Explore topic-wise InterviewSolutions in Current Affairs.

lim α→β (sin²α-sin²β/α²-β²)we are getting 0/0 form so using L hospital rulelim α→β(2sinα/2α)=sinβ/β

lim α→β (sin²α-sin²β/α²-β²)we are getting 0/0 form so using L hospital rulelim α→β(2sinα/2α)=sinβ/β

please like my answer if you find it useful

Let PQ and RS are two direct common tangents and EF is the transverse common tangent.

We know that tangent line segments are equal in length from an external point to circle.

Therefore, EP = EC and EQ = EC ⇒ EP = EQ

and FR = FC, FS = FC ⇒ FR = FS

Hence, ECF bisects PQ and ECF bisects RS.

∴ The common tangent bisects the other two common tangents

u can see the theorem 10.2 in ncert

I think option 2 is the correct answer to this question.

option (ii) is the correct answer

the Second one is the correct answer

please like my answer if you find it useful

let p(x) = x³-6x²+9x+3we will use remainder theorem so reaminder will be p(1) = 1³-6×1²+9×1+3 = 1-6+9+3 = 7

We are wealthy but not happy.

nhi aa RHA hai

(x / x+ 1)^2-5(x/ x+1) + 6 =0; ( x^2 / x^2+2(x)( x+1))-5x / x+1+6 = = ( x^2/ x^2 + 2x^2+x)-5x/x +7 =(x^2/3x^2+x - 5x / x+7)= =( x^2/3x^2+x -5x/ x+7)+6=0 = x^2( x+7)-5x(3x^2+x)+6= x^3 +7x^2 - 15x^3+5x^2 +6= -14x^3+12x^2+6=2(-7x^3+6x^2+3) = (-2, 3/2)

Let x/(x+1) be y.Therefore,y^2-5y+6=0y^2-2y-3y+6=0y(y-2)-3(y-2)=0(y-2)(y-3)=0Therefore,y=2 or y =3.Putting the value of y,x/(x+1)=2x=2x+2x=-2.x/(x+1)=3x=3x+3.-2x=3.x=-3/2.

2x-x=3+2x=5is the correct answer

2x - 3 = x + 22x - x = 2 + 3x = 5

=2x-3=x+2=2x-x=2+3=x=5

x=5 is the answer ok

According to angle sum property, the sum of all the angles should be 180°.So here, the sum of 90°, 20° and x should be 180°90° + 20° + x = 180°110° + x = 180°x = 180 -110 x = 70°Therefore, x is equal to 70°

thankyou so much

The number is 84+108/2

= 192/2

= 96

First read the question then answer idiot

2/3 is greater than 5/9 and by 1/9 factor.

2115×41=86,7152015×27=54,405

86,715 is greater

sin30=sin(90-30)=cos60; cos(40+x)=sin30; cos(40+x)=cos60; 40+x=60; x=60-40=20; x=20

cos(40+x)=sin30cos(40+x)= cos (90-30)40+x= 60x= 60-40=20

cos(40+x)= sin30; cos(40+x)= cos(90-30); 40+x=60; x=60-40=20

When a, b, c are in AP

2b = a + c

So, here we have

2(2k - 1) = k + 9 + 2k + 7

4k - 2 = 3k + 16

4k - 3k = 16 + 2

k = 18

✓32=5.65 is the correct answer

applying formula (a+b)²=a²+2ab+b² we get32²=(30+2)²=(30)²+2(30)(2)+2²=900+120+4=1024

1024 is the correct answer

x+2=x=-2 p(x) =(-2)^3-(-2)^2+(-2)-2 =8-4-2-2 =8-8 =0 And

Rong answer

please specify your questions

Radius (r1) of 1stcircle = 19 cm

Radius (r2) or 2ndcircle = 9 cm

Let the radius of 3rdcircle ber.

Circumference of 1stcircle = 2πr1= 2π (19) = 38π

Circumference of 2ndcircle = 2πr2= 2π (9) = 18π

Circumference of 3rdcircle = 2πr

Given that,

Circumference of 3rdcircle = Circumference of 1stcircle + Circumference of 2ndcircle

2πr= 38π + 18π = 56π

→ r = 28 CM ________answer

Ans :- विद्युत धारा का मात्रक एम्पीयर (ampere)होता है.

x²+5x-4x-20=0x(x+5)-4(x+5)=0(x-4)(x+5)=0x=4,-5

100 paise =1 rupees70 paise= (1/100)*70= 0.7 rupees

Radius of cylinderical pillars = r = 28cm =0.28mheight ,h=4mcurved surface area of a cylinder =2πrhcurved surface area of a pillar= 2×22\6×0.28×4=7.04metresq.Curved surface area of 24 such pillars = 7.04×24=168.96metresq.cost of cementing an area of 1metresq. =Rs8so,cost of painting 1689.6metresq.=168.96×8=Rs1,351.68

THANKS

HCF of 3/5 ,5/7,1/8= hcf of 3,5,1/lcm of 5,7,8= 1/280

Let the angle be xsum of all the angles of triangles = 18x + 50 + 90 = 180 x + 140 = 180 x = 180 - 140 x = 40

This question of answer is 30

a) = 3x^2(3×-1) -1(3×-1) = (3×^2-1) (3×-1)....ans....

20 /21× 27/25=20×27/21×25=540/525=1.02