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Area = 0.5*d1*d2= 0.5*15*24= 12*15= 180 sq cm

है 8th stander च आहे ना

You need to know that a dozen is 12 and a score is 20

Find the unit price in each case.

12 articles cost Rs45 →4512=Rs3.75 each

20 articles sold at Rs85 →8520=Rs4.25 each

The amount of gain per article is Rs4.25−Rs3.75=Rs0.50

The percent gain is: 0.53.75×100%=1313%

ax + bx - ay - by = ax - ay + bx - by = a(x - y) + b(x - y) = (x - y)(a + b)

answer in the book was 27.50

ax+bx+ay+by=x(a+b)+y(a+b)=(a+b)(x+y), is right answer

(a+b)(x+y) is answer

ax + by = c.............. (1)bx + ay = 1 + c........ (2)

Multiply eq(1) by b and eq(2) by a

abx + b^2y = bc................(3)abx + a^2y = a(1 + c)....... (4)

Subtract eq(3) from eq(4)(a^2 - b^2)y = a(1 + c) - bc

y = (a + ac - bc)/a^2 - b^2

Multiply eq(1) by b and eq(2) by a

a^2x + aby = ac.......... (5)b^2x + aby= bc+b.......(6)

Subtract eq(6) from eq(5)(a^2- b^2)x = (ac-bc -b)

x = (ac-bc -b)/(a^2- b^2)

∴ the values of x and y arex = (ac-bc -b)/(a^2- b^2)y = (a + ac - bc )/( a^2 - b^2) .

Elimination method se please

ax+by=1................(i)bx+ay=2ab/a² + b² .......................(ii)

ax+by=1 .................. *aa²x + aby = a ....................(iii)

bx+ay=2ab/(a² + b²).......................* bb ² x +aby = 2ab² /(a² + b²)..........(iv)

subtracting eq (iv) from (iii)a²x + aby = a - [b ² x +aby = 2ab² /(a² + b²)]a²x - b²x + aby - aby = a -2ab² /(a² + b²)(a² - b²)x = [a(a² + b²) - 2ab²]/(a² + b²)(a² - b²)x = [a³ + ab² - 2ab²]/(a² + b²)(a² - b²)x = [a³ - ab²]/(a² + b²)x = [a(a² - b²)] / (a² + b²)(a² - b²)x = a/ (a² + b²)

putting x in (i)ax+by=1a [a/ (a² + b²)] +by = 1a²/(a² + b²) + by = 1by = 1 - a²/(a² + b²)by = [a² + b² - a²]/a² + b²by = b²/a² + b²y = b²/(a² + b²)(b)y = b/a² + b²

x = a/ (a² + b²)y = b/a² + b²

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75% of 540= (75/100)*540= (3/4)*540= 3*135= 405

1 2/5 of 405= (7/5)*405 = 7*81 = 567

(2/3) of 567= 2/3 * 567 = 2*189= 378

(a) is correct option

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answer

2/3 * (7/5 * 75/100 * (540))

= 2/3 * (7/5 * (3/4 * 540))

= 2/3 * (7/5 * (3 * 135))

= 2/3 * (7/5 * 405)

= 2/3 * (7 * 81)

= 2/3 * (567)

= 2 * 189

= 378

(a) is correct option

(4root5 - 3root2)(4root5 + 3root2)

[using (a - b)(a + b) = a^2 - b^2]

= (4root5)^2 - (3root2)^2

= 16*5 - 9*2

= 80 - 18

= 62

a3=4, a8=-8, a+2d=4; a+8d=-8/-6d=12; d=12/-6=-2; a=4-2d=4-2(-2)=4+4=8; an=0; 8+( x-1)×-2=8+2x-2: -10=2x; x=-10/2=-5

It is weight height measurement checker. It tells weight according to height and vice versa

in 6 2/3 hrs he cover 15kmand 6 2/3 =(6×3 +2)/3 =20/3in 1 hr he will cover = 15/(20/3)=9/4 km

Let ACB be the given arc subtending at an angle of 60° at the centre.

Now, we have:

r=7 cm and θ=60°

∴ Length of the arc ACB=2πr*(60/360)

=2×22/7×7×60/360=7.33 cm

Arc length = 2πr (theta/360°)2π*7(60/360)2*22/7*7*1/644/6cm7.3cm

-s^4t^3u^3 is correct answer of following question

5^n+3 - 6*5^n+1/9*5^n - 2^2*5^n

= 5^n(5^3 - 6*5)/ 5^n(9 - 4)

= (125 - 30)/5

= 95/5 = 19

Answer:B)1explanation :if sin A=cos Athen A=45 degreetherefore tan A=tan 45 degree=1

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8 : 25 :: 56 : x8/25=56/xx=56×25/8x=7×25x=175

z²-z+2z-2=(z+2)(z-1)

z²-z+2z-2=(z+2)(z-1

=> (cos^2A + cos^2B + 2cosAcosB) + (sin^2A + sin^2B - 2sinAsinB)

=> cos^2A + cos^2B + sin^2A + sin^2B + 2cosAcosB - 2sinAsinB

=> cos^2A + sin^2A + cos^2B + sin^2B + 2(cosA*cosB - sinA*sinB)

=> 1 + 1 + 2(cosA*cosB - sinA*sinB)

=> 2 + 2(cosA*cosB - sinA*sinB)

=> 2 (1 + (cosA*cosB + sinA*sinB))

=> 2 * (1 + cos(A-B))

{Because: cosA*cosB - sinA*sinB = cos(A+B)}

=> 2 * 2cos^2 ((A+B)/2)

=> 4cos^2 (A+B)/2

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cos tan

sincos × ---------- cos

sin

tan=sin/cos

cos.tan=sincos.sin/cos=sinsin=sin

hence, proved

=================================cosA - sinA = √2 sinA

cosA = sinA + √2 sinA

cosA = sinA (√2 + 1)

sinA/cosA = 1/(√2 + 1)

sinA/cosA = 1/(√2 + 1) * (√2 - 1)/(√2 - 1)

sinA/cosA = (√2 - 1)/(2 - 1)

sinA/cosA = √2 - 1

sinA = (√2 - 1)cosA

sinA = √2 cosA - cosA

sinA + cosA = √2 cosA

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Tan is + ve& sin is - vemeans it must be 3 rd quadrant...where cos is -veperpendicular= 5base= 12hencehypotenuse= 13

hence cos (thita )=(-12/13 )sin (thita)=5/13

(x+3)(x-3)सरल.कीजिए.ा