22)
∴ a3 − 3a −110 = (a − 5)(a2+5a+22) = 0
Hence (a − 5) = 0 and (a2+5a+22) ≠ 0
⇒(x+1/x) − 5=0
∴ (x+1/x) = 5
Posted by Saniya Chokshi 5 years, 3 months ago
- 1 answers
Yogita Ingle 5 years, 3 months ago
The square of a number is the number multiplied by itself. So, 22 is 2 × 2 , or two square and is equal to 4 .
In the same way, a number to the power of three is called the cube of the number. So <nobr aria-hidden="true">33</nobr> is <nobr aria-hidden="true">3×3×3</nobr>, or three cubed and is equal to <nobr aria-hidden="true">27</nobr>.
Posted by Saniya Chokshi 5 years, 3 months ago
- 4 answers
Sia ? 5 years, 3 months ago
Any integer that can be divided exactly by 2 is an even number.
The last digit is 0, 2, 4, 6 or 8
Any integer that cannot be divided exactly by 2 is an odd number.
The last digit is 1, 3, 5, 7 or 9
Ritu Sahu 5 years, 3 months ago
Ritu Sahu 5 years, 3 months ago
Yogita Ingle 5 years, 3 months ago
Even numbers always end with a digit of 0, 2, 4, 6 or 8.
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30 are even numbers.
Odd numbers always end with a digit of 1, 3, 5, 7, or 9.
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31 are odd numbers.
Posted by Saniya Chokshi 5 years, 3 months ago
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Posted by Dumpala Jaswanth Reddy 5 years, 3 months ago
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Khushi Sharma 5 years, 3 months ago
Ankit Patel 5 years, 3 months ago
Posted by Shiv Pawadashetti 5 years, 3 months ago
- 2 answers
Sia ? 5 years, 3 months ago
{tex}( a + b + c ) ^ { 3 } = [ a + ( b + c ) ] ^ { 3 }{/tex}
{tex}= a ^ { 3 } + 3 a ^ { 2 } ( b + c ) + 3 a ( b + c ) ^ { 2 } + ( b + c ) ^ { 3 }{/tex}
{tex}= a ^ { 3 } + 3 a ^ { 2 } b + 3 a ^ { 2 } c + 3 a[ \left( b ^ { 2 } + 2 b c + c ^ { 2 } \right)] + \left( b ^ { 3 } + 3 b ^ { 2 } c + 3 b c ^ { 2 } + c ^ { 3 } \right){/tex}
{tex}= a ^ { 3 } + 3 a ^ { 2 } b + 3 a ^ { 2 } c + 3 a b ^ { 2 } + 6 a b c + 3 a c ^ { 2 } + b ^ { 3 } + 3 b ^ { 2 } c + 3 b c ^ { 2 } + c ^ { 3 }{/tex}
{tex}= a ^ { 3 } + b ^ { 3 } + c ^ { 3 } + 3 a ^ { 2 } b + 3 a ^ { 2 } c + 3 b ^ { 2 } c + 3 b ^ { 2 } a + 3 c ^ { 2 } a + 3 c ^ { 2 } b + 6 a b c{/tex}
{tex}= a ^ { 3 } + b ^ { 3 } + c ^ { 3 } + 3 a ^ { 2 } ( b + c ) + = a ^ { 3 } + b ^ { 3 } + c ^ { 3 } + 3 a ^ { 2 } ( b + c ){/tex}
{tex}\therefore{/tex}{tex}( a + b + c ) ^ { 3 } = a ^ { 3 } + b ^ { 3 } + c ^ { 3 } + 3 ( a + b ) ( b + c ) ( c + a ){/tex}
Posted by Yash Gupta 5 years, 3 months ago
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Posted by Mujeeb Ul Nisa 5 years, 3 months ago
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Surbhii Mittal 5 years, 3 months ago
A Gautam 5 years, 3 months ago
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Posted by Himanshu Kumawat 5 years, 3 months ago
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Ritu Sahu 5 years, 3 months ago
Posted by Chaitanya Madiwalar 5 years, 3 months ago
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Ria Rockstar 5 years, 3 months ago
Posted by Manpreet Singh 5 years, 3 months ago
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Posted by Laxmi Kannolli 5 years, 3 months ago
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Posted by ꧁༒ Shaurya༻ 5 years, 3 months ago
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Laxmi Kannolli 5 years, 3 months ago
Posted by Aryan Yadav 5 years, 3 months ago
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Yogita Ingle 5 years, 3 months ago
The pairs of angles on one side of the transversal but inside the two lines are called Consecutive Interior Angles.
Posted by Rajat Deshmukh 5 years, 3 months ago
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Posted by Shivansh Dubey 5 years, 3 months ago
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Posted by Shofia Jha 5 years, 3 months ago
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Sia ? 5 years, 3 months ago
Let {tex}\sqrt 5{/tex} is a rational number.
{tex}\sqrt { 5 } = \frac { a } { b }{/tex}(a, b are co-primes and b{tex}\neq{/tex}0)
or, {tex}a = b \sqrt { 5 }{/tex}
On squaring both the sides, we get
a2=5b2 ---------------------------------(1)
Hence 5 is a factor of a2
so 5 is a factor of a
Let a = 5c, (c is some integer)
{tex}\therefore{/tex} a2 = 25c2
From equation(1) putting the value of a2
or, 5b2 = 25c2
or b2=5c2
so 5 is a factor of b2
or 5 is a factor of b
Hence 5 is a common factor of a and b
But this contradicts the fact that a and b are co-primes.
This is because we assumed that {tex}\sqrt 5{/tex} is rational
{tex}\therefore{/tex} {tex}\sqrt 5{/tex} is irrational.
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Madhuri Lakhera 5 years, 3 months ago
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