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let ,|=x-y=8 ||=3x-3y=16

|×3=3x-3y=24 ||×1=3x-3y=16___________ y=40

y=40 in number 1_ x-y=8 =)x-40=8 =)x =8 by 40 =)x =1by 5

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As ABQ = 180 - angle B ACP = 180 - angle Cgiven ABQ = ACPso180 - angle B = 180 - angle Cso we getangle B = angle C

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2a^2 ×a^3 ×b×b=2×a^(2+3)b^(1+1)=2a^5 b^2

exponential form will be3^3*2^3

2 raise to the power -7 is correct answer

2raise to the power-7 is correct

2 raise to the power -7 is correct answer

256=(16)^2=(4^2)^2=2^2)^2)^2

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3*3*3*3=81 is the answer

y=2x-2y=4x-44x-4=2x-24x-2x=-2+42x=2so x=1y=2-2=0

Prime Factors of 1728 =2, 2, 2, 2, 2, 2, 3, 3, 3

so Exponential form is 2^6 × 3³

10/512

1/512

-5(xy-y)-2y(-11x+7)

= -5xy-5y+22xy+14y

= 17xy+9y.

let x1, y1 be the point of intersection for both the curves

y^2 = 4ax => dy/dx = 2a/y = 2a/y1 = m1 (say)

from second

xy = c^2 => dy/dx = -y/x = -y1/x1 = m2 (say)

Now, m1.m2 = -1 (since they intersect orthogonally)

=> x1 = 2a

Again from second,

x1y1 = c^2 => y1 = c^2/x [since x1, y1 lies on it]

putting in first we get

c^4 = 4a*x^3 = 4a * 8a^3 = 32a^4

or, c4 = 32a4

nahi

Area of door =3×2=6area of wall=10×10=100area of remaining part for painting =area of wall -area of door. =100-6=94Labour charge for painting 94sq metres =94×2.5=235so rs 235 is required to paint the wall.

if sign changed when you solve the q. eg..+×- =-

(2.5×1.5)x*x + 2.5×3y - 2.5×3 - 3×1.5xy - 3*3y*y + 3*3y = 3.75x*x + 7.5y - 7.5 - 4.5xy - 9y*y + 9y = 3.75x*x + 16.5y - 4.5xy - 9y*y - 7.5

Thnxxx

(i) 4532655=Fourty five lakh, thirty two thousand,six hundred fifty five.(ii) 673429=Six lakh,seventy three thousand, four hundred twenty nine.

f(2)=(-2)^2-(-2)-6 =4+2-6 =6-6 =0

Given, 3PB = ABAB/PB = 1/3AB - PB/PB = 3 - 1/1 = 2/1AP/PB = 2/1

Using section formula coordinates of P areP(x, y) = P(2*10 + 1*20)/2+1, 2*-20 + 1*0/2+1)= P(40/3, - 40/3)

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log(x³-3) = 0

⇒ log(x³-3) = log1 [ since log1 = 0 ]

Taking anti log, we get

⇒ x³-3 = 1

⇒ x³ = 4

⇒ x = 4^(1/3)

Hence x= { 4^(1/3)}

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Hence ABCD is a rhombus.

CosA-sinA+1/cosA+sinA-1=(cosA-sinA+1)(cosA+sinA+1)/(cosA+sinA-1)(cosA+sinA+1)=(cos²A-cosAsinA+cosA+cosAsinA-sin²A+sinA+cosA-sinA+1)/{(cosA+sinA)²-(1)²}=(cos²A-sin²A+2cosA+1)/(cos²A+2cosAsinA+sin²A-1)={cos²A+2cosA+(1-sin²A)}/(1+2cosAsinA-1) [∵, sin²A+cos²A=1]=(cos²A+2cosA+cos²A)/2cosAsinA=(2cos²A+2cosA)/2cosAsinA=2cosA(cosA+1)/2cosAsinA=(cosA+1)/sinA=cosA/sinA+1/sinA=cotA+cosecA=cosecA+cotA (Proved)

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3m - 8n = 273m = 27 + 8nm = (27 + 8n)/3

Taking different values of n, we can calculate the corresponding values of mFor example, when n = 0, m=9. when n=3, m=17 and when n=6, m=25

There can be several sets of other solutions

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14 + 6√5

(√5)² + 3² + 2 × 3 × √5

(√5 + 3)²

To 14 + 6√5 ka vargmul hai

√((√5 + 3)²) = + (√5 + 3) aur - (√5 + 3)

2x + 1 = 52x = 5 - 1 = 4x = 2

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y Ka v

Solution:-Let the four consecutive numbers in AP be (a - 3d), (a - d), (a + d) and (a + 3d)So, according to the question.a-3d + a - d + a + d + a + 3d = 324a = 32a = 32/4a = 8 ......(1)Now, (a - 3d)(a + 3d)/(a - d)(a + d) = 7/1515(a² - 9d²) = 7(a² - d²)15a² - 135d² = 7a² - 7d²15a² - 7a² = 135d² - 7d²8a² = 128d²Putting the value of a = 8 in above we get.8(8)² = 128d²128d² = 512d² = 512/128d² = 4d = 2So, the four consecutive numbers are 8 - (3*2)8 - 6 = 28 - 2 = 68 + 2 = 108 + (3*2)8 + 6 = 14Four consecutive numbers are 2, 6, 10 and 14

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