Three digit number which leaves remainder 2 when divided by 5 is
102,107,112,117......997
This form an AP whose first term is =102, d=5
Let 997 is the n th term of AP i.e. an​=997
a_n​=a+(n−1)d
997=102+(n−1)5
5n=900
n=180
Sum of all three digits which leaves remainder 2 when divided by 5 is
S_n​=n/2 ​[2a+(n−1)d]
=180/2 ​[2×102(180−1)5]
=90[204+179×5]
=90[204+895]
=90×1099

=98910

HRD Minister Ramesh Nishank announced a major CBSE syllabus reduction for the new academic year 2020-21 on July 7 which was soon followed by an official notification by CBSE on the same.

Considering the loss of classroom teaching time due to the Covid-19 pandemic and lockdown, CBSE reduced the syllabus of classes 9 to 12 with the help of suggestions from NCERT.

The CBSE syllabus has been rationalized keeping intact the learning outcomes so that the core concepts of students can be retained.

Deleted syllabus of CBSE Class 10 Mathematics

Tan theta=4/3
tan theta=p/b
p=4,and b=3

by paithagoras theorem,
H=root under P^2+B^2
H=root under 4^2+3^2
H=root under 16+9
H=root under 25
H=5

sin theta=p/h
sin theta=4/5
cos theta=b/h
cos theta=3/5

sin theta+cos theta/sin theta-cos theta
=4/5+3/5 by 4/5-3/5
=4+3/5by4-3/5
=7/5by1/5
=7/5×5/1
=7/1
=7

According to Euclid’s Division Lemma if we have two positive integers a and b, then there exist unique integers q and r which satisfies the condition a = bq + r where 0 ≤ r ≤ b.

HCF is the largest number which exactly divides two or more positive integers.

Since 12576 > 4052

12576 = (4052 × 3) + 420

420 is a reminder which is not equal to zero (420 ≠ 0).

4052 = (420 × 9) + 272

271 is a reminder which is not equal to zero (272 ≠ 0).

Now consider the new divisor 272 and the new remainder 148.

272 = (148 × 1) + 124

Now consider the new divisor 148 and the new remainder 124.

148 = (124 × 1) + 24

Now consider the new divisor 124 and the new remainder 24.

124 = (24 × 5) + 4

Now consider the new divisor 24 and the new remainder 4.

24 = (4 × 6) + 0

Reminder = 0

Divisor = 4

HCF of 12576 and 4052 = 4.

Let the first number be x and the second number is 27 - x.
Therefore, their product = x (27 - x)
It is given that the product of these numbers is 182.
Therefore, x(27 - x) = 182
⇒ x² – 27x + 182 = 0
⇒ x² – 13x - 14x + 182 = 0
⇒ x(x - 13) -14(x - 13) = 0
⇒ (x - 13)(x -14) = 0
Either x = -13 = 0 or x - 14 = 0
⇒ x = 13 or x = 14
If first number = 13, then
Other number = 27 - 13 = 14
If first number = 14, then
Other number = 27 - 14 = 13
Therefore, the numbers are 13 and 14.

Applying Euclid 's division algorithm,

H.C.F of 420 and 272
 

Now, The remainder becomes 0.

So, HCF of 420 and 272 is 4.

Applying fundamental theorem of arithmetic for verifying,

The prime factors are
 

Therefore, HCF of 420 and 272 is 4.

The constant plynomial f(x) = 0 is called zero polynomial.

A real number 'a' is called a zero of polynomial p(x) = 0

Triangle ABC is Isosceles

and it is given

AC = BC

AB² = 2 AC²

= AC² + AC²

AC = BC so we can write,

AB² = BC² + AC²

This is similar to Pythagoras Theorem

AB = hypotenuse

BC = Perpendicular or Base

AC = Base or Perpendicular

Therefore

C = 90°

Given equations are ,

10/( x + y ) + 2/( x - y ) = 4 ---( 1 )

15/( x + y ) - 5/( x - y ) = -2 --( 2 )

Let ,

1/(x+y) = a , 1/(x-y) = b

10a + 2b = 4

Divide each term with 2 , we get

5a + b = 2

=> b = 2 - 5a ----( 3 )

15a - 5b = -2 -----( 4 )

Substitute b = 2 - 5a in equation

( 4 ) , we get

15a - 5( 2 - 5a ) = -2

=> 15a - 10 + 25a = -2

=> 40a = -2 + 10

=> 40a = 8

=> a = 8/40

=> a = 1/5

Put a = 1/5 in equation ( 3 ) , we

get

b = 2 - 5 × 1/5

=> b = 2 - 1

b = 1

Therefore ,

1/( x + y ) = 1/5 => x + y = 5 --( 5 )

1/(x-y) = 1/1 => x - y = 1 ---( 6 )

Add equations ( 5 ) and ( 6 ) ,

We get

2x = 6

=> x = 6/2 = 3 ,

Put x = 3 in equation ( 5 ) , we get

3 + y = 5

=> y = 5 - 3 = 2

Therefore ,

x = 3 , y = 2

The angle of a quadrilateral are in Arithmetic Progression, now as there are four angles, the greatest of all the angles is double to that of smallest angles. Then to find the 4 angles of the quadrilateral,  the angles can be written as (x-3d)(x-d)(x+d)(x+3d)

The common difference in A.P. = d

Therefore, the sum of the four angles should be equal to that of 360°

Hence, (x-3d)+(x-d)+(x+d)+(x+3d)=360°

4x=360°

x=90°

Let (x+3d) be more than (x-3d)

Putting x = 90 degree in the equation x+3d = 2 (x-3d)

Then, 90+3d = 2 (90-3d)

9d=90;d=10

Hence, the common difference is 10

Therefore putting the value of  x=90° and  d=10 in (x-3d)+(x-d)+(x+d)+(x+3d)=360°

The angles are 60, 80, 100, and 120 of the quadrilateral. which are in AP.

If a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio. Given : In ∆ABC , DE || BC and intersects AB in D and AC in E. ... between the same || lines. The intercept theorem, also known as Thales' theorem (not to be confused with another theorem with the same name) or basic proportionality theorem, is about ratios of various line segments that are created if two intersecting lines are intercepted by a pair of parallels.

Given, sum of first m terms of an A.P. is n.

Also, sum of first n terms is m.

Subtracting (2) from (1),

Hence, the sum of its first (m + n) terms is -(m + n).

Let us assume that 4-5√2 is be rational.

=> 4-5√2 = a/b [ where, a,b belongs to Z]

=> -5√2 = a/b -4

=> √2 = 4/5 -a/b

L.H.S = irrational number as √2 is an irrational number .

R.H.S = rational number

Hence, R.H.S ≠ L.H.S

Therefore, our assumption is wrong & 4-5√2 is an irrational number.

Given f(x) = x^4 - 11x^2 + 34x - 12.

Given g(x) = x - 2.

Now, we need to divide f(x) by g(x).


          x^3 + 2x^2 - 7x + 20
        -----------------------------------
x - 2) x^4          - 11x^2 + 34x - 12

         x^4 - 2x^3

         ---------------------------------------

                  2x^3 - 11x^2 + 34x - 12

                  2x^3 - 4x^2

         -------------------------------------------

                             -7x^2 + 34x - 12

                             - 7x^2 + 14x - 12

             -------------------------------------

                                           20x - 12

                                           20x -40

                 ------------------------------------

                                                   28


We know that Dividend = Divisor * Quotient + Remainder

 = > (x - 2) * (x^3 + 2x^2 - 7x + 20) + 28

= > x^4 + 2x^3 - 7x^2 + 20x - 2x^3 - 4x^2 + 14x - 40 + 28

= > x^4 - 11x^2 + 34x - 12

= > Dividend

Given f(x) = x^4 - 11x^2 + 34x - 12.

Given g(x) = x - 2.

Now, we need to divide f(x) by g(x).


          x^3 + 2x^2 - 7x + 20
        -----------------------------------
x - 2) x^4          - 11x^2 + 34x - 12

         x^4 - 2x^3

         ---------------------------------------

                  2x^3 - 11x^2 + 34x - 12

                  2x^3 - 4x^2

         -------------------------------------------

                             -7x^2 + 34x - 12

                             - 7x^2 + 14x - 12

             -------------------------------------

                                           20x - 12

                                           20x -40

                 ------------------------------------

                                                   28


We know that Dividend = Divisor * Quotient + Remainder

 = > (x - 2) * (x^3 + 2x^2 - 7x + 20) + 28

= > x^4 + 2x^3 - 7x^2 + 20x - 2x^3 - 4x^2 + 14x - 40 + 28

= > x^4 - 11x^2 + 34x - 12

= > Dividend

( sin ² 63° + sin ² 27° ) / ( cos ² 17° + cos ² 73° )
= { sin ² 63° + sin ² ( 90 - 63 ) ° } / { cos ² 17° + cos ² ( 90 - 17 ) ° }
= ( sin ² 63° + cos ² 63° ) / ( sin ² 17° + cos ² 17° )
= 1 / 1
= 1

Given:

sec θ + tan θ = x

To find:

Sec θ = ?

Solution:

sec θ + tan θ = x --- eq 1

sec θ = x - tan  θ

Squaring both sides -  

sec θ = x² - 2x² tan θ + tan²θ

sec² θ - tan² θ = x - 2xtan  θ

1 = x²  - 2xtan  θ

tan θ = x²  - 1/2x

Substituting value in eq 1  

sec θ + x² -1/2x = x

sec θ = x²  - x-1/2x  

= 2x² - x²  +1/2x

sec θ = x²  + 1/2x

Answer : The value of sec θ = x²  + 1/2x

HRD Minister Ramesh Nishank announced a major CBSE syllabus reduction for the new academic year 2020-21 on July 7 which was soon followed by an official notification by CBSE on the same.

Considering the loss of classroom teaching time due to the Covid-19 pandemic and lockdown, CBSE reduced the syllabus of classes 9 to 12 with the help of suggestions from NCERT.

Deleted syllabus of CBSE Class 10 Mathematics

X²+20x+100

= X² + 10x + 10x +100

= x ( x+ 10) + 10 (x + 10)

= (x +10) (x +10)

= ( x + 10)²