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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

hello this is a very tough question. I am also following this.

Where is the data provided?

(3y - 8) × ( 5y - 1)when x = 0 and y = -2(-6 - 8) × (-10 -1)(-14) × (-11)154

and is 4.....like my answer

a) 0 is the correct answer

B) is the correct answer

option d is the correct answer of the given question

the correct answer is 4

option (D) 4 is the right answer.

speed = 200 m/36 s

= 200*18/36*5 km/hr

= 20 km/hr

A leap year has 366 days or 52 weeks and 2 odd days.

The two odd days can be {Sunday,Monday},{Monday,Tuesday},{Tuesday,Wednesday},{Wednesday,Thursday},{Thursday,Friday},{Friday,Saturday},{Saturday,Sunday}.

So there are 7 possibilities out of which 2 have a Sunday.

So the probability of 53 Sundays is 2/7.

(b) is correct option

option 2 is right . leap year ka 2/7 hota hai aur normal year ka 1/7 hota hai.

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6:11=11:6 hoga NMD ah kk

option (D) 30for 6:116 = 2^2+211 = 2^3 + 3similarlyfor 11:? 11= 3^2 + 2? = 3^3 + 3 = 30hence, ? = 30

(sin5x-2sin3x+sinx)/(cos5x-cosx)=[{2sin(5x+x)/2cos(5x-x)/2}-2sin3x]/{2sin(5x+x)/2sin(x-5x)/2}=(2sin3xcos2x-2sin3x)/{2sin3xsin(-2x)}={2sin3x(cos2x-1)}/{-2sin3xsin2x}=-(cos2x-1)/sin2x=(1-cos2x)/sin2x=2sin²x/2sinxcosx=sinx/cosx=tanx (Proved)

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A)<B]>C]>D]> this is the right answer

P-8=7P=8+7P=15

f÷11=6f=11×6f=66

LHS = (sinθ - cosθ + 1)/(sinθ + cosθ - 1) dividing by cosθ both Numerator and denominator = (sinθ/cosθ - cosθ/cosθ + 1/cosθ)/(sinθ/cosθ + cosθ/cosθ - 1/cosθ)= (tanθ + secθ - 1)/(tanθ - secθ + 1)

Multiply (tanθ - secθ) with both Numerator and denominator

= (tanθ + secθ - 1)(tanθ - secθ)/(tanθ - secθ + 1)(tanθ - secθ)

= {(tan²θ - sec²θ) - (tanθ - secθ)}/(tanθ - secθ + 1)(tanθ - secθ)= (-1 - tanθ + secθ)/(tanθ - secθ + 1)(tanθ - secθ) [ ∵ sec²x - tan²x = 1 ]= -1/(tanθ - secθ) = 1/(secθ - tanθ) = RHS

Let two parts of number are x and (34 - x)

As per given condition(4/7)x = (2/5)(34 - x)4x/7 = 68/5 - 2x/54x/7 + 2x/5 = 68/5(20x + 14x)/35 = 68/534x/35 = 68/5x = 68*35/34*5x = 2*7 = 14

Therefore two parts of number are 14 and 20

cos7x+cos5x = 2cos6xcosx sin7x-sin5x = 2cos6xsinx (cos7x+cos5x)/(sin7x-sin5x) = 2cos6xcosx/2cos6xsinx = cotx

tan theta = 1theta= 45°then 2tan45°+ cos^2(45°)2+1/23/2

Here insufficient info is given. value of lenth or breadth is essential.

sorry I didn't know

Subtract area of strip surrounding pond from total area .You obtain area of pond.l×b = area solve the question ahead

option C is right answer

A triangle is only possible when sum of angles are 180°a) 45° + 60° + 73° = 178°, not possibleb) 50° + 68° + 62° = 180°, possiblec) 70° + 48° + 62° = 180°, possibled) 90° + 30° + 90° = 210°, not possible

1because all angles have more 1angles and measure in the compase

Given,:a/(x-a) + b/(x-b) = 2c/(x-c)[a(x-b)+b(x-a)]/(x-a)(x-b) = 2c/(x-c)(x-c)[a(x-b)+b(x-a)] = 2c(x-a)(x-b)

ax2 – 2abx + bx2 - acx + 2abc – bcx = 2cx2 – 2bcx – 2acx + 2abcax2+ bx2 - 2cx2 = 2abx – acx – bcx(a+b-2c)x2 = x(2ab – ac – bc)(a+b-2c)x2 - x(2ab – ac – bc) = 0x[(a+b-2c)x - (2ab – ac – bc)] = 0x = 0 or (a+b-2c)x - (2ab – ac – bc) =0x = 0 or (a+b-2c)x = (2ab – ac – bc)x = 0 or x = (2ab – ac – bc) / (a+b-2c)

2 roots of the given equation are x = 0 and x = (2ab – ac – bc) / (a+b-2c)

ratio of A =1:3,B=4:15,C I dont now, D=1:4

First arrange data in ascending order19, 25, 30, 31, 32, 35, 48, 51, 59

Total number of terms are 9So median = (9 + 1)/2 term

Median is 5th term = 32

If we replace 25 by 52Then, data in ascending order19, 30, 31, 32, 35, 48, 51, 52, 59

New median is 5th term = 35

A / (x - a) + b / (x - b) = 2c / (x - c)=> [a(x - b) + b (x - a)] / (x - a) (x - b) = 2c / (x - c)=> [(a + b) x - 2ab] (x - c) = 2c [x^2 - (a + b)x + ab]=> (a + b) x^2 - [2ab + c(a + b)]x + 2abc = 2cx^2 - 2c(a + b)x + 2abc=> (a + b - 2c) x^2 = (2ab + ac + bc - 2ca - 2bc) x=> (a + b - 2c) x^2 = (2ab - ca - bc) x=> x = 0or x = (2ab - ca - bc) / (a + b - 2c)

thanks bro......

plzzz..... solve my next math

x²dx + 1/y dy = 0

Integrating

∫x²dx + ∫ 1/y dy = 0

x³/3 + log y = C

We know

HCF × LCM = product of number

HL = AB

So, B = HL/A

A / (x - a) + b / (x - b) = 2c / (x - c)=> [a(x - b) + b (x - a)] / (x - a) (x - b) = 2c / (x - c)=> [(a + b) x - 2ab] (x - c) = 2c [x^2 - (a + b)x + ab]=> (a + b) x^2 - [2ab + c(a + b)]x + 2abc = 2cx^2 - 2c(a + b)x + 2abc=> (a + b - 2c) x^2 = (2ab + ac + bc - 2ca - 2bc) x=> (a + b - 2c) x^2 = (2ab - ca - bc) x=> x = 0or x = (2ab - ca - bc) / (a + b - 2c)

A+B=9/4 + (-1/4)=9/4-1/4=2