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X+5×X=420X+5X=4206X=420X=420/6=70

x+x×5=420x+5x=4206x=420x=70

Ans :- p(x) = 2x³ + 6x² – 4x + 9

Define α,β and γ

Let α,β and γ be the 3 roots

αβ = 3 (Given)

Find the products of all the zeros:

αβγ = - d/a

αβγ = - 9/2

Find the third zeros:

Given αβ = 3

3γ = -9/2

γ = -9/2 ÷3

γ = -3/2

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2x²+46x+420=2x²+60x-14x+420=2x(x+30)-14(x+30)=(x+30)(2x-14)

2x^2+46x+420I feel there is something wrong in coping question.please check carefully.

x=7 is the correct answer was

Rational number= √2+√7/21.41+2.674.05/22.02

a rational number between √2 and √7 is 2 because √4=2 and √4 is situated between √2 and √7

The Rotational No Is= a bit

If x=4 y=5x+y =4+5=9x-y=4-5=-1

the natural no. are ≥ 1eg. 1,2,3,4,5.... so for x+y+z =6

the the minimum value of x,y,z can be 1 and max value would be 4. because others have to 1 so for either of x,y,z = 4 the no. of solutions are 3for either of x,y,z =3 , no. of solutions 2*3 = 6for either of x,y,z = 2 no. of solutions will be 1 , when all are 2,2,2..so total solutions are 6+3+1 = 10

Two zeroes = 0, 0Let the third zero be P.The, using relation between zeroes and coefficient of polynomial, we have:P + 0 + 0 = -b/aThird zero = P = -b/a

Given polynomial2x^3 - x^2 - 5x - 2

Let p, q and r are three zeroes of polynomialp*q = - 1*2 = - 2.......(1)

We know,p*q*r = - d/a = -(-2)/2 = 1p*q*r = 1......(2)

Put value of p*q from eq(1) in eq(2)r = 1/-2 = - 1/2

Therefore, third zero is - 1/2

—1

x% of 420=63x/100×420=63x=63÷420×100x=15

let the required no. be x and y

so according to the question

=> x+ y = 50 -------(1)

and => xy = 525 -------(2)

so put y = 525/x in eqn (1)

we get ,

=> x + 525/x = 50

=> x^2 + 525 = 50x

=> x^2 -50x +525 = 0

=> x^2 -35x -15x + 525 =0

=> x(x-35) -15(x-35) =0

=> (x-35)(x-15) = 0

=> x-35 = 0 or x-15 = 0

=> x = 35 , 15

# so the required number is 35 and 15

select any five number less than the given number thwn subtract it from the given number. addition of those two numbers would be your answer

the pair is 5,20 fulfill qbove

since these two nos have no common factor other than 1 so these are co prime numbers

385,621 =The common factor of 385 are:- 1,5,7,55,77,and 385 = The common factor of 621 are:- 1,3,9,27,69and 621

Here the common factor is 1.so,H.C.F is 1.Here the given numbers co-primes. *******thanks and I hok understand anything you .

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31 is the correct answer

31 is the right answer

avg.=17total sum of three no. =17(3)=51two no. are 8 and 12sum of given no. 8+12=20then Third no. =51-20=31

31 is the correct answer of the given question

31 is the correct answer

थिटा=120हमको ज्ञात हे किसम्पूर्ण ब्रत का परिधि =2πr होती हेअतः 2πr= 360°तब 1 डिग्री की मान 2πr/360°तथा 120° मान 2πr/360°×120°2πr/32*3.14*12/325.12 cm

Let their present ages be x and 2xThe ratio of their ages five years before is 1/3

their ages 5 yrs before are x -5 and 2x - 5

i.e.,

x-5/2x- 5 = 1/3

cross multiply ,

2x - 5 = 3x - 15

x = 10

2x = 20

their present ages are 10 and 20 respectively.Their ages 5 years later are 15 and 25 years.

Ratio of ages 5 yrs later = 15/25 = 3/5 = 3:5

Let the height of the first floor of qutub minar be AB=EC=25m

And DEC (DC) be the height of the tower.

BC=AE= Distance between their bases

Angle DAE be the angle of elevation of the top of the building.

Angle EAC be the angle of depression of the base of the building.

In triangle EAC,

Tan angle EAC=EC/AE

=> √3=25/AE

=> AE=25/ √3 meters

In triangle ADE,

Tan angle DAE=DE/AE

=> 1/ √3= DE/25/√3

=> 3 DE=25

=> DE=25/3

=> DE=8.33 approx

Height of the building= DE+EC = 8.33+25=33.33 meters approx.

Thus, the height of the building=33.33 metres.

let 3 no's be x,x+2,x+4

x+ x+2+ x+4. = 36

3x = 30

x= 10

so middle 10+2 = 12

Let the three numbers be x , x+2 , x+4Then, (x)+(x+2)+(x+4)=36Or, x + x + 2 + x + 4 = 36Or, 3x + 6 = 36Or, 3x = 36 - 6Or, 3x = 30Or, x = 30/3Or, x = 10So numbers = 10 , 10+2 , 10+4= 10, 12 , 14So the middle term is 12(1) 12 Ans.

12.because 10+12+14=36.

3/8 6/8 4/8 1/81/8<3/8<4/8<6/8

Given that the zeros are 3, 1/2 and -1

⇒ The factors are (x - 3), (x - 1/2) and (x + 1)

Find the cubic polynomial:

(x - 3) (x - 1/2) (x + 1) = 0

(x - 3) (x ² + x - x/2 - 1/2) = 0

(x - 3) (2x² + x - 1) = 0

2x³ + x² - x - 6x² - 3x + 3 = 0

2x³ - 5x² - 4x + 3 = 0

Answer: 2x³ - 5x² - 4x + 3 = 0

Suppose numbers are x and x+3 x(x+3) = 54 x(x+3) = 6×9 x = 6 Numbers are 6 and 9.

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9 and 6. there is a difference of 3 btw the numbers and their product comes to be 54.

40= 2*2*2*556= 2*2*2*760= 2*2*3*5HCF= 2*2= 4

Here, the highest power of x is 1. Therefore, the given polynomial has only one zero.

2x-3=0

=> x= 3/2

Hence, 3/2 is the zero of the polynomial 2x-3.

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zero of polynomial is2x -3 = 0x = 3/2

discriminatant=b^2-4ac=9-32=-23

hencediscriminatant is less than zerohence no real roots

Coffecient of x⁴ is 2

Coffecient of x² is - 1

coffecient of x is 8

cubic polynomial has degree 3 which is higher value of xlike x^3+x^2+x+1

cubic polynomial has a degree 3

cubic polynomial has degree 3

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3/7-6/11-8/12+5/2233-42/77-2/3+5/229/77-44+15/229/77-29/221/11(9/7-29/2)1/11(18-203/14)-185/154

x - y = 2, x y = 360; x^2 -20x +18x - 360; x( x - 20) + 18( x - 20); ( x + 18)( x - 20);, x = -18 & 20;

let 1st no be X , 2nd no = x+3x(x+2)=360x^2+2x-360=0x^2+20x-18x-360=0x(x+20)-18(x+20)=0(x+20)(x-18)=0either x+20=0 , x = -20or x-18=0, x= 18so when x= -20 (1st no)2nd no = x+2= -20+2= -18when x=18(1st no )so 2nd no = x+2= 18+2=20so nos may be ( -20, -18) or (18,20)

the numbers are 18 and 20