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25a^2-35a+12=25a^2-20a-15a+12=5a(5a-4)-3(5a-4)=(5a-3)(5a-4)so length and breadth will be (5a-3) and (5a-4)

3x^3-12x =3x(x^2-4); ii) 12y^2+8y^2-20= 12y^2-12y+20y-20= 12y(y-1)+20( y-1)= (12y+20)(y-1)

Proof . – Let ABCD be a rhombus whose diagonal AC and BD intersect at the point O.We know that the diagonals of a parallelogram bisect each other.Also we know that every rhombus is a parallelogram.Therefore OA=OC and OB=OD.From triangle(COB) and triangle (COD), we have:CB=CD sides of rhombus.CO=CO. CommonOB=OD provedTherefore tri(COB)~tri(COD) by SSS congruence.=> But Thus, Hence, the diagonals of a rhombus bisect each other at right angles.

(1)^2(2)+(1)^2(2)^2-(1)(2)^2=2+4-4=2

take 135 outside as the no to be divided

(I) 135,225

Since 225 > 135, we apply the division lemma to 225 and 135 to obtain

225 = 135 × 1 + 90 Since remainder 90 ≠ 0

, we apply the division lemma to 135 and 90 to obtain 135 = 90 × 1 + 45

We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain 90 = 2 × 45 + 0 Since the remainder is zero, the process stops. Since the divisor at this stage is 45,

Therefore, the HCF of 135 and 225 is 45.

(ii) 196 and 38220

Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain

38220 = 196 × 195 + 0

Since the remainder is zero, the process stops. Since the divisor at this stage is 196,

Therefore, HCF of 196 and 38220 is 196.

use Euclid's division algorithm to find the HCF of 135and 225

Curved surface area = πrl

the curved surface of area is=πrl

curved surface area = πrl

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area of square Hai 2nd may to triangle cube

second me area of square Hai To area of triangle kyu hai please Bataye

Taking the middle of the class we get

(25*10+75*15+125*30+175*35+225*25+275*15)/(10+15+30+35+25+15)

= 21000/(130) = 161.53

option c is correct answer

C is a correct answer

(tanx + tany)^2 - (Tanx- tany)^2= tanx^2+ tany^2+2 tanxtany-tanx^2-tany^2+ 2tanxtany=4Vmn

option c is the correct answer of the given question

Diameter= 10.5 cm

Radius (r) = 10.5/2 cm = 5.25 cm

Slant height (l) = 10 cm

Curved surface area of the cone = (πrl)

=(22/7 × 5.25 × 10)=165 cm²

Hence, the curved surface area of a cone

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Let 'a' be the 1st term(a₁) and 'd' the common difference of the A.P.s.

Ans1=> a₉ = 19. a+(9-1)d = 19. a+8d=19 .....(1)

Now, a₁₉ = 3×a₆ a+(19-1)d=3[a+(6-1)d] a+18d=3(a+5d) simplifying, we get, a=3d/2. ......(2)Putting value of a in (1) 3d/2 + 8d=19Solving we get d=2.Putting d=2 in (1) we get a=3.

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(Mon+Tue+Wed)/3=370==>Mon+Tue+Wed=1110 eqn 1(Tue+Wed+Thu)/3=340==>Tue+Wed+Thu=1020 eqn 2On subtraction Mon+Thu=90....As we know that Thu=(4/5)Mon ....On solving we will get answer as 360

SP=1550profit=8.33%so.CP=1550/1.0833=1430(approx)so per radio =1430/4=357.5

71) E scores the maximum runs.

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the next is correct

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tan⁻¹√3 - cot⁻¹√3 = tan⁻¹(tan (π/3)) - cot⁻¹(-cot(π/6)) = tan⁻¹(tan (π/3)) - cot⁻¹(cot(π - π/6)) = (π/3) - (5π/6) = -π/2

AS we knowthat tan 60=underroot3so

We know that tan¯¹ √3 is at the angle π/3.Now,

⇒ tan¯¹ √3 - cot-1 (-√3)⇒ π/3 - (π-π/3)⇒ π/3 - 5π/6⇒ -3π/6⇒ -π/2

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In triangular pyramid mber of faces = 4Numer of vertices = 4Numer of edges = 6

Let the number between 10 and 100 = 10x + yIt is eight times the sum of its digits⇒ 10x + y = 8(x + y)⇒ 2x - 7y = 0 ....(1)

If 45 is subtracted from the number, the digits are reversed.(10x + y) - 45 = (10y + x)⇒ 9x - 9y = 45⇒ x - y = 5 .....(2)By solving equations (1) and (2), 7y/2 - y = 55y/2 = 5y = 2So x = 7we getx = 7 and y = 2.Therefore, the original number is 72.