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Posted by Aditya Sharma 7 years, 5 months ago

CBSE > Class 10 > Mathematics

0 answers

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What is the formal (a-b)

Posted by Arman Kanojia 7 years, 5 months ago

CBSE > Class 10 > Mathematics

0 answers

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Solve ax/b -by/a= a+b,ax-by=2ab

Posted by Vanshika Mohit 6 years, 6 months ago

CBSE > Class 10 > Mathematics

2 answers

Sia ? 6 years, 6 months ago

{tex}\frac { a x } { b } - \frac { b y } { a } = a + b{/tex}
{tex}\Rightarrow{/tex} {tex}a^2x - b^2y = a^2b + ab^2{/tex}............(i)
{tex}ax - by = 2ab{/tex}..............................(ii)
Multiply (ii) by -b, we get
{tex}-abx + b^2y = -2ab^2{/tex}.......................(iii)
Adding (i) and (iii), we get
{tex}a^2x - abx = a^2b + ab^2 - 2ab^2{/tex}
{tex}\Rightarrow{/tex} {tex}x(a^2 - ab) = a^2b - ab^2{/tex}
^{{tex}\Rightarrow x = \frac { a b ( a - b ) } { a ( a - b ) }{/tex}

{tex}\Rightarrow x = b{/tex}}
Substituting x = b in (ii), we get
a(b) - by = 2ab
{tex}\Rightarrow{/tex} -by = ab
{tex}\Rightarrow{/tex} y = -a
So, x = b and y = -a

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Vandan Dev Barnwal 4 years, 6 months ago

axb−bya=a+b ⇒ a2x−b2y=a2b+ab2............(i) ax−by=2ab..............................(ii) Multiply (ii) by -b, we get −abx+b2y=−2ab2.......................(iii) Adding (i) and (iii), we get a2x−abx=a2b+ab2−2ab2 ⇒ x(a2−ab)=a2b−ab2 ⇒x=ab(a−b)a(a−b) ⇒x=b Substituting x = b in (ii), we get a(b) - by = 2ab ⇒ -by = ab ⇒ y = -a So, x = b and y = -a

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X+y=33 and x+y=19

Posted by Yugam .. 7 years, 5 months ago

CBSE > Class 10 > Mathematics

2 answers

Shiv Kumar Chaudhary 7 years, 5 months ago

X=5 and. Y = 14

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Sachin Agrawal 7 years, 5 months ago

X=33-y Put value of x in 2nd eq. 33-y=19 -y=19-33 -y=-14 Y=14 Put value of y in any eq. X+14=19 X=19-14 X=5

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×2+1/×2=7 find the value of ×3+1/×3

Posted by Dinesh Biswa 7 years, 5 months ago

CBSE > Class 10 > Mathematics

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1 minus 10 squared theta is equal to 30 degrees then find sin theta minus cos square theta

Posted by Rao D 7 years, 5 months ago

CBSE > Class 10 > Mathematics

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Draw the graph of the equation x»3,x»5 and 2x-y-4»0 also find the area of the quadreilateral formed by the lines and the x-axis

Posted by Anuj Gautam 5 years, 8 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

The given equations are
{tex}x = 3{/tex} …(i)
{tex}x = 5{/tex} …(ii)
{tex}2x - y - 4 = 0{/tex}
⇒ {tex}y = 2x - 4{/tex}
{tex}\left. \begin{gathered} \begin{array}{*{20}{c}} {x = 0,then}&{y = 2(0) - 4 = 0 - 4 = - 4} \end{array} \hfill \\ \begin{array}{*{20}{c}} {x = 1,then}&{y = 2(1) - 4 = 2 - 4 = - 2} \end{array} \hfill \\ \begin{array}{*{20}{c}} {x = 2,then}&{y = 2(2) - 4 = 4 - 4 = 0} \end{array} \hfill \\ \begin{array}{*{20}{c}} {x = 3,then}&{y = 2(3) - 4 = 6 - 4 = 2} \end{array} \hfill \\ \begin{array}{*{20}{c}} {\mathop {x = 4,then}\limits_{x = 3} }&{\mathop {y = 2(4) - 4 = 8 - 4 = 4}\limits_{x = 5} } \end{array} \hfill \\ \end{gathered} \right\} \Rightarrow{/tex}

x	0	1	2	3	4
y	-4	-2	0	2	4
III	G	H	J	K	L

x	3	3	3	x	5	5	5
y	1	2	3	y	3	4	5
I	A	B	C	II	D	E	F

The coordinates of the vertices of the required quadrilateral {tex}PMNB{/tex} are {tex}P(3, 0),\ M(5, 0),\ N(5, 6)\ and\ B(3, 2){/tex}
The quadrilateral formed by these given three lines and x-axis is {tex}PMNB{/tex}. It is trapezium. So, area of the required trapezium
{tex}= \frac { 1 } { 2 } ( B P + M N ) \times P M{/tex}
{tex}= \frac { 1 } { 2 } [ ( 2 - 0 ) + ( 6 - 0 ) ] ( 5 - 3 ){/tex}
{tex}= \frac { 1 } { 2 } \times 8 \times 2 = 8{/tex}square units
{tex}\therefore{/tex} the area of required {tex}PMNB{/tex} = {tex}8\ square\ units.{/tex}

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How solve The root is irrational number

Posted by Pankaj Kumar 7 years, 5 months ago

CBSE > Class 10 > Mathematics

2 answers

Shivanshu Ladgotra 7 years, 5 months ago

Refer ncert book

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Aman Singh 7 years, 5 months ago

Can't be explain . Refer NCERT

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2x+y=-12 2x-y=7

Posted by Anil Kushwah 7 years, 5 months ago

CBSE > Class 10 > Mathematics

0 answers

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Class 10 cbse pattern 2018-19 pleas tell

Posted by Priyanshu Kumar 7 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

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

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How to prove Thale's therome

Posted by Amit Kumar Rawat 7 years, 5 months ago

CBSE > Class 10 > Mathematics

2 answers

Shiv Kumar Chaudhary 7 years, 5 months ago

Let be ∆ABC in which D and E are the mid points of AB and AC respectivily. BC parrallel to DE. in triangle ADE draw the two perpendicular on AD and AE. And D point meet with C and point E meet with B . And DM and NE are the two perpendicular on side AE And AD respectively. In ∆ADE, Ar ∆ ADE = 1/2 X AD X NE. (1) AND, AR ∆ ADE= 1/2 X AE X MD (2) And in AR. ∆ BCD = 1/2 X NE X BD (3) IN AR. ∆ DCE = 1/2 X MD X EC. (4) FROM 1 and 3 AR ∆ ADE = AD _________________ AR. ∆ BCD = BD. FR0M 2and 4 AR.∆ ADE = AE ________________ Ar. ∆ DCE = EC. Ar.∆ BCD = Ar∆ DCE. (Both triangles lie on same base and parallel lines) AR. ∆ ADE. AR.∆ ADE ___________ Equals to (=)____________ Ar. ∆BCD. AR.∆ BCE Hence, AD. AE ____. = ______ BD. CE

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Sayantika Roy 7 years, 5 months ago

By BPT theorem.

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(m²+n²)x² -2(mp+np)x + (b² + c²) =0

Posted by Ayush Jain Nkps Rajawas 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

We have the following equation,

{tex}(m^2+n^2)x^2-2(mp+nq)x+p^2+q^2=0{/tex}
where, a = (m² + n²), b = -2(mp + nq), and c = p² + q²
The given equation will have equal roots,if D = 0
{tex}\Rightarrow{/tex} {tex}b^2-4ac=0{/tex}
{tex}\Rightarrow{/tex}{tex}[-2(mp+nq)]^2-4(m^2+n^2)(p^2+q^2)=0{/tex}
{tex}\Rightarrow{/tex}{tex}4m^2p^2+4n^2q^2+8mnpq-4m^2p^2-4m^2q^2-4n^2q^2-4n^2p^2=0{/tex}
{tex}\Rightarrow{/tex} {tex}-4m^2q^2-4p^2n^2+8mnpq=0{/tex}

{tex}\Rightarrow{/tex}{tex}-4(m^2q^2+p^2n^2-2mnpq)=0{/tex}
{tex}\Rightarrow{/tex} {tex}m^2q^2+p^2n^2-2mnpq=0{/tex}

{tex}\Rightarrow{/tex} {tex}(mq-pn)^2=0{/tex}
{tex}\Rightarrow{/tex} {tex}mq-pn=0{/tex}

{tex}\Rightarrow{/tex}{tex}mq=pn{/tex}

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Determine algebraically the vertices of the triangle formed by the lines 3x-y=3 , 2x-3y=2 , x+2y=8

Posted by Chinmoy Das 7 years, 5 months ago

CBSE > Class 10 > Mathematics

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what is iota

Posted by Kavya Mahajan 7 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Shalu Gupta 7 years, 5 months ago

IOTA is a cryptocurrency designed for the Internet of Things.

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3/2x+3y+4/2x-3y=7/2

Posted by Shatakshi Chandel 7 years, 5 months ago

CBSE > Class 10 > Mathematics

0 answers

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How to get full mark in maths

Posted by Senthil Balusamy 7 years, 5 months ago

CBSE > Class 10 > Mathematics

2 answers

Asmita Kumari 7 years, 5 months ago

1.We practise about 150 questions daily in math and learn basic concept for get full marks in math . 2. We also strong our calculation power

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Reetika Srivastava 7 years, 5 months ago

Practice maths daily for atleast an hour,clear doubts fully..donot hesitate, revise going on concepts daily

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0.4x + 0.3y = 1.7 0.7x - 0.2y = 0.8 Solve it using substitution method

Posted by Sunidhi Baghel 7 years, 5 months ago

CBSE > Class 10 > Mathematics

6 answers

Shivani Sreenivas 7 years, 5 months ago

U just gotta practice a bit more if ur still not confident enough and you'll do great!

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Shivani Sreenivas 7 years, 5 months ago

And ur ans is correct ?

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Shivani Sreenivas 7 years, 5 months ago

Yes Alpna I can

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Shivani Sreenivas 7 years, 5 months ago

4/10x + 3/10y = 17/10 7/10x - 2/10y = 8/10 = 4x + 3y /10 = 17/10 4x + 3y = 17 -1st eq Similarly 7x - 2y = 8 - 2nd eq x = 8+ 2y / 7 (from 2 ) Putting value in eq 1 4 (8+2y /7 ) +3y =17 32+8y /7 +3y = 17 Solve and get the value of y Then put the value of y in equation no. 1 And get the value of x Easey peasy dude ? don't freak out because of the decimal and make it easy by dividing it by 10 or 100 or 1000 And then solve according to the question and you'll do fine .??

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Alpna Suthar 7 years, 5 months ago

Are u able to understand it????

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Alpna Suthar 7 years, 5 months ago

On taking LCM 4x+3y=17 7x-2y=8 Now on multiplying 28x+21y=119 28x-8y. =32 y=3 and x=2

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Pythagoras theorem formula

Posted by Nabajit Biswas 7 years, 5 months ago

CBSE > Class 10 > Mathematics

4 answers

Vino Dhini 7 years, 5 months ago

hypotenus square =side square+side square

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Sujanka Talukdar 7 years, 5 months ago

H^2=p^2+b^2

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Vinayak Shukla 7 years, 5 months ago

Hypotenuse square is equal to base square + perpendicular square

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Alpna Suthar 7 years, 5 months ago

(Hypotenuse)^2 = (Perpendicular)^2+ (Base)^2

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Q. If a right triangle ABC, right-angled at B, if tan A=1 then verify that 2sin A × cos A =1 .

Posted by Dheeraj Verma 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

In {tex} \triangle A B C{/tex},
{tex}\tan A = 1{/tex}
{tex}\Rightarrow \quad \frac { B C } { A C } = 1{/tex}
{tex}\Rightarrow {/tex} BC = x and AC = x
Using Pythagoras theorem,
{tex}\Rightarrow A B ^ { 2 } = A C ^ { 2 } + B C ^ { 2 }{/tex}
{tex}\Rightarrow \quad A B ^ { 2 } = x ^ { 2 } + x ^ { 2 }{/tex}
{tex}\Rightarrow \quad A B = \sqrt { 2 } x{/tex}
{tex}\therefore \quad \sin A = \frac { B C } { A B } = \frac { x } { \sqrt { 2 } x } = \frac { 1 } { \sqrt { 2 } } \text { and } \cos A = \frac { A C } { \sqrt { 2 } x } = \frac { x } { \sqrt { 2 } x } = \frac { 1 } { \sqrt { 2 } } {/tex}

2 sin A cos A {tex}= 2 \times \frac { 1 } { \sqrt { 2 } } \times \frac { 1 } { \sqrt { 2 } } = 1{/tex}

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Ch 3 mein hum x and y ko 3 time put kyon karte h or konse sums mein

Posted by Manvi Munpariya 7 years, 5 months ago

CBSE > Class 10 > Mathematics

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(a+b)^3=?

Posted by Anmol Bawaskar 7 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Alpna Suthar 7 years, 5 months ago

(a)^3 + (b)^3 + 3(ab)(a+b)

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The 17th term of an AP exeeds its 10th term by 7.find the common difference

Posted by Jijith Jijith 7 years, 5 months ago

CBSE > Class 10 > Mathematics

2 answers

Alpna Suthar 7 years, 5 months ago

We are given a(17) = a+(10-1)d a(17)= a+9d +7 so a+(17-1)d = a+9d+7 16d=9d+7. {a is cancelled) 16d-9d= 7 7d=7 d=1

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Surya Thambirajan 7 years, 5 months ago

17th=10th+7 a+(n-1)d=a+(n-1)d a+16d=a+9d+7 16d-9d=7 7d=7 d=1

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5+√3 it is irrational number prove that

Posted by Vishal Prasad 7 years, 5 months ago

CBSE > Class 10 > Mathematics

3 answers

Shivani Sreenivas 7 years, 5 months ago

Your Welcome ?

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Shivani Sreenivas 7 years, 5 months ago

Let ✓3 be rational Therefore ✓3 = a/b where a and b are co primes From the above eq ✓3b = a This shows that ✓3 is a. factor of a This contradicts the fact that a is a co prime -1st statement b= a /✓3 This also shows that ✓3 is a factor of b This contradicts the fact that b is a co prime -2nd statement From the statements 1 and 2 we can conclude that ✓ 3 is not a rational number and therefore our assumption is wrong. Hence ✓ 3 is irrational The addition of a rational and an irrational number is always irrational So therefore 5+✓3 is also irrational Hence proved

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Prince Kumar 7 years, 5 months ago

Sssssu

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The ratio of income of two persons is 9:7and the ratio of their expenditure is4:3 if each of them manages to save rs.2000 per month find their monthly income

Posted by Shruti Patil 7 years, 5 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 months ago

Let the common ratio term of income be x and expenditure be y.
So, the income of first person is Rs.9x and the income of second person is Rs.7x.
And the expenditures of first and second person is 4y and 3y respectively.
Then, Saving of first person =9x - 4y
and saving of second person = 7x - 3y
As per given condition
9x - 4y = 200
{tex}\Rightarrow{/tex} 9x - 4y - 200=0 ... (i)
and, 7x - 3y = 200
{tex}\Rightarrow{/tex} 7x - 3y - 200 =0 ..... (ii)
Solving equation (i) and (ii) by cross-multiplication, we have
{tex}\frac { x } { 800 - 600 } = \frac { - y } { - 1800 + 1400 } = \frac { 1 } { - 27 + 28 }{/tex}
{tex}\frac { x } { 200 } = \frac { - y } { - 400 } = \frac { 1 } { 1 }{/tex}
{tex}\Rightarrow{/tex} x =200 and y =400
So, the solution of equations is x = 200 and y = 400.
Thus, monthly income of first person = Rs.9x = Rs.(9 {tex}\times{/tex} 200)= Rs.1800
and, monthly income of second person = Rs.7x = Rs.(7{tex}\times{/tex} 200)= Rs. 1400