No products in the cart.

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

Ask Question
  • 2 answers

........ ...... 5 years, 1 month ago

Here, a=2, b=k, c=3 D=b^2-4ac/2a =k^2-4×2×3/2×2 =K^2-24/4=0 k^2=24 K=√24 .......It may helpful to you.

Yogita Ingle 5 years, 1 month ago

We have to find the values of k for quadratic equations 2x² + kx + 3 = 0 so that they have two equal roots.

we know, quadratic equation will be equal only when

discriminant, D = b² - 4ac = 0

on comparing 2x² + kx + 3 = 0 with general form of quadratic equation , ax² + bx + c = 0 we get, a = 2, b = k and c = 3

so Discriminant , D = (k)² - 4(2)(3) = 0

or, k² - 24 = 0

or, k = ± √24 = ±2√6

hence, the value of k = 2√6 or -2√6

  • 0 answers
  • 1 answers

........ ...... 5 years, 1 month ago

Here, a=1,b=5,c=6 alpha+beta=-b/a =-5/1 Alpha+beta=-5,.......(i) And alpha×beta=c/a=6/1 Alpha×beta=6.......(ii) Now, 1/alpha+1beta+alpha×beta= Beta+alpha/alpha×beta+alpha×beta =-5/6+6 (from equation i and ii) =31/6 .......It may helpful to you
  • 5 answers

Manisha 5 years, 1 month ago

16/9-7/9 9/9 = 1 Right answer.....
16/9-7/9=9/9=1 hi hoga ?✌

Shashank Sen 5 years, 1 month ago

16/9-7/9=9/9=1 hai

Priyanshu ?✌??? 5 years, 1 month ago

16/9-7/9=9/9=1

........ ...... 5 years, 1 month ago

16/9-7/9=9/9=1
  • 5 answers

Abhinav Lal Karn 5 years, 1 month ago

-7

Priyanshu ?✌??? 5 years, 1 month ago

-7

........ ...... 5 years, 1 month ago

= -7
Khud hi question ask kiya aur khud hi answer de diya??????

Vishal Kumar Kurrey Maths Xi A 5 years, 1 month ago

=7
  • 3 answers
Yes leave it

Yogesh Thakur 5 years, 1 month ago

Leave it .

Sonam Kumari Paswan 5 years, 1 month ago

yes
  • 1 answers

Gaurav Seth 3 years, 7 months ago

Given : x is a rational number whose decimal expansion terminates . p&q are two integers in which prime Factorisation of q is of the form 2^m5^n where p&q are co-prime & non negative integer

 

To Find : How x can be expressed

 

Solution :

• Consider the theorm ,

Let x be a rational number whose decimal expansion terminates.

Then x can be expressed in the form of p/q , where p and q are coprime and the prime factorisation of q is of the form 2^n5^m

, where n, m are non-negative integers.

 

•According to theorm

X can be expressed in the form of p/q

 

•Hence , X can be expressed in the form of p/q

  • 1 answers

Gaurav Seth 5 years ago

sin theta=p/h=4/5 so we should find b from this equation

b=✓h²-p²

✓5²-4²

✓25-16

✓9=3

 

 

 

4tan theta - 5 cos theta by sec theta + 4 cot theta

4×p/b - 5×b/h divide by h/b + 4×b/p

4×4/3 - 5×3/5 divide by 5/3 + 4×3/4

16/3-3 by 5/3+ 3

7/3 by 14/3

3_3cancel

7/14=1/2

  • 5 answers

Shubham Saxena ???? 5 years, 1 month ago

2x - 3 = 25 ,, 2x = 25 + 3 ,, x = 28 / 2 ,, x = 14 ......
2x-3=25 =》2x=25+3 =》x=28/2 =》x=14

Abhinav Lal Karn 5 years, 1 month ago

2x=25+3=》2x=28=》x=28/2=》x=14

Priyanshu ?✌??? 5 years, 1 month ago

(2x-3)=25 2x=25+3 =x=28/2 =x=14

........ ...... 5 years, 1 month ago

2x-3=25 2x=25+3 x=28/2 x=14
  • 5 answers

Shubham Saxena ???? 5 years, 1 month ago

34² = 1156 .

Abhinav Lal Karn 5 years, 1 month ago

(34)^2=1156

Priyanshu ?✌??? 5 years, 1 month ago

1156

Kartik Yadav 5 years, 1 month ago

1156

Soham Shah 5 years, 1 month ago

1156
  • 0 answers
  • 1 answers

Hiya Suthar 5 years, 1 month ago

0
  • 5 answers

Sharma G 5 years, 1 month ago

Yes I know CBSE will release sample in the end of year .. Probably November...

Anushka Mittal 5 years, 1 month ago

There is no such information given by CBSE .

Priyanshu ?✌??? 5 years, 1 month ago

Don't know
Sorry but I don't know ??

Shreya Raj 5 years, 1 month ago

No
  • 2 answers

Manpreet Yadav 5 years, 1 month ago

Let A=45°,B=30° (so that sum of A and B =75° and we can get Sin74°) (put the value of A and B in equation) - Sin75°=Sin45°*Cos30° + Cos 45°*Sin30° - Sin75° = 1/√2*√3/2 + 1/√2*1/2 - Sin75° = √3/2√2 + 1/2√2 - Sin75° = (√3+1)/2√2 Sin75° = (√6+√2)/4

Utsav Kumar 5 years, 1 month ago

sin ( A+ B) = sin75° Let A = 45° and B = 30° Then sin75° = sin45° cos30° + cos45° sin30° = 1/√2 * √3/2 + 1/√2 * 1/2 = √3 / 2√2 + 1/2√2 = √3 + 1/2√2 You can rationalise the denominator.
  • 3 answers

Parth Gupta 5 years, 1 month ago

2x^2-5x-7x=0 2x^2-5x-7 (2x^2-7x)+(2x-7) x(2x-7)+1(2x-7) (x+1)(2x-7)

Gaurav Seth 5 years, 1 month ago

Answer:the roots are not real

Step-by-step explanation:

comparing the equation in ax^2+bx+c=0

a=2

b=-5

c=7

Delta =b^2-4ac

         =25-4 x 2 x 7

         =25-56

         =   -31

It describes that the roots are imaginary or do not exist .

Gaurav Seth 5 years, 1 month ago

Answer:the roots are real and equal

Step-by-step explanation:

comparing the equation in ax^2+bx+c=0

a=2

b=-5

c=7

Delta =b^2-4ac

         =25-4 x 2 x 7

         =25-56

         =   -31

It describes that the roots are imaginary or do not exist .

  • 3 answers

Utsav Kumar 5 years, 1 month ago

Sorry for wrong answer the answer is. √3/2

Utsav Kumar 5 years, 1 month ago

Sin60° = √3/3

........ ...... 5 years, 1 month ago

Sin 60°=√3/2
  • 1 answers

Gaurav Seth 5 years, 1 month ago

ANSWER
Empirical relationship between mean, median and mode is:
  Mode = 3 Median - 2 Mean
⇒ Mode - Mean = 3 Median - 2 Mean - Mean
⇒ Mode - Mean = 3 Median - 3 Mean
⇒ Mode - Mean = 3 [Median - Mean]

  • 2 answers

Ansh Tyagi 5 years, 1 month ago

It will end after 3 decimal places u should use the trick to do it instead of dividing

Gaurav Seth 5 years, 1 month ago

Q u e s t i o n :After how many palces will the rational number 1251/125 terminate
A n s w e r :
1251 / 1250 = 1.0008

After 4 decimal places the rational number end.

  • 3 answers

Priyanshu ?✌??? 5 years, 1 month ago

4x^2+3x+7 =4x^2+7x-4x+7 =4x(x+7/4)-4(x+7/4) =(4x-4) or (x+7/4)=0 =x=1 or x=-7/4

Manpreet Yadav 5 years, 1 month ago

1,-7/4

Parth Gupta 5 years, 1 month ago

4x^2 +3x+7 (4x^2+4x)+(7x+7) 4x(x+1)+7(x+1) (4x+7)(x+1)
  • 1 answers

Gaurav Seth 5 years, 1 month ago

<font face="Times New Roman, serif"><font style="box-sizing: border-box;"><font size="3"><font style="box-sizing: border-box;">(i) We have,</font></font></font></font>

<font color="#808080"><font style="box-sizing: border-box;"></font></font>

<font face="Times New Roman, serif"><font style="box-sizing: border-box;"><font size="3"><font style="box-sizing: border-box;">Theorem states: </font></font></font></font>

<font face="Times New Roman, serif"><font style="box-sizing: border-box;"><font size="3"><font style="box-sizing: border-box;">Let </font></font></font></font><font face="Times New Roman, serif"><font style="box-sizing: border-box;"><font size="3"><font style="box-sizing: border-box;">be a rational number, such that the prime factorization of q is not of the form</font></font>, where <font size="3"><font style="box-sizing: border-box;">m and n are non-negative integers.</font></font></font></font>

<font face="Times New Roman, serif"><font style="box-sizing: border-box;"><font size="3"><font style="box-sizing: border-box;">Then, x has a decimal expression which does not have terminating decimal.</font></font></font></font>

<font face="Times New Roman, serif"><font style="box-sizing: border-box;"><font size="3"><font style="box-sizing: border-box;">(ii) We have,</font></font></font></font>

<font color="#808080"><font style="box-sizing: border-box;"></font></font>

<font face="Times New Roman, serif"><font style="box-sizing: border-box;"><font size="3"><font style="box-sizing: border-box;">Theorem states: </font></font></font></font>

<font face="Times New Roman, serif"><font style="box-sizing: border-box;"><font size="3"><font style="box-sizing: border-box;">Let </font></font></font></font><font face="Times New Roman, serif"><font style="box-sizing: border-box;"><font size="3"><font style="box-sizing: border-box;">be a rational number, such that the prime factorization of q is not of the form</font></font>, where <font size="3"><font style="box-sizing: border-box;">m and n are non-negative integers.</font></font></font></font>

<font face="Times New Roman, serif"><font style="box-sizing: border-box;"><font size="3"><font style="box-sizing: border-box;">Then, x has a decimal expression which does not have terminating decimal.</font></font></font></font>

<font face="Times New Roman, serif"><font style="box-sizing: border-box;"><font size="3"><font style="box-sizing: border-box;">(iii) We have,</font></font></font></font>

<font color="#808080"><font style="box-sizing: border-box;"></font></font>

<font face="Times New Roman, serif"><font style="box-sizing: border-box;"><font size="3"><font style="box-sizing: border-box;">Theorem states: </font></font></font></font>

<font face="Times New Roman, serif"><font style="box-sizing: border-box;"><font size="3"><font style="box-sizing: border-box;">Let </font></font></font></font><font face="Times New Roman, serif"><font style="box-sizing: border-box;"><font size="3"><font style="box-sizing: border-box;">be a rational number, such that the prime factorization of q is not of the form</font></font>, where <font size="3"><font style="box-sizing: border-box;">m and n are non-negative integers.</font></font></font></font>

<font face="Times New Roman, serif"><font style="box-sizing: border-box;"><font size="3"><font style="box-sizing: border-box;">Then, x has a decimal expression which does not have terminating decimal.</font></font></font></font>

<font face="Times New Roman, serif"><font style="box-sizing: border-box;"><font size="3"><font style="box-sizing: border-box;">(iv) We have,</font></font></font></font>

<font color="#808080"><font style="box-sizing: border-box;"></font></font>

<font face="Times New Roman, serif"><font style="box-sizing: border-box;"><font size="3"><font style="box-sizing: border-box;">Theorem states: </font></font></font></font>

<font face="Times New Roman, serif"><font style="box-sizing: border-box;"><font size="3"><font style="box-sizing: border-box;">Let </font></font></font></font><font face="Times New Roman, serif"><font style="box-sizing: border-box;"><font size="3"><font style="box-sizing: border-box;">be a rational number, such that the prime factorization of q is of the form</font></font>, where <font size="3"><font style="box-sizing: border-box;">m and n are non-negative integers.</font></font></font></font>

<font face="Times New Roman, serif"><font style="box-sizing: border-box;"><font size="3"><font style="box-sizing: border-box;">Then, x has a decimal expression which terminates after k places of decimals, where k is the larger of m and n.</font></font></font></font>

<font face="Times New Roman, serif"><font style="box-sizing: border-box;"><font size="3"><font style="box-sizing: border-box;">Then, x has a decimal expression which will have terminating decimal after 3 places of decimal.</font></font></font></font>

 

  • 1 answers

Mk Sharma 5 years, 1 month ago

0
  • 1 answers

........ ...... 5 years, 1 month ago

Let the no at tenths place= x And the no at ones place=y :. the no will be (10x+y) A/Q 7(10x+y)=4(10y+x) 70x+7y=40y+4x 66x=33y x=y/2 Given the difference between them is 3. Since x is less then y-x=3 Put the value of x y-y/2=3 y=6 : . x=3 Hence, the original no is 36 and the reverse no is 63.

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