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  • 1 answers

Sia ? 6 years ago

Let PM be the light house of height 200 m and let A and B be two ships on the either sides of lighthouse such that the angles of depression of A and B are 60o and 45o respectively.

Let, AM = x m and BM=y m

Then, XPB=MBP=45o [alternate angle]

and YPA=MAP=60o [alternate angles]

In right-angled AMP, tan60o=PB=PMAM

3=200x

x=2003m

In right-angled BMP, tan45o=PMBM

1=200y

y=200m

Now, distance between the two ships = AB = x+y=200+2003=315.48m

  • 2 answers

Gaurav Chilkoti 6 years, 3 months ago

I know

Mansi ??⚘?? 6 years, 6 months ago

Search on cbsesqp 2018-19 (CBSE Sample Question Paper)
  • 4 answers

Abc . 6 years, 6 months ago

CBSE is not announced yet.

Riyam Arora 6 years, 6 months ago

6 march

Bhoomi Gupta 6 years, 6 months ago

From 5march to 4 april

S Sihag Ji 6 years, 6 months ago

In March
  • 1 answers

Sia ? 6 years ago

Since, α and β are the zeroes of the quadratic polynomial
f(x) = x2 - p (x+1) - c=x2-px-(p+c)
So A=1 B=-p,C=-(p+c)
Sum of the zeroes α + β = BA=P
Product of the zeroes αβ =CA=- (p + c)
(α + 1)(β + 1)
= αβ + α + β + 1
= αβ + (α + β) + 1
=-(p+c)+p+1
=-p-c-p+1
=1-c
Hence proved

  • 0 answers
  • 3 answers

Puja Sahoo 6 years, 6 months ago

To tum meri party ke ho....

Abc . 6 years, 6 months ago

Me ?

Santosh Parida 6 years, 6 months ago

Ham. Chahte hain par sab jaate hain isliye hame bhi jana pad raha hai
  • 2 answers

Puja Sahoo 6 years, 6 months ago

It is called as emperical formula.....

Soham Deshpande 6 years, 6 months ago

3MEDIAN=MODE+2MEAN
  • 1 answers

S Sihag Ji 6 years, 6 months ago

See in NCERT Book
  • 3 answers

Riyam Arora 6 years, 6 months ago

ax2+bx+c=0

Pratibha Sahu 6 years, 6 months ago

ax²+bx+c=0

Sumant Adky 6 years, 6 months ago

ax^2+c=0
  • 2 answers

Abhinav Yadav 6 years, 6 months ago

135 cm2

Aditya Gupta 6 years, 6 months ago

Area of the given Triangle =1/2×base×height =( 1/2×25×10.8 )cm2 =135cm2. Answer
  • 2 answers

Jatin Gera 6 years, 6 months ago

Area of equilateral triangal = root3/4 a² Root3/4a² = 36root3 Root 3 and root3 will cancel. a²=36x4 a²=144 a=12 Perimeter of equilateral triangle = 3a P=3x12 = 36.

Abhinav Yadav 6 years, 6 months ago

36 cm
  • 2 answers

Kalash Maan 6 years, 6 months ago

Thanks Abhishek

Abhishek Kandari 6 years, 6 months ago

π(R+r)l
  • 4 answers

Akshita Wadhwa 6 years, 6 months ago

4/√3

Shaurya Gupta 6 years, 6 months ago

4√3/3

Kalash Maan 6 years, 6 months ago

2/√3

Chētnà Pandey✌️ 6 years, 6 months ago

3/√3
  • 2 answers

Samridhi Yadav 6 years, 6 months ago

X = a+b Aur X= a-b

Shaurya Gupta 6 years, 6 months ago

x=a-b and x = a+b
  • 3 answers

Shaurya Gupta 6 years, 6 months ago

K=5

Abhishek Kandari 6 years, 6 months ago

K(2)² - 14(2)+8=0 4k-28+8=0 4k-20=0 4k=20 K=5

Puja Sahoo 6 years, 6 months ago

K=24.......
  • 1 answers

Riya Tulsani 6 years, 6 months ago

Let unit digit no. Be X Let tens digiti no. Be Y Since the digits differ by 3 Therefore , x-y=3 .......(1) And let the original number is 10y + x Now, Number obtained by reversing this , 10x+y According to ques , 10y + x + 10x + y =165 11x + 11y = 165 x+y= 15........(2) Then , by eliminating (1) & (2) x=6 & y=3 So the number is 36 or 63
  • 1 answers

Jatin Gera 6 years, 6 months ago

THE FUNDAMENTAL THEOREM OF ARITHMETICS STATES THAT EACH AND EVERY COMPOSITE NUMBER CAN BE WRITTEN AS THE PRODUCT OF PRIME NUMBERS IN ONE AND ONLY ONE WAY EXCEPT THE ORDER OF NUMBERS.
  • 5 answers

Pappu?? Shikari?? 6 years, 6 months ago

Gall ni kadni

Puja Sahoo 6 years, 6 months ago

Confidence is good but overconfidence is not good , anyways wishing all of you best of luck for your exams...do your best????

Rosie Ruby 6 years, 6 months ago

In sha allah

Amira Roy❤❤ 6 years, 6 months ago

I think it's me ?? ?

Abc . 6 years, 6 months ago

Inshallah we all try our best..
  • 1 answers

Karuna Kashyap 6 years, 6 months ago

See in ncert book page no.182
  • 1 answers

Neha Jangir 6 years, 6 months ago

under root.x^2 +y^2
  • 0 answers
  • 1 answers

Abhinav Yadav 6 years, 6 months ago

Start with ax^2 + bx + c=0 Divide the equation by a x^2 + bx/a + c/a = 0 Put c/a on other side x^2 + bx/a = -c/a Add (b/2a)2 to both sides x^2 + bx/a + (b/2a)^2 = -c/a + (b/2a)^2 The left hand side is now in the x2 + 2dx + d2 format, where "d" is "b/2a" So we can re-write it this way: "Complete the Square" (x+b/2a)^2 = -c/a + (b/2a)^2 Now x only appears once and we are making progress. Now Solve For "x" Now we just need to rearrange the equation to leave "x" on the left Start with (x+b/2a)^2 = -c/a + (b/2a)^2 Square root (x+b/2a) = (+-) sqrt(-c/a+(b/2a)^2) Move b/2a to right x = -b/2a (+-) sqrt(-c/a+(b/2a)^2) That is actually solved! But let's simplify it a bit: Multiply right by 2a/2a x = [ -b (+-) sqrt(-(2a)^2 c/a + (2a)^2(b/2a)^2) ] / 2a Simplify: x = [ -b (+-) sqrt(-4ac + b^2) ] / 2a
  • 0 answers

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