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sin35/cos55=sin35/sin(90-55)=sin35/sin35=1

(cosecA-sinA)(secA-cosA)=(1/sinA – sinA )(1/cosA – cosA)=[(1- sin2A)/sinA][(1-cos2A)/cosA]=[(cos2A)/sinA][(sin2A)/cosA]=sinA*cosAsinA. cosA/sin²a+cos²a1/sin²a/sinA cosa+cos²a/sinA cosa1/tana+cota

using BODMAS,

=550-400-150+100

=650-550

=100

Half turn is the angle name for half a revolution.

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10 is the answer of the question

10 is the correct answer

10 is the correct answer of the given question

10 is the correct answer

12 ÷ 6 - 2 + 10= 2 - 2 + 10= 2 - 12= 10 is the correct answer of the following question.

-10 is the right answer of this question

12÷6-2+10=2-2+10=2-12=10 ANS.....SO,10 is the correct answer

10 is the correct answer

12÷6-2+102+8=10 This is your answer.

10 is the correct answer

10 is the right answer

10 is the answer of this question

10 is the correct answer

your answer is 10 use the formula -BODMAS

12÷6=22-2=00+10=10

Applying BODMAS rule we get 10 as correct answer

12%6-2+102-2+100+1010is the answer

10 is the correct answer

10 is the best and right answer

BODMAS12÷6-2+10=2-2+10=2-12=1010 IS THE CORRECT ANSWER

10 is the right answer

10 is the right answer

=2-2+10=0+10=10Answer-10=10

10 is the right answer

10 is correct answer

(12:-6)-2+10(2-2)+100+10=1010 is the correct answer

10 is the correct answer

x*x*x+2x*x-x*x-2x-12x-24 =x*x(x+2)-x(x+2)-12(x+2) =(x+2)(x*x-x-12) =(x+2)(x-4)(x+3)

2x³ - 5x² - 14x + 8 = 0put x = -2 we get 0 hence x = -2 is a root of the given equation2x²(x+2) -9x(x+2) +4(x+2) = 0(x + 2)(2x² -9x +4) = 0(x + 2)(2x² -8x - x + 4) = 0(x +2){2x(x-4) -1(x-4)} = 0(x+2)(x-4)(2x-1) =0x = -2, 4, 1/2

correct answer is x= -1

2x-1X=1/24(1/2)^3-12(1/2)^2+14(1/2)-34*1/8-12/4+14/2-3=1/2-3+7-3=1/2-6+7=1/2+1=3/2

The percentage is given by (225/550)×100=9/22×100 =40.91%

First term = aCommon diff. = d6th term = a + 5da + 5d = 0 a = -5d

21st term = a + 20d = -5d + 20d = 15d11th term = a + 10d = -5d + 10d = 5d Therefore,21st term = 3 * 11th term = 3 * 5d = 15d (proved)

2a+2b-4-5+a3a+2b-9put a=-1 and b=-2hence3(-1)+2(-2)-9=-3-4-9=-7-9=-16

Let the roots of the required quation be M and N let the roots of the equation 2x²+2(p+q)x+p²+q²=0 be a and ba + b = -(p+q)ab = (p^2 + q^2) / 2(a+b)^2 = (p+q)^2(a-b)^2 = (a+b)^2 - 4ab(a-b)^2 = -(p - q)^2we wanted the values of square of sum of the roots and square of difference of the rootsNow M = (a+b)^2 = (p+q)^2 and N = (a-b)^2 = -(p - q)^2M + N = 4pqMN = (p+q)^2 [-(p - q)^2]MN= -(p^2 - q^2)^2hence the required equation isx^2 - (4pq)x - (p^2 - q^2)^2 = 0Hope this helps!!!

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We have to maximize Z = 60x+ 15yFirst, we will convert the given inequations into equations, we obtain the following equations:x+y= 50, 3x+y= 90,x= 0 andy= 0

Region represented byx+y≤ 50:The linex+y = 50 meets the coordinate axes atA50,050,0andB0,500,50respectively. By joining these points we obtain the line 3x+ 5y= 15.Clearly0,00,0satisfies the inequationx+y≤ 50. So,the region containing the origin represents the solution set of the inequationx+y≤ 50.

Region represented by 3x+y≤ 90:The line 3x+y= 90 meets the coordinate axes atC30,030,0andD0,900,90respectively. By joining these points we obtain the line 3x+y= 90.Clearly0,00,0satisfies the inequation 3x+y≤ 90. So,the region containing the origin represents the solution set of the inequation 3x+y≤ 90.

Region represented byx≥ 0 andy≥ 0:Since, every point in the first quadrant satisfies these inequations. So, the first quadrant is the region represented by the inequationsx≥ 0, andy≥ 0.

The feasible region determined by the system of constraints,x+y≤ 50, 3x+y≤ 90,x≥ 0, andy≥ 0, are as follows.



The corner points of the feasible region areO0,00,0,C30,030,0,E(20,30)E20,30andB0,500,50.

The values of Z at these corner points are as follows.

Corner pointZ = 60x+ 15yO0,00,060 × 0 + 15 × 0 = 0C30,030,060 × 30 + 15 × 0 = 1800E(20,30)E20,3060 × 20 + 15 × 30 =1650B0,500,5060 × 0 + 15 × 50 = 750

Therefore, the maximum value of Z is1800atthepoint(30,0)1800atthepoint30,0.Hence,x= 30 andy= 0 is the optimal solution of the given LPP.Thus, the optimal value of Z is 1800.

sum of root a+b = -(-1/2) = 1/2 product of roots = ab = -2/2 = -1

so, a²+b² = (a+b)²-2ab = (1/2)²-2(-1) = 1/4+2 = 9/4

a/b +b/a = (a²+b²)/ab = -9/4

a⁴+b⁴ = (a²+b²)²-2a²b² = (9/4)²-2(a²b²)² = 81/16-2(1)= 67/16

Given X and Y are two circles touch each other externally at P. AB is the common tangent to the circles X and Y at point A and B respectively.

To find : ∠APB

Proof: let ∠CAP = α and ∠CBP = β.

CA = CP [lengths of the tangents from an external point C]

In a triangle PAC, ∠CAP = ∠APC = α

similarly CB = CP and ∠CPB = ∠PBC = β

now in the triangle APB,

∠PAB + ∠PBA + ∠APB = 180° [sum of the interior angles in a triangle]

α + β + (α + β) = 180°

2α + 2β = 180°

α + β = 90°

∴ ∠APB = α + β = 90°

Distance between centres when the circles touch externally = r1+ r2 = 8 + 5 = 13cm

2x^2-6x+k=0; b^2-4ac = (-6)^2-4(2)(k); 36-8k=0; 36=8k ; k=36/8=4.5

condition for real and equal root b^2-4ac=0(-6)^2-4(2)(k)=036-8k=08k= 36k= 36/8= 9/2

2x^2-6x+k, formula b^2-4ac ; (-6)^2-4(2)(k); 36 = 8k; k= 36 / 8 = 4.5

2x²-5x+3=02x²-3x-2x+3=0x(2x-3)-1(2x-3)=0(x-1)(2x-3)=0x=1,3/2

2x²-5x+3=02x²-3x-2x+3=0x(2x-3)-1(2x-3)=0(x-1)(2x-3)=0x=1,3/2

Number of books in the library for primary section=5365Number of books in the library for senior section=4370:. Total number of books in the school library=5365+4370=9735.

books in primary section is 5365books in senior section is 4370=. 5365 +.4370=. 9730

1 )Sec ( 90 - A ) = cosecA

2 ) Tan( 90 - A ) = cot A

3 )Cos 25 = cos ( 90 - 65 ) = sin65

4 )Tan 27 = tan ( 90 - 63 ) = cot 63

5 ) cosec²A - cot² A = 1

6 ) sin² A + cos² A = 1

7 ) cotA tanA = 1

Now ,

Sec(90-A)cosecA - tan( 90-A)cotA+ cos² 25 + cos² 65/ 3tan27tan63

= CosecAcosecA - cotAcotA + sin²65+cos²65/3cot63tan63

=( Cosec² A - cot² A)+ ( sin²A+cos²A)/ 3cot63tan63

= ( 1 + 1 )/ 3

= 2/3

2x² - 5x + 3 = 0

2x² - 2x - 3x + 3 = 0

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

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

So, x = 3/2 and 1

TheWheatstone bridgeis a circuit which is used to measure accurately an unknown resistance

PRINCIPLE:Wheatstone bridge principle statesthat when thebridgeis balanced, the product of the resistance of the opposite arms are equal.

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We know HCF × LCM = product of numberslet other number qso 2× 208 = 16× qq = 416/16 = 26 so other number is 26

(x1+x2+...+xN)/N = mean = X. Mean of new values = a(x1+x2+...+xN)/N = aX.

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Given, difference between two number is 26 and one number is 3 times the other.

Let the number are x and y. then,

x-y = 26

and

x=3y

If we solve these two equations, then we get, x = 39 and y = 13

And, x = -39 and y = -13 is not the solution of the above two equation. substitute x= -39 and y = -13 in equation (1), we have,

-39 - ( -13) = 26

-26 = 26 , which is not possible.

Hence, x = 39 and y = 13 is the solution of the equations.

1005 × 168= (1000+5)×168= 1000 ×168+168×5= 168000 + 840= 168840

Sin²25°+sin²65°+√3(tan5°tan15°tan30°tan75°tan85°)=sin²25°+sin²(90°-25°)+√3{tan5°tan15°×(1/√3)×tan(90°-15°)tan(90°-5°)}=sin²25°+cos²25°+√3×1/√3(tan5°tan15°cot15°cot5°)=1+1=2 Ans.