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  • 1 answers

Aangi Mehta 5 years, 4 months ago

In first x+5=0 x=-5 In second x+6=-1 x=-7
  • 1 answers

Aangi Mehta 5 years, 4 months ago

CotA=8/15 O=15k. A=8k By Pythagoras the theorem,h=√225+64 =√289 =17k So, sinA=15/17 SecA=17/8
  • 3 answers

Gaurav Seth 5 years, 4 months ago

3x-y=3........1
9x-3y=9..(divided by 3)
3x-y=3......2

3x=3-y
X=3-y/3

sub X in 1
3x-y=3
3(3-y/3)-y=3
3-y-y=3
-2y=3-3

y=0

sub y in 2

3x-y=3
3x-0=3
3x=3

X=1

Yogita Ingle 5 years, 4 months ago


3x-y=3........(i)
9x-3y=9..(divided by 3)
3x-y=3......(i)

3x=3-y
X=3-y/3

sub X in (i)
3x-y=3
3(3-y/3)-y=3
3-y-y=3
-2y=3-3
y=0
sub y in (ii)
3x-y=3
3x-0=3
3x=3
X=1

Lucifer?? Morningstar?? 5 years, 4 months ago

Put the value,Y=3x-3 in eq (2) 9x-3(3x-3)=9 => 9x-9x+9=9.. hence, proved
  • 1 answers

Gaurav Seth 5 years, 4 months ago

Let the zeros be a,b,c and d

 

Then,

 

 

 

..........(1)

 

 

 

 

..........(2)

 

Using (1) and (2), we get

 

 

 

 

 

 

when a=-1, b=-2

 

when a=-2, b=-1

 

  • 0 answers
  • 1 answers

Madhura Dighikar 5 years, 3 months ago

Hii
  • 1 answers

Yogita Ingle 5 years, 4 months ago

et the sum of first n terms of the A.P.=Sn
 Given: Sn = 3n2 – 4n    ...(i)
 Now,
Replacing n by (n –1) in (i), we get,
 Sn – 1 = 3(n – 1)2 – 4(n – 1)
 nth term of the A.P. an = Sn – Sn – 1
 ∴ an = (3n2 – 4n) – [3(n – 1)2 – 4(n – 1)]
⇒ an = 3 [ n2 – (n – 1)2] – 4 [n – (n – 1)]
⇒ an = 3 (n2 – n2 + 2n – 1) – 4 (n – n + 1)
⇒ an = 3(2n –1) – 4
⇒ an = 6n – 3 – 4
⇒ an= 6n – 7
Thus, the nth term of the A.P = 6n – 7.

  • 2 answers

Student ✍️✍️✍️ 5 years, 4 months ago

Ya thanks ???

Palak ? 5 years, 4 months ago

Baby, check on YouTube rashmi maths teacher channel u will get a great help from there..... Nd u will understand the ch Or questions in which u r getting confused..... ??HOPE MY ADVICE WILL HELP U.... ?
  • 5 answers

Palak ? 5 years, 4 months ago

Baby, if u r very week in maths then give near about an hour for solving one exercise of ncert nd do extra questions too of that ch...... HOPE YOU WILL SCORE 100 IN MATHS IN THIS CLASS BY FOLLOWING MY THESE INSTRUCTIONS...... ??

Riyana D'Cunha 5 years, 4 months ago

Practice daily for 30 minutes and practice solved example from NCERT textbook.

Student ✍️✍️✍️ 5 years, 4 months ago

Really how many hours

Dev Varshney 5 years, 4 months ago

Pado or padao

Ritesh Sharma 5 years, 4 months ago

In practice
  • 0 answers
  • 5 answers

Tarannum Jahan 5 years, 3 months ago

Maths topics are deleted I-NUMBER SYSTEMS REAL NUMBERS  Euclid’s division lemma UNIT II-ALGEBRA POLYNOMIALS  Statement and simple problems on division algorithm for polynomials with real coefficients. PAIR OF LINEAR EQUATIONS IN TWO VARIABLES  cross multiplication method QUADRATIC EQUATIONS Situational problems based on equations reducible to quadratic equations ARITHMETIC PROGRESSIONS Application in solving daily life problems based on sum to n terms UNIT III-COORDINATE GEOMETRY COORDINATE GEOMETRY  Area of a triangle. UNIT IV-GEOMETRY TRIANGLES Proof of the following theorems are deleted The ratio of the areas of two similar triangles is equal to the ratio of the squares of their corresponding sides.  In a triangle, if the square on one side is equal to sum of the squares on the other two sides, the angle opposite to the first side is a right angle. CIRCLES No deletion CONSTRUCTIONS  Construction of a triangle similar to a given triangle. UNIT V- TRIGONOMETRY INTRODUCTION TO TRIGONOMETRY  motivate the ratios whichever are defined at 0o and 90o TRIGONOMETRIC IDENTITIES Trigonometric ratios of complementary angles. HEIGHTS AND DISTANCES No deletion UNIT VI-MENSURATION AREAS RELATED TO CIRCLES  Problems on central angle of 120° SURFACE AREAS AND VOLUMES  Frustum of a cone. UNIT VI-STATISTICS & PROBABILITY STATISTICS  Step deviation Method for finding the mean  Cumulative Frequency graph PROBABILITY No deletion

Riyana D'Cunha 5 years, 4 months ago

http://cbseacademic.nic.in/Revisedcurriculum_2021.html Go on this link and select Math.

Sonu Rana 5 years, 4 months ago

Sir math ka syllabus kya board exam ka 2020 2021

Arpit Bhardwaj 5 years, 4 months ago

Search on youtube

Abhishek Singh 5 years, 4 months ago

30% topic Kam hua hai
  • 2 answers

Mansee Sharma 5 years, 4 months ago

Hum.....

Tilak Ladi 5 years, 4 months ago

Topic reomved for this year
  • 2 answers

Atul Gupta 5 years, 4 months ago

x^2+x-1

Student ✍️✍️✍️ 5 years, 4 months ago

x square +x-1
  • 1 answers

Gaurav Seth 5 years, 4 months ago

Given points are  A(-2, k), B(-2k , -3) and P(2, 2)
It is given that point P is the equidistant from the points A and B
So, PA = PB

Now,


Squaring on both sides
⇒ (- 2 - 2)2 + (k - 2)2 = (-2k - 2)2 + (-3 - 2)2
⇒ 16 + k2 + 4 - 4k = 4k2 + 4 + 8k + 25            [∵ ( a - b)2 = a2 - 2ab + b2  ]
⇒  4k2 + 4 + 8k + 25 -16 - k2 - 4 + 4k = 0
⇒ 3k2 + 12k + 9 = 0
⇒ k2 + 4k + 3 = 0
⇒ k2 + 3k + k + 3 = 0
⇒ k(k + 3) + 1(k + 3) = 0
⇒ (k + 3) (k + 1) = 0
⇒ (k + 3) = 0     or    (k + 1) = 0

∴  k = -1 or  -3.

  • 1 answers

Gaurav Seth 5 years, 4 months ago

Step-by-step explanation:

cos =  b/h and sin = b/h

LHS

Sin2 +cos2

={p/h}2 +[b/h]2

=p2 +b2 /h2

=h2/h2

=1

=RHS.......{...HENCE PROVED}

  • 2 answers

Tarannum Jahan 5 years, 4 months ago

X=-1,y=3

Pranali Solanke 5 years, 4 months ago

Get the variable with the smaller coefficient by itself: 2x+5y−13=02x+5y−13=0 ⟹2x=−5y+13⟹2x=−5y+13 ⟹2x=−4y+12+1−y⟹2x=−4y+12+1−y ⟹2x2=−4y 2+122+1−y2⟹2x2=−4y 2+122+1−y2 ⟹x=−2y+6+1−y2.⟹x=−2y+6+1−y2. Next letting t=1−y2t=1−y2 ⟹2t=1−y⟹2t=1−y ⟹y=1−2t.⟹y=1−2t. Let tt be an integer, which means that yy is an integer. Then x=−2y+6+1−y2x=−2y+6+1−y2 =−2(1−2t)+6+t=−2(1−2t)+6+t =−2+4t+6+t=−2+4t+6+t =4+5t.=4+5t. Check these two answers: 2x+5y−13=02x+5y−13=0 ⟹2(4+5t)+5(1−2t)−13⟹2(4+5t)+5(1−2t)−13 =8+10t+5−10t−13=8+10t+5−10t−13 =13+10t−10t−13=0✓=13+10t−10t−13=0✓ ∴∴ integer solutions to the given equation have the form x=4+5t, y=1−2tx=4+5t, y=1−2t where tt is an integer. So there is a countable infinitude of integer solutions to the given equation and more than 4.4. Choose integer values of tt to obtain specific answers. For example, t=10t=10 yields the answer x=4+5(10)=4+50=54, y=1−2(10)=1−20=−19.x=4+5(10)=4+50=54, y=1−2(10)=1−20=−19. Select four different values of tt to get four distinct solutions. To obtain positive answers for xx and y,y, must have x=4+5t>0 and y=1−2t>0x=4+5t>0 and y=1−2t>0 ⟹5t>−4 and 1>2t⟹5t>−4 and 1>2t ⟹t>−45 and t<12⟹t>−45 and t<12 ⟹−45<t<12⟹−45<t<12 ⟹t=0⟹t=0 since tt is an integer. Then the only positive solution is x=4+5(0)=4, y=1−2(0)=1.
  • 1 answers

Yogita Ingle 5 years, 4 months ago

Given 3 points A(2,3), B(5,k) and C(6,7) which are collinear we have to find the value of k.

Points are collinear if the slopes of any two pairs are equal.
Slope of AB = y2 -y2/x2 - x1= k-3/5-2
Slop of BC= y2 -y2/x2 - x1= 7-k/6-5

As lines are collinear slopes are equal

⇒k-3/5-2= 7-k/6-5

⇒ k- 3= 2(7-k)

⇒ k= 6

The value of k is 6.

 

  • 2 answers

Navami Mahato 5 years, 4 months ago

x²+5x+6=0 or, x²+3x+2x+6=0 or, x(x+3)+2(x+3)=0 or, (x+3)(x+2)=0 or, roots are, x+3=0 x= -3 And x-2=0 x=2 ? @ns...

Chandresh Tiwari 5 years, 4 months ago

X² +5x+6 X²+(6-1)x+6 X²+6x-1x+6 X(x+6)-1(x+6) X+6=0 & X-1=0 X=-6 & X= 1
  • 1 answers

Riyana D'Cunha 5 years, 4 months ago

Most of the questions come from ncert text.
  • 5 answers

Palak ? 5 years, 4 months ago

Ok baby....

Riyana D'Cunha 5 years, 4 months ago

Substitution method is not deleted, Cross multiplication method is deleted.

Gaurav Seth 5 years, 4 months ago

 

7x-2y=20          .... (1)

11x+15y = -23  .... (2)

Multiply equation (1) by 15 and (2) by 2, we get

105x - 30y = 300   ... (3)

22x  + 30y = -46   ... (4)

Adding (3) and (4), we get

127x = 254

→   x = 2

Substitute the value of x = 2 in (1), we get

7(2)-2y=20

→ 2y = -6

→   y = -3

 

Therefore x = 2 and y = -3.

Palak ? 5 years, 4 months ago

Dear,according to me this method has been deleted by cbse

Abhishek Rajawat 5 years, 4 months ago

Take equ. No 1
  • 1 answers

Riyana D'Cunha 5 years, 4 months ago

Let us assume, to the contrary, that 2-√3 is rational. That is , we can find coprime a and b (b≠0) such that 2-√3= a/b. Therefore 2- a/b= √3 Rearranging this equation, we get √3=2-a/b= 2b-a/b Since a and b are integers, we get 2-a/b is rational, and so √3 is rational. But this contradicts the fact that √3 is rational. This contradiction has arisen because of our wrong assumption that 2- √3 is rational. So, we conclude that 2-√3 is irrational. Hope it works.
  • 1 answers

Riya Mol 5 years, 4 months ago

X+y=7-------(1) 5x+12y=7-------(2) From equation (1) -----> X+y=7 =X=7-y Substitute equation (1) in equation (2) =5(7-y)+12y=7 =35 - 5y + 12y = 7 7y = 7 - 35 7y = -28 Y= -28/7 Y= -4 Substitute y= -4 in equation (1) X+(-4) =7 X -4 =7 X=7+4 X=11 X=(11) and y=(-4)
  • 2 answers

Riyana D'Cunha 5 years, 4 months ago

150= 2×5×3×5 175= 5×5×7 180= 2×2×3×3×5 HCF (150,175,180) = 5

Palak ? 5 years, 3 months ago

Dear, this is A question of 5th class nd u r in merit.... U must know how to find HCF nd if you don't know ask from child of 5th standard
  • 1 answers

Gaurav Seth 5 years, 4 months ago

We know, if A and B lies on circle and C is the centre of circle , then line joining of points A and B is intersected by Centre of circle.
e.g., O lies on midpoint of AB. means, we have to use midpoint section formula.

here, A (-1, y) and B(5,7) lies on circle with centre C(2,-3y) .
A--------------C------------------B
now, use midpoint section formula,
[if (x,y) is the midpoint of (x1,y1) and (x2, y2) then, x = (x1 + x2)/2 and y = (y1+y2)/2 ]
so, -3y = (y + 7)/2
=>-6y = y + 7
=> -7y = 7
=> y = -1

  • 1 answers

Gaurav Seth 5 years, 4 months ago

Let AB be the pole and BC be its shadow.

Given that AB/BC = 1/√3

Let ACB be the angle of elevation.

In triangle ABC,

X is equal to theta

tan x = AB/BC

tan x = 1/√3

tan x = tan 30

X = 30

 

Hence angle of elevation =30∘

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