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a) 15+5=2020+7=2727+9=36

b)same as (a) 15+5=20

a) 15/20= 27/x3/4= 27/x x= 36

b) 15/x= 27/3615/x= 9/1215/x= 3/4x= 20

A = 15x = 540X=540/15=36

B= 27x =540=540/27=20

2x²+3x+1]4x³+2x²+5x-6[2x-2 4x³+6x²+2x —————— -4x²+3x-6 -4x²-6x-2 ————— 9x+8so quotient=2x-2 remainder=9x+8

2x-2 is the quotient of the given polynomial

quotient = 2x-2remainder = 9x+8

Answer is may be 22.

22 is the correct answer

22 is right banswer

if 27 added to number of 5 answer is 32

22 is the right answer

22 is the right answer

as per the question we get, 5+X =27X=27-5X=22

5+X=27X=27-5X=22. is the correct answer

5+x = 27X = 27-5X=22

17 + 6p = 96p = 9 - 176p = -8p = -8/6 = -4/3

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For unit's digit to be 0, then 4n should have 2 and 5 as its prime factors, but 4n =( 2²)n. it does not contain 5 as one of its prime factors.hence, 4n will not end with digit 0

(-3,2) lie in second quadrant

(-3,2) lies in second quadrant

(-3, 2) lie in second quadrant as -3 is in negitive x axis and 2 is in positive y axis

24p^2 is the correct answer of the given question

(-6p)×(-4p)= 24p^2 is the answerhere (-)×(-)=+p×p=p^2

-6×-4=-24,; pxp= p^2; 25 p^2

New origin is (-3,2)hencenew points will beX=x-h Y=y-kX=3-(-3)=6 Y=1-2=-1hence (6,-1)

Intersecting linesIntersecting linesare two lines thatshare exactly one point.This shared point is called thepoint of intersection

what is parallel lines

For a number to end with the digit 0 it's prime factorization should have 2 and 5 as a common factor.

here 8^n = (2*4)^n doesn't have 5 in its prime factorization.

Therefore 8^n cannot end with the digit 0.

Factorization :6p - 12q6 ( p - 2q)

3pq+6p= 3p (q + 2) answer

6p = 9-17

6p = -8

p = -4/3

They are generally planted in November and harvested in April it's call Rabi crop

ex.pea,gram,wheat and barely

Rabi crops or rabi harvest are agricultural crops that are sown in winter and harvested in the spring in India.

suppose digits are x and ydifference is one so x-y=1 so x=y+1now original number is 10x+yand reverse of number is 10y+xwhen digits are reversed number becomes 5/6 of original10y+x=(5/6)(10x+y)60y+6x=50x+5y55y=44x5y=4xnow x=y+15y=4y+4so y=4so x=5so original number is 54

The centralcarbonatom ofalleneforms two sigma bonds and two pi bonds. The centralcarbonis sp-hybridized, and the two terminalcarbonatoms are sp2-hybridized. The bond angle formed by thethree carbonatoms is 180°, indicating linear geometry for thecarbonatoms ofallene.

Given units digit is x and tens digit is yHence the two digit number = 10y + xNumber obtained by reversing the digits = 10x + yGiven that sum of a two digit number and the number obtained by reversing the order of its digits is 121.Hence (10y + x) + (10x + y) = 121⇒ 11x + 11y = 121∴ x + y = 11Thus the required linear equation is x + y = 11.

148 is the right answer

148 is the answer of sum

148 is the right answer

Let,

Unit's place digit be x

Ten's place digit be ( x + 1 )

Number = 10( x + 1 ) + x

Given, product of the number and the number obtained by reversing the digits is 252.

Original number = 10( x + 1 ) + x

Number when digit are reversed = 10x + ( x + 1 )

Product = [ [10( x + 1 ) + x ] [ 10x + ( x + 1 ) ]

= > 252 = [ 10x + 10 + x ] [ 10x + x + 1 ]

= > 252 = [ 11x + 10 ] [ 11x + 1 ]

= > 252 = 121x^2 + 11x + 110x + 10

= > 252 = 121x^2 + 121x + 10

= > 121x^2 + 121x + 10 - 252 = 0

= > 121x^2 + 121x - 242 = 0

= > 121( x^2 + x - 2 ) = 0

= > x^2 + x - 2 = 0

= > x^2 + ( 2 - 1 ) x - 2 = 0

= > x^2 + 2x - x - 2 = 0

= > x( x + 2 ) - ( x + 2 ) = 0

= > ( x + 2 ) ( x - 1 ) = 0

= > x = - 2 or x = 1

It is given that the required number is a natural number, so x can't be equal to - 2. Therefore, x = 1 .

Then,

Unit's place of the number = x

Unit's place of the number = 1

Ten's place of the number = x + 1

Ten's place of the number = 1 + 1

Ten's place of the number = 2

Therefore, required natural number = xy = 21

Required number = 21 .

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Given Principle amount = Rs . 6000Rate of interest = r = 5% compounded annuallyInterest after 1 year = 6000 × 5 × 1100 = 60× 5 = Rs . 300Total money owed after 1 year = 6000 + 300 = 6300Rs. 1200 paid , SoTotal money starting of second year = 6300 - 1200 = Rs.5100Interest after 2 year = 5100 × 5 × 1100 = 51× 5 = Rs . 255money owned after 2 year = 5100 + 255 = 5355AndRs. 1200 paid , Total money outstanding starting of Third year = 5355 - 1200 = Rs.4155

9999 is largest 4 digit number, and100^2 is10000, or 1 greater than the largest 4 digit number. This means that the closest square root of the largest perfect square is most likely99. So99^2 (9801)is the largest perfect square of four digits

thanks

P=12000, R=8% p.a., T=7/2 =3.5 Years S.I. = (P×R×T)/100= (12000×8×3.5)/100 =12×8×35=3360

answer:::::::::::162

676 is perfect square , 525 is not perfect square , 625 is perfect square , 1089 is also a perfect square

625 is A perfect dquare

Let the ten’s and the unit’s digits in the first number be x and y, respectively. So, the first numbercan be written as 10x + y in the expanded form.

When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s digit. This number, in the expanded notation is 10y + x.

According to the given condition.

(10x + y) + (10y + x) = 66

11(x + y) = 66

x + y = 6 ... (1)

Youare also given that the digits differ by 2. Therefore,

either x – y = 2 ... (2)

or y – x = 2 ... (3)

If x – y = 2, then solving (1) and (2) by elimination, you get x = 4 and y = 2.In this case, the number is 42.

If y – x = 2, then solving (1) and (3) by elimination, you get x = 2 and y = 4.In this case, the number is 24.

Thus, there are two such numbers 42 and 24.

Bhai iska answer Shi bnaya gya h

2x^2-4xy+10is the correct answer

2x²-4xy+10 is correct answer

the answer is (xy)=5

5x² – 6y² + 8y – 5 from 7x² – 5xy + 10y² + 5x – 4y

(7x² – 5xy + 10y² + 5x – 4y) - (5x² – 6y² + 8y – 5)

= 7x² – 5xy + 10y² + 5x – 4y - 5x² + 6y² - 8y + 5

= 7x² - 5x² - 5xy + 10y² + 6y² + 5x - 4y - 8y + 5

= 2x² - 5xy + 16y² + 6y² + 5x - 12y + 5

5√5x² + 30x + 8√5

splitting the middle term

= 5√5x² + 20x + 10x + 8√5

= 5√5x² + 4 × 5x + (2√5 ) × √5 x + 2√5 × 4

= 5x ( √5x + 4 ) + ( 2√5 ) [ √5x + 4 ]

= ( √5x + 4 ) ( 5x + 2√5 )

I hope this helps you.

let the two digit number be 10a+b10a+b-10b-a=9(a-b)

option 2 is the answer

-3x² + 7x - 4

-(3x² - 7x + 4)

- ( 3x² -3x - 4x + 4)

- ( 3x(x - 1) -4(x - 1))

- ( 3x - 4) ( x - 1)

( 1 - x) ( 3x - 4)