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  • 2 answers

Ketan Bapna 4 years, 4 months ago

Jskekwksjsbzbxjxskekwkskkwjsjdjdjkrjrkejwjsjjs...... Is your answer

Avneet Kaur 4 years, 4 months ago

What's it....don't know english or what ???
  • 1 answers

Avneet Kaur 4 years, 4 months ago

Idk
  • 0 answers
  • 0 answers
  • 1 answers

Sumesh Sharma 4 years, 4 months ago

There is no option for pdf,
  • 1 answers

Sia ? 4 years, 3 months ago

In algebra, polynomial long division is an algorithm for dividing a polynomial by another polynomial of the same or lower degree, a generalized version of the familiar arithmetic technique called long division. It can be done easily by hand, because it separates an otherwise complex division problem into smaller ones.

  • 1 answers

Preeti Dabral 4 years, 4 months ago

suppose √8 = a/b with integers a, b
and gcd(a,b) = 1 (meaning the ratio is simplified)

then 8 = a²/b²
and 8b² = a²

this implies 8 divides a² which also means 8 divides a.

so there exists a p within the integers such that:
a = 8p
and thus,
√8 = 8p/b
which implies
8 = 64p²/b²
which is:
1/8 = p²/b²
or:
b²/p² = 8
which implies
b² = 8p²
which implies 8 divides b² which means 8 divides b.

8 divides a, and 8 divides b, which is a contradiction because gcd (a, b) = 1
therefore, the square root of 8 is irrational.

  • 3 answers

Preeti Dabral 4 years, 4 months ago

second number = 153

Explanation:

Let first number = a,

second number = b

a = 27 ( given )

HCF(a,b) = 9

and

LCM (a,b) = 459

To find:

Value of b

solution:

We know that,

$\implies b = \frac{9\times 459}{27}$

After cancellation, we get

$\implies b = 153$

Therefore,

second number = b = 153

Anjali Malviya 4 years, 4 months ago

The other number is 153.

Shamim Banu 4 years, 4 months ago

Lesson 15 exercise 15.1
  • 2 answers

Sia ? 4 years, 4 months ago

let a be any positive integer
a=6q+r where 0< or equal to r <6
put r=1: a=6q+1 =odd integer
r=3: a=6q+3=odd integer
r=5: a=6q+5=odd integer
therefore,any positive odd integer is of the form 6q+1,6q+3 or 6q+5,where q is some integer

Leitanthem Denel [Youtube] 4 years, 4 months ago

r=1
  • 2 answers

Sia ? 4 years, 4 months ago

a sequence of numbers in which each differs from the preceding one by a constant quantity (e.g. 1, 2, 3, 4, etc.; 9, 7, 5, 3, etc.).

Aditya Sharma 4 years, 4 months ago

An Arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. ... The sum of a finite arithmetic progression is called an arithmetic series.
  • 1 answers

Leitanthem Denel [Youtube] 4 years, 4 months ago

In any quadratic polynomial: The sum of the zeroes is equal to the negative of the coefficient of x by the coefficient of x2. The product of the zeroes is equal to the constant term by the coefficient of x2
  • 2 answers

Sia ? 4 years, 4 months ago

95

Anmol Preet 4 years, 4 months ago

95
  • 2 answers

Avneet Kaur 4 years, 4 months ago

Thanks

Mann Goyal 4 years, 4 months ago

All the best ???
  • 1 answers

Sia ? 4 years, 4 months ago

Let assume that 7-2root3 be rational and rational numbers are in the form of p/q form.

We know that root 3 is irrational and not p/q form. It contradicts the statement that it is irrational as it is p/q form.
Subtraction and multiplication of irrational number form irrational number so 7-2root3 is irrational.

  • 1 answers

Sia ? 4 years, 4 months ago

root 50, root 72

  • 1 answers

Bhavesh B Nair 4 years, 4 months ago

S
  • 1 answers

Sia ? 4 years, 4 months ago

Lets take assumed mean, A as 16

Class interval =2

∴ui​=hxi​−A​=2xi​−16​

Now, we have A=16,xˉ=18 and h=2

We know that,

Mean, xˉ=A+h(N1​∑fi​ui​)

⇒18=16+2(f+442f+24​)

⇒f+442f+24​=1

⇒f+44=2f+24

⇒f=20

Hence, the value of missing frequency, f=20

  • 2 answers

Pooja Bora 4 years, 4 months ago

SinA= cotA/cosA. SecA= cot A×SineA Tan A =1/cotA

Gouri Yaswanth Kumar 4 years, 4 months ago

To convert the given trigonometric ratios in terms of cot functions, use trigonometric formulas We know that, cosec2A – cot2A = 1 cosec2A = 1 + cot2A Since cosec function is the inverse of sin function, it is written as 1/sin2A = 1 + cot2A Now, rearrange the terms, it becomes sin2A = 1/(1+cot2A) Now, take square roots on both sides, we get sin A = ±1/(√(1+cot2A) The above equation defines the sin function in terms of cot function Now, to express sec function in terms of cot function, use this formula sin2A = 1/ (1+cot2A) Now, represent the sin function as cos function 1 – cos2A = 1/ (1+cot2A) Rearrange the terms, cos2A = 1 – 1/(1+cot2A) ⇒cos2A = (1-1+cot2A)/(1+cot2A) Since sec function is the inverse of cos function, ⇒ 1/sec2A = cot2A/(1+cot2A) Take the reciprocal and square roots on both sides, we get ⇒ sec A = ±√ (1+cot2A)/cotA Now, to express tan function in terms of cot function tan A = sin A/cos A and cot A = cos A/sin A Since cot function is the inverse of tan function, it is rewritten as tan A = 1/cot A
  • 3 answers

Preeti Dabral 4 years, 4 months ago

2x+3y=-5 (1)

3x-2y=12 (2)

Add eq 1 and 2 we get

5x+1y=7 (3)

Now

Subtract 1 from 2,we get

X-5y =17 (4)

Multiply 3 by 5 we get,

25x+5y=35 (5)

Now ,

Add 4 and 5 we get,

26x=52

X=2

now ,

Substitute x=2 in eq3

5x+1y=7

5×2+1y=7

10+y=7

y=7-10

y=-3

(x,y)=(2,-3)

Gaurav Singh 4 years, 4 months ago

Thanks

Ashish Raj 4 years, 4 months ago

https://doubtnut.app.link/krNEmWdGxhb
  • 1 answers

Preeti Dabral 4 years, 4 months ago

p(x)=3x²-2k+1

zeroes=α,7α

sum of zeroes= - b/a

α+7α= -(2k/3)

8α=2k/3

8α×3=2k

24∝=2k

k=12∝    (1)

product of zeroes= c/a

∝×7∝=1/3

7∝²=1/3

∝²=1/(3×7)

∝²=1/21

∝=±1/√21

∵∝=±1/√21

∴k=12(±1/√21)

k = ±12/√21

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