No products in the cart.

Ask questions which are clear, concise and easy to understand.

Ask Question
  • 5 answers

Seenu B 4 years, 9 months ago

2

Rajneesh Payal 4 years, 9 months ago

2

H. K 4 years, 9 months ago

3x-5x^5+8y

Kap Sharma 4 years, 9 months ago

Tan theta + cot theta = 2

Samarth Suryavamshi 4 years, 9 months ago

2
  • 2 answers

Kap Sharma 4 years, 9 months ago

P/2= -3/2 (cross multiply) -2p = 2×3 -2p = 6 P = 6/-2 P = -3 ?

Kap Sharma 4 years, 9 months ago

P = -3
  • 2 answers

Ansh Rathore 4 years, 9 months ago

N = 10

Yogita Ingle 4 years, 9 months ago

27,24,21....

d=24-27=-3 a=27

let nth term=0

a+(n-1)d=0

27+(n-1)(-3) =0

(n-1)(-3)=-27

n-1=9

n=10

  • 1 answers

Keshav Yadav 4 years, 9 months ago

5PX = 2QX so, PX/QX = 2/5 As QR || XY, So, PX/QX = PY/RY Thus, 2/5 = 6/RY By cross multiplication : 2RY = 30 RY = 15 cm ( I hope this is correct ✔️)
  • 2 answers

Angad Deep 4 years, 9 months ago

But how

Kap Sharma 4 years, 9 months ago

More than 3
  • 1 answers

Yogita Ingle 4 years, 9 months ago

Let ΔABC and ΔPQR be two similar triangles. AD and PM are the medians of ΔABC and ΔPQR respectively.  

  • 1 answers

Sumesh ☺️☺️☺️ 4 years, 9 months ago

What
  • 2 answers

Himanshu Kumar 4 years, 9 months ago

Write full question please

Simran Naaz 4 years, 9 months ago

Cbse has remove this topic
  • 0 answers
  • 2 answers

Alisha. Afreen 4 years, 9 months ago

Give me solutiin

Jadala Krithika 4 years, 9 months ago

Ok
  • 0 answers
  • 2 answers

Alisha. Afreen 4 years, 9 months ago

Thanks ?

Gaurav Seth 4 years, 9 months ago

Width of the track = 10 m

Distance between two parallel lines = 60 m

Length of parallel tracks = 106 m

DE = CF = 60 m

Radius of inner semicircle, r = OD = O’C

= 60/2 m = 30 m

Radius of outer semicircle, R = OA = O’B

= 30+10 m = 40 m

Also, AB = CD = EF = GH = 106 m

Distance around the track along its inner edge = CD+EF+2×(Circumference of inner semicircle)

= 106+106+(2×πr) m = 212+(2×22/7×30) m

= 212+1320/7 m = 2804/7 m

Area of the track = Area of ABCD + Area EFGH + 2 × (area of outer semicircle) – 2 × (area of inner semicircle)

= (AB×CD)+(EF×GH)+2×(πr2/2) -2×(πR2/2) m2

= (106×10)+(106×10)+2×π/2(r2-R2) m2

= 2120+22/7×70×10 m2

= 4320 m2

  • 2 answers

Rajneesh Payal 4 years, 9 months ago

Rewrite the equation as: (x + 1)(x + 4)(x + 3)(x + 2) = 120 Multiply the first 2 and last 2 expressions:  ..... (1) Let  (1) becomes, (y + 4)(y + 6) = 120 y2 + 10x + 24 = 120 y2 + 10x - 96 = 0 y2 + 16x - 6y - 96 = 0 y(y + 16) - 6(y + 16) = 0 (y + 16)(y - 6) = 0 y = -16 or 6 y2+ 5x = 16 or 6 When x2 + 5x = 6 x2 + 5x - 6 = 0 x2 + 6x - x - 6 = 0 x(x + 6) - 1(x + 6) = 0 (x + 6)(x - 1) = 0 x = -6, 1 When x^2 + 5x = -16 x^2 + 5x + 16 = 0 Using quadratic These are all the possible solutions. But the only real solutions are -6 and 1.

Yogita Ingle 4 years, 9 months ago

Rewrite the equation as:

(x + 1)(x + 4)(x + 3)(x + 2) = 120

Multiply the first 2 and last 2 expressions:

{tex}(x^2 + 5x + 4)(x^2 + 5x + 6) = 120{/tex} ..... (1)

Let {tex}x^2 + 5x = y{/tex}

(1) becomes,

(y + 4)(y + 6) = 120

y2 + 10x + 24 = 120

y2 + 10x - 96 = 0

y2 + 16x - 6y - 96 = 0

y(y + 16) - 6(y + 16) = 0

(y + 16)(y - 6) = 0

y = -16 or 6

y2+ 5x = 16 or 6

When x2 + 5x = 6

x2 + 5x - 6 = 0

x2 + 6x - x - 6 = 0

x(x + 6) - 1(x + 6) = 0

(x + 6)(x - 1) = 0

x = -6, 1

When x^2 + 5x = -16

x^2 + 5x + 16 = 0

Using quadratic formula:

{tex}x = [-5 +/- sqrt (25 - 64)]/2{/tex}

The solutions are, -6, 1, [-5 - sqrt (25 - 64)]/2 and [-5 + sqrt (25 - 64)]/2

These are all the possible solutions. But the only real solutions are -6 and 1.

  • 1 answers

Yogita Ingle 4 years, 9 months ago

Given : Line segment joining the points (6,4)and(1,-7) is divided by x axis.

To find : The ratio in the line segment and the coordinates of point of division?

Solution :

Let the line segment points A=(6,4) and B=(1,-7)

Let the line segment divide by x-axis with point P=(x,0)

Let the ratio in which line segment divide be m:n=k : 1

Applying section formula,

Substitute the values,

Compare the y-coordinate,

So, The ratio is 4 :7.

Compare the x-coordinate,

Put the value of k,

So, The coordinate of point of division is 

  • 2 answers

Raja Gupta 4 years, 9 months ago

Yes , right answer

Yogita Ingle 4 years, 9 months ago

Given:- A circle with center O,PA and PB are tangents drawn at ends A and B on chord AB.

To prove:- ∠PAB=∠PBA

Construction:- Join OA and OB

Proof:- In △AOB, we have

OA=OB(Radii of the same circle)

∠OAB=∠OBA.....(1)(Angles opposite to equal sides)

∠OAP=∠OBP=90(∵Radius⊥Tangent)

⇒∠OAB+∠PAB=∠OBA+∠PBA

⇒∠OAB+∠PAB=∠OAB+∠PBA(From (1))

⇒∠PAB=∠PBA

Hence proved.

  • 1 answers

Gaurav Seth 4 years, 9 months ago

Polynomial   P(x) = x² - 6 x + a
Given  α and  β  are the roots.     To find a , if 3 α + 2 β = 20.  ---(1)

From the quadratic expression:
    α + β = 6       ---(2)
    and   α β = a  --- (3)

Multiply equation (2) by 2 and subtract from (1) to get:
    α = 20 -12 = 8

Substitute this value in(2) to get:    
     β = 6-8 = -2   

Substitute these in (3) to get:    a = α β = -16.

  • 1 answers

Gaurav Seth 4 years, 9 months ago

Let the tens place digit be x.

And the units place digit be y.

According to the Question,

⇒ xy = 18

⇒ y = 18/x .....(i)

And, (10x + y) - 63 = 10y + x

⇒ 9x - 9y = 63

⇒ x - y = 7 .... (ii)

Putting y's value in Eq (ii), we get

⇒ x - 18/x = 7

⇒ x² - 18 = 7x

⇒ x² - 7x - 18 = 0

⇒ x² - 9x + 2x - 18 = 0

⇒x(x - 9) + 2(x - 9) = 0

⇒ (x - 9) (x + 2) = 0

⇒ x - 9 = 0 or x + 2 = 0

⇒ x = 9, - 2 (As x can't be negative)

⇒ x = 9

Putting x's value in Eq (i), we get

⇒ xy = 18

⇒ 9y = 18

⇒ y = 18/9

⇒ y = 2

Number = 92

Hence, the required number is 92.

  • 0 answers
  • 2 answers

Vivek Pandey 4 years, 9 months ago

Thank you very much

Yogita Ingle 4 years, 9 months ago

Let the length and breadth of the park be l and b.
Perimeter = 2 (l + b) = 80
l + b = 40
Or, b = 40 - l
Area = l×b = l(40 - l) = 40l - l240l -  l2 = 400
l2 -  40l + 400 = 0
Comparing this equation with al2 + bl + c = 0, we get
a = 1, b = -40, c = 400
Discriminant = b2 - 4ac
(-40)2 - 4 × 400
= 1600 - 1600 = 0
b2 - 4ac = 0
Therefore, this equation has equal real roots. And hence, this situation is possible.
Root of this equation,l = -b/2a
l = (40)/2(1) = 40/2 = 20
Therefore, length of park, l = 20 m
And breadth of park, b = 40 - l = 40 - 20 = 20 m.

  • 0 answers
  • 2 answers

Tanmay Nagar 4 years, 9 months ago

A=0°

Yogita Ingle 4 years, 9 months ago

sin2A=2sinA

2sinA.cosA-2sinA =0

2sinA(cosA-1)=0

either 2sinA=0

sinA=0

sinA=sin0°

A= 0°

or cosA -1=0

cosA=1

cosA=cos 0°

A=0° , Answer

  • 1 answers

Tanmay Nagar 4 years, 9 months ago

5×593
  • 0 answers
  • 1 answers

Yogita Ingle 4 years, 9 months ago

Odd number always end in one of the digits 

1, 3, 5, 7, 9

  • 2 answers

Bhaskar Gusain 4 years, 9 months ago

150 degree

Bhoomika Gugnani 4 years, 9 months ago

60°

myCBSEguide App

myCBSEguide

Trusted by 1 Crore+ Students

Test Generator

Test Generator

Create papers online. It's FREE.

CUET Mock Tests

CUET Mock Tests

75,000+ questions to practice only on myCBSEguide app

Download myCBSEguide App