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

{tex}abx^2 + (b^2 - ac)x - bc = 0 {/tex}
{tex}\Rightarrow{/tex} {tex}abx^2 + b^2x - c(ax + b) = 0 {/tex}
{tex}\Rightarrow{/tex} {tex}bx(ax + b) - c(ax + b) = 0 {/tex}
{tex}\Rightarrow{/tex}{tex}(bx - c)(ax + b) = 0 {/tex}
x = {tex}\frac{c} {b}{/tex}, {tex}\frac{-b}{a}{/tex}

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(CosecA -cot A)2=1-cos A÷1+cosA

Posted by Hardev Singh Khosa 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

LHS=(cosecA - cotA)² = {tex}\left( \frac { 1 } { \sin A } - \frac { \cos A } { \sin A } \right) ^ { 2 }{/tex}
{tex}= \left( \frac { 1 - \cos A } { \sin A } \right) ^ { 2 }{/tex}
{tex}= \frac { ( 1 - \cos A ) ^ { 2 } } { \sin ^ { 2 } A }{/tex}
{tex}= \frac { ( 1 - \cos A ) ^ { 2 } } { 1 - \cos ^ { 2 } A }{/tex}
{tex}= \frac { ( 1 - \cos A ) ^ { 2 } } { ( 1 - \cos A ) ( 1 + \cos A ) }{/tex}
{tex}= \frac { 1 - \cos A } { 1 + \cos A }{/tex}

Hence proved

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Hii 10th kesi ja rhi h sbki??

Posted by Bindass Girls ?? 6 years, 8 months ago

CBSE > Class 10 > Mathematics

5 answers

??Bestie Ki Bestie?? 6 years, 8 months ago

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Bindass Girls ?? 6 years, 8 months ago

Meri bhi mst ek dm?

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??Bestie Ki Bestie?? 6 years, 8 months ago

Badia chal rhi hai khushi???

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??.Dil Lagi Kudi Agra Di.?? 6 years, 8 months ago

Mst

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Gungun Lalwani ? 6 years, 8 months ago

Now we are in 11th

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How to know where we have to find hcf and where lcm?

Posted by Bindass Girls ?? 6 years, 8 months ago

CBSE > Class 10 > Mathematics

0 answers

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what is the simplest form of justification

Posted by Vicky Garg 6 years, 8 months ago

CBSE > Class 10 > Mathematics

0 answers

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Explaination of practice paper 3 level 1 question 5

Posted by Purvesh Patil 6 years, 8 months ago

CBSE > Class 10 > Mathematics

0 answers

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Question of 3.3 of question 1 ka 3rd in Linear equation in two variable

Posted by Dheeraj Kumar 6 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Anurag Singh 6 years, 8 months ago

Abe kutte tujhe dekh ke samajh Nahi aa raha Kya saale

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What is arithmetic progression?

Posted by Sanjana Sahu 6 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Deepak Mishra 6 years, 8 months ago

A sequence in which each term differs from its preceding term by a constant is called an arithmetic progression.

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The angle of depression of the top and the bottom of an 8 m tall buildings from the top of multi storyed building are 30. And 45. Respectively. Find the hight of multi storyed building and the distance between two buildings

Posted by Keshari Unstoppableking 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Let h is height of big building,here as per the diagram.
AE = CD = 8 m (Given)
BE = AB-AE = (h - 8) m
Let AC = DE = x
Also, {tex}\angle F B D = \angle B D E = 30 ^ { \circ }{/tex}
{tex}\angle F B C = \angle B C A = 45 ^ { \circ }{/tex}

In {tex}\triangle {/tex}ACB, {tex}\angle A = 90 ^ { \circ }{/tex}
{tex}\tan 45 ^ { \circ } = \frac { A B } { A C }{/tex}
{tex}\Rightarrow {/tex} x = h, ...(i)
In {tex}\vartriangle {/tex}BDE, {tex}\angle E = 90 ^ { \circ }{/tex}
{tex}\tan 30 ^ { \circ } = \frac { B E } { E D }{/tex}
{tex}\Rightarrow \quad x = \sqrt { 3 } ( h - 8 ){/tex} .(ii)
From (i) and (ii), we get
{tex}h = \sqrt { 3 } h - 8 \sqrt { 3 }{/tex}
h(√3 - 1) = 8√3
h = {tex}\frac{8\sqrt3}{\sqrt3-1}=\frac{8\sqrt3}{\sqrt3-1}×\frac{\sqrt3+1}{\sqrt3+1}{/tex}
= {tex}\frac{1}{2}×(24+8√3)=\frac{1}{2}×(24+13.84)=18.92 m{/tex}
Hence height of the multistory building is 18.92 m and the distance between two buildings is 18.92 m.

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Which of the following pairs of linear equations are consistent/inconsistent? If consistent obtain the solution graphically

Posted by Bharani Sivanandam 6 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Bharani Sivanandam 6 years, 8 months ago

3x + 2y = o

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Exercise 15.1

Posted by Malkit Singh 6 years, 8 months ago

CBSE > Class 10 > Mathematics

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-x+3=5

Posted by Pooja Elligari 6 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Pooja Elligari 6 years, 7 months ago

X=-3=2

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Ye 3696 kon h bhai ... Ya bhen ... Thanks hi mil rhe h ...yaha se..

Posted by ??.Dil Lagi Kudi Agra Di.?? 6 years, 8 months ago

CBSE > Class 10 > Mathematics

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1÷7x+1÷6y=3 and 1÷2x-1÷3y=5 solve by reducing method

Posted by Piyush Parekh 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

{tex}\frac{1}{7x}{/tex} + {tex}\frac{1}{6y}{/tex} = 3 .......(i)
and {tex}\frac{1}{2x}{/tex} - {tex}\frac{1}{3y}{/tex} = 5 ...........(ii)
Multiplying equation (ii) by {tex}\frac{1}{2}{/tex}, we get
{tex}\frac{1}{4x}{/tex} - {tex}\frac{1}{6y}{/tex} = {tex}\frac{5}{2}{/tex} ..........(iii)
Adding eq. (i) and (iii), we get
{tex}\frac{1}{4x}{/tex} + {tex}\frac{1}{7x}{/tex} = {tex}\frac{5}{2}{/tex} + 3
{tex}\Rightarrow{/tex} {tex}\frac{7 + 4}{28x}{/tex} = {tex}\frac{11}{2}{/tex}
{tex}\Rightarrow{/tex}{tex}\frac{11}{28x}{/tex} = {tex}\frac{11}{2}{/tex}
{tex}\Rightarrow{/tex} 28x = 2
{tex}\Rightarrow{/tex} x = {tex}\frac{1}{14}{/tex}
Putting the value of x in eq.(i), we get
{tex}\frac{1}{7(\frac1{14})}{/tex} + {tex}\frac{1}{6y}{/tex} = 3
y = {tex}\frac{1}{6}{/tex}
Hence x = {tex}\frac{1}{14}{/tex} and y = {tex}\frac{1}{6}{/tex} is the solution of given system of equations.

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Byee.... thodi der mai aati hu...????????????????

Posted by Amira ❤? 6 years, 8 months ago

CBSE > Class 10 > Mathematics

2 answers

Kuldeep Bestie Sahiba?? 6 years, 8 months ago

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Yash ? 6 years, 8 months ago

Bilkul..bbye

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Find the largest number which divides 245 and 1245 leaving remainder 5 in each case

Posted by Radhya Agarwal 6 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Suruthi Suruthi 6 years, 8 months ago

Why kharif crops cannot be grown in rabi season

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1/x - 1/x - 3 =3 : slove the polynomial by middle team splitting method ???

Posted by Aravind Chidambaram 6 years, 8 months ago

CBSE > Class 10 > Mathematics

0 answers

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13/3125 Question 1.4 1

Posted by Raj Jaiswa 6 years, 8 months ago

CBSE > Class 10 > Mathematics

0 answers

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If -2 is a zero of polynomial 3x2+4x+2k,find the value of k

Posted by Jyoti Singh 6 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 7 months ago

Let f(x)=3x² + 4x +2k
If -2 is zero of f(x) then f(-2)=0 then f(-2)=0
{tex}\Rightarrow{/tex} 3 × (-2)² + 4 × -2 + 2k = 0
{tex}\Rightarrow{/tex} 12 - 8 + 2k = 0
{tex}\Rightarrow{/tex} 4 + 2k = 0
{tex}\Rightarrow{/tex} 2k = -4
{tex}\Rightarrow{/tex} k = {tex}\frac{{ - 4}}{2}{/tex}
{tex}\Rightarrow{/tex} k = -2

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It being given that 1 is zero of the polynomial 7x-x3-6 find its other zeroes.

Posted by Jyoti Singh 6 years, 7 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 7 months ago

The given polynomial is -x³ + 7x - 6 and let f (x) = -x³ + 7x - 6 .
Since 1 is a zero of f(x), so (x - 1) is a factor of f(x).
Now we divide f(x) = -x³ + 7x - 6 by (x - 1), we obtain

Where quotient = (-x²- x + 6)
{tex}\therefore{/tex} f(x) = (-x³+ 7x - 6) = (x - 1)(-x²- x + 6)
{tex}= - ( x - 1 ) \left( x ^ { 2 } + x - 6 \right) = - ( x - 1 ) \left( x ^ { 2 } + 3 x - 2 x - 6 \right){/tex}
{tex}= ( 1 - x ) [ x ( x + 3 ) - 2 ( x + 3 ) ] = ( 1 - x ) ( x + 3 ) ( x - 2 ){/tex}
{tex}\therefore{/tex} f(x) = 0 {tex}\Rightarrow{/tex} {tex}( 1 - x ) ( x + 3 ) ( x - 2 ){/tex} = 0
{tex}\Rightarrow{/tex} (1 - x ) = 0 or (x + 3) = 0 or (x - 2) = 0
{tex}\Rightarrow{/tex} x = 1 or x = -3 or x = 2.
Thus, the other zeros are -3 and 2

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Factorise X square + 11 x + 13

Posted by Ansh Raman 6 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Arjun Bhatty 6 years, 8 months ago

b² - 4ac = (11)² - 4(1)(13) = 121 - 52 = 69 x = -b +/- √69/2a. =. -11+/-√69/2 So x = -11 + √69/2. Or. x = -11-√69/2

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Prove that cube of any positive integer can be written as 3m,3m+1 or 3m+2

Posted by Abhishek Kumar 5 years, 9 months ago

CBSE > Class 10 > Mathematics

2 answers

Saptarshi Ghosh 6 years, 8 months ago

Let, 'n' be a positive integer. By using Euclid's Division Lemma, 'n' can be written in the form of— 3q+r, where 0 <= r <3. Therefore, r = {0,1,2} Case 1:r = 0 n = 3q =>n³ = 27q³ = 3m (where m = 9q³) Case 2:r = 1 n = 3q+1 =>n³ = 27q³+27q²+9q+1 = 3m +1(where m = 9q³+9q²+3q) Case 3:r = 2 n = 3q+2 =>n³ = 27q³+54q²+36q+8 = 3m+2(where m = 9q³+18q²+12q+2) Hence, it is proved that the cube of any positive integer can be written in the form of 3m or 3m+1 or 3m+2. OR We know that the cube of any positive integer is a positive integer. On applying Euclid's division lemma, n³ = 3m+r, where 0 ≤ r < 3 Therefore, r = {0,1,2} Therefore, n³ = 3m or 3m+1 or 3m+2

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Tanmayee? Hariyan? 6 years, 8 months ago

Ex: agar tum 3m le rahe ho vaha pe eg taur par 3q lena aur use square kar 'q' se aage jo bhi tumhe milega use 'm' consider karo aur doubt katam

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Prime factors of 213478

Posted by Prangik Kakati 6 years, 8 months ago

CBSE > Class 10 > Mathematics