CBSE > Class 10 > Mathematics | CBSE Board | Homework Help

It cannot end with digit zero because in 16 n (2×2×2×2) n is there and for ending with digit zero the number must contain 2 and 5 in its prime factors and 16 only contain 2 but not 5 that is why it cannot end with 0

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?Niharika Sharma ? 6 years, 4 months ago

16ⁿ = (2×2×2×2)ⁿ Prime factorisation of 16ⁿ does not contain the prime factor 5. Therefore, 16ⁿ cannot end with digit 0 for any natural number n

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Ex6.3

Posted by Sougata Mandal 6 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 3 months ago

You can check NCERT Solutions here : https://mycbseguide.com/ncert-solutions.html

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Prove that √1+cos/1-cos=1+cos/sin

Posted by Vivek Singh 6 years, 4 months ago

CBSE > Class 10 > Mathematics

2 answers

Vinayak Chaudhary 6 years, 4 months ago

Multiply both numerater and denominator by 1+cos You will get √(1+cos)^2/√1-cos^2 1-cos^2 = sin^2 √(1+ cos)^2/√sin^2 1+cos/sin

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Dheeraj Upraity 6 years, 4 months ago

√1+cos 1. --------------- √1-cos 2. √1+cos x1+cos -------------------------- √1-cos x 1+cos 3. √(1+cos)*2 ------------------ √ 1-cos*2 4. √1+cos ---------- √ Sin*2 5. 1+cos ------- Sin

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Hii...?

Posted by Chetna ❤️ 6 years, 4 months ago

CBSE > Class 10 > Mathematics

2 answers

Aishreet ... 6 years, 4 months ago

Hlo bestie

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Janu Likki 6 years, 4 months ago

Hlo

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ANSWER

Solve √2

Posted by Naman Sharma 5 years, 8 months ago

CBSE > Class 10 > Mathematics

2 answers

Vivek Kumar Verma 6 years, 4 months ago

1.4142135623730

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Dheeraj Upraity 6 years, 4 months ago

1.414

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ANSWER

Find the irrational number between √5 and √7

Posted by Cbse Student 6 years, 4 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

X+y/2 +y-1/3=9 X-1/3 + y+1/2 =8

Posted by Prem Jadhav 6 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 3 months ago

The given equations may be written as
{tex} \frac { x + 1 } { 2 } + \frac { y - 1 } { 3 } = 8 {/tex}
⇒ 3(x + 1) + 2 (y - 1) = 48
⇒ 3x + 3 + 2y - 2 = 48
⇒ 3x + 2y + 1= 48
⇒ 3x+2y = 47 ... (i)
{tex}\frac { x - 1 } { 3 } + \frac { y + 1 } { 2 } = 9{/tex}
⇒ 2(x -1 ) + 3(y +1) = 54
⇒ 2x - 2 + 3y + 3 = 54
⇒ 2x + 3y + 1 = 54
⇒ 2x + 3y = 53. ... (ii)
Multiplying (i) by 2 and (ii) by 3 and subtracting, we get
(4 - 9)y = 94-159
{tex} \Rightarrow - 5 y = - 65 {/tex}
{tex}\Rightarrow y = \frac { - 65 } { - 5 } {/tex}

{tex}\Rightarrow y = 13{/tex}
Putting y = 13 in (i), we get
3x + (2 {tex} \times{/tex} 13) = 47
{tex} \Rightarrow{/tex} 3x + 26 = 47
{tex} \Rightarrow{/tex} 3x = (47 - 26)
{tex} \Rightarrow{/tex}3x = 21
{tex} \Rightarrow \quad x = \frac { 21 } { 3 } = 7{/tex}
So, x = 7
Hence, x = 7 and y = 13

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ANSWER

In a lottery there are 8 prizes and 16 blanks.What is the probability of getting a price?

Posted by Sushmita Sen 6 years, 4 months ago

CBSE > Class 10 > Mathematics

3 answers

Janu Likki 6 years, 4 months ago

8+16= 24 8/24== 1/3

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$N€H!L J@@T ? 6 years, 4 months ago

Go ahead and use youtube

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Yashwath Calidas 6 years, 4 months ago

50%

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ANSWER

Send me paper format

Posted by Ojus Gupta 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Check here : https://mycbseguide.com/cbse-syllabus.html

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ANSWER

how to proof thales theorem

Posted by Bablu Kumar Sharma 6 years, 4 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Prove that 3+2 root 5 is irrational

Posted by Sahib Ansari 6 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 3 months ago

We will prove this by contradiction.
Let us suppose that (3+2 {tex}\sqrt { 5 }{/tex}) is rational.
It means that we have co-prime integers a and b such that
{tex}\frac { a } { b } = 3 + 2 \sqrt { 5 } \quad \frac { a } { b } - 3 = 2 \sqrt { 5 }{/tex}

{tex}\Rightarrow \frac{{a - 3b}}{b} = 2{\sqrt 5 \,{ \Rightarrow }}\frac{{a - 3b}}{{2b}} = \sqrt 5 {/tex} ....(1)
a and b are integers.
It means L.H.S of (1) is rational but we know that {tex}\sqrt { 5 }{/tex} is irrational. It is not possible. Therefore, our supposition is wrong. (3+2 {tex}\sqrt { 5 }{/tex}) cannot be rational.
Hence, (3+2 {tex}\sqrt { 5 }{/tex}) is irrational.

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Prove that 3+2

Posted by Sahib Ansari 6 years, 4 months ago

CBSE > Class 10 > Mathematics

3 answers

Vivek Kumar Verma 6 years, 4 months ago

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Khushi Mehandiratta 6 years, 4 months ago

5 is the right answer

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Pk . 6 years, 4 months ago

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Any body prove that LHS=RHS

Posted by Ayushman Nautiyal 6 years, 4 months ago

CBSE > Class 10 > Mathematics

4 answers

Vivek Kumar Verma 6 years, 4 months ago

LHS=0. andRHS=0. Hence LHS=RHS Proved

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Abhay Singh 6 years, 4 months ago

Hence LHS=RHS

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Pk . 6 years, 4 months ago

This is the right way to answer isn't it?

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#Sareef Balak 6 years, 4 months ago

LHS=RHS

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ANSWER

If perimeter and area of a circle is equal and find radius

Posted by Sujal Chaudhary 6 years, 4 months ago

CBSE > Class 10 > Mathematics

3 answers

Soumya Ranjan Patra 6 years, 4 months ago

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Rohit Yadav 6 years, 4 months ago

Radius is 2

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Rohit Yadav 6 years, 4 months ago

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ANSWER

X2+5x-(a2+a-6)=0

Posted by Yashavant Kumar Sharma 6 years, 4 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Two opposite vertices of a square (-1,2) and (3,2). Find the coordinate of remaining vertices.

Posted by Aadya Singh 6 years, 3 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 3 months ago

Let ABCD be a square and B (x, y) be the unknown vertex.
AB = BC
{tex} \Rightarrow {/tex} AB² = BC²

{tex} \Rightarrow {/tex} (x + 1)² + (y - 2)² = (x - 3)² + (y - 2)²
{tex} \Rightarrow {/tex} x² + 1 + 2x + y² + 4 - 4x = x² - 6x + 9 + y² + 4 - 4x
{tex} \Rightarrow {/tex} 2x + 1 = - 6x + 9
{tex} \Rightarrow {/tex} 8x = 8
{tex} \Rightarrow {/tex} x = 1 ........ (i)
In {tex}\triangle{/tex}ABC, AB² + BC² = AC²
{tex} \Rightarrow {/tex} (x + 1)² + (y - 2)² + (x - 3)² + (y - 2)² = (3 + 1)² + (2 - 2)²
{tex} \Rightarrow {/tex}x² + 1 + 2x + y² - 4x + 4 + x² + 9 - 6x + y² + 4 - 2y = 16 + 0
{tex} \Rightarrow {/tex} 2x² + 2y² + 2x - 4y - 6x - 4y + 1 + 4 + 9 + 4 = 16
{tex} \Rightarrow {/tex} 2x² + 2y² - 4x - 8y + 2 = 0
{tex} \Rightarrow {/tex} x² + y² - 2x - 4y + 1 = 0 ..... (ii)
Putting the value of x in eq. (ii),
1 + y² - 2 - 4y + 1 = 0
{tex} \Rightarrow {/tex} y² - 4y = 0
{tex} \Rightarrow {/tex} y(y - 4) = 0
{tex} \Rightarrow {/tex} y = 0 or 4
Hence the other vertices are (1, 0) and (1, 4).

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Trignometry 9chapter 11question i need answer

Posted by Jerlin Christy 6 years, 4 months ago

CBSE > Class 10 > Mathematics

0 answers

ANSWER

Sin18°/cos72° evaluate this.

Posted by Suniksha Bhatia 6 years, 4 months ago

CBSE > Class 10 > Mathematics

2 answers

Sia ? 6 years, 4 months ago

Sin 18/cos72 cos(90-72)/cos72
Cos72/sin72=1
Tan26/tan(90-74)=1
Cos48-sin42=sin(90-42)-sin42=0
Cosec31-Cosec(90-69) Cose31-Cosec31=0

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Tanisha Yadav 6 years, 4 months ago

Sin theata = cos (90 - theata) So, Sin 18° =cos (90°- 18°) =cos 72° Cos 72°/cos72° =1

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ANSWER

Given seco=13/12, calculate all trigonometrc ratios.

Posted by Suniksha Bhatia 6 years, 4 months ago

CBSE > Class 10 > Mathematics

3 answers

Sia ? 6 years, 4 months ago

Consider a triangle ABC in which {tex}\angle A = \theta {/tex} and {tex}\angle B = {90^o}{/tex}

Let AB = 12k and AC = 13k
Then, using Pythagoras theorem,
{tex}BC=\sqrt { ( \mathrm { AC } ) ^ { 2 } - ( \mathrm { AB } ) ^ { 2 } } = \sqrt { ( 13 k ) ^ { 2 } - ( 12 k ) ^ { 2 } }{/tex}
{tex}= \sqrt { 169 k ^ { 2 } - 144 k ^ { 2 } } = \sqrt { 25 k ^ { 2 } } = 5 k{/tex}
{tex}\therefore {/tex} {tex}\sin \theta = \frac { B C } { A C } = \frac { 5 k } { 13 k } = \frac { 5 } { 13 }{/tex}
{tex}\cos \theta = \frac { A B } { A C } = \frac { 12 k } { 13 k } = \frac { 12 } { 13 } \tan \theta = \frac { B C } { A B } = \frac { 5 k } { 12 k } = \frac { 5 } { 12 }{/tex}
{tex}\cot \theta = \frac { A B } { B C } = \frac { 12 k } { 5 k } = \frac { 12 } { 5 } \cos e c \theta = \frac { A C } { B C } = \frac { 13 k } { 5 k } = \frac { 13 } { 5 }{/tex}

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Tanisha Yadav 6 years, 4 months ago

Sec theata = hypotenuse/ base So, base =12 Hypotenuse=13 Prependicular =√( hypotenuse)2 - ( base)2 √(13)2-(12)2=√169-144=√25 =5 Sin theata =P/H =5/13 Cos theata=B/H =12/13 Tan theata=P/B =5/12 Cot theata=B/P =12/5 Cosec theata= H/P = 13 /5

1Thank You

Anshuman Singh Chouhan 6 years, 4 months ago

SecA=H/B, Here secA=13/12 So, H=13 and B=12 We know that, P2=H2-B2 P2=164-144 P2=25 P=5 Now, All trignometric ratios SinA=P/H=5/13 CosA=B/H=12/13 TanA=P/B=5/12 CosecA=H/P=13/5 CotA=B/P=12/5

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ANSWER

7 years ago Sparsh's age was five times the square of the age of Arya. 3 years hence, age of Arya will be two-fifth of Sparsh. Find their present ages.

Posted by Anna Shivani 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Let seven years ago arya's age be x years.
As per given condition
Seven years ago sparsh's age was five times the square of arya's age.
Then, seven years ago sparsh's age was 5x² years.
Therefore, arya's present age = (x + 7) years
And sparsh's present age = (5x² + 7) years
Three years hence arya's age = (x + 7 + 3) years = (x + 10) years
And three years hence sparsh's age = (5x² + 7 + 3) years = (5x² + 10) years
As per given condition
Three years hence arya's age will be {tex}\frac{2}{5}{/tex} of sparsh's age.
Therefore, x + 10 = {tex}\frac{2}{5}{/tex}(5x² + 10)
{tex}\Rightarrow{/tex} x + 10 = 2(x² + 2)
{tex}\Rightarrow{/tex} x + 10 = 2x² + 4
{tex}\Rightarrow{/tex} 2x² - x - 6 = 0
{tex}\Rightarrow{/tex} 2x² - 4x + 3x - 6 = 0
{tex}\Rightarrow{/tex} 2 x ( x - 2 ) + 3 ( x - 2 ) = 0
{tex}\Rightarrow{/tex} (2x + 3) (x - 2) = 0
{tex}\Rightarrow{/tex} x - 2 = 0 [{tex}\because{/tex} 2x + 3 {tex}\neq{/tex} 0 as x > 0]
{tex}\Rightarrow{/tex} x = 2
Hence, arya's present age = (2 + 7) years = 9 years
sparsh's present age = (5 {tex}\times{/tex}2² + 7) years = 27 years