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average speed=(total distance)/(total time)=(50+60)/(2+3)=110/5=22km/h

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360° = 2π rad 1° = 2π/360radso, 10° = 10× 2π/360 rad =π/18 rad

1' ( it means minute )60 minutes (60') = 1°1 minute = 1/60°but 1° =π/180 radso, 1/60° =π/60×180 =π/10800 rad

1'' = ? rad60'' = 1 minuteand 60' = 1° so, 3600'' = 1°so, 1'' =π/180×3600 =π/648000 rad

Total time taken = 8 hrTotal distance covered = 555 kmInitial speed of car = 60 km/hr

Let time taken at speed 60 km/hr = tLet distance covered at speed 60 km/hr = d

As per given conditiond/75 + (555 - d)/60 = 8(4d + 5(555-d))/300 = 84d + 2775 - 5d = 24005d - 4d = 2775 - 2400d = 375 km

Now,60*t = 375t = 375/60 = 6.25 hrs

Therefore,At speed 60 km/hr time taken = 6.25 hrsAt speed 60 km/hr distance covered = 375 km

first you can write your name than address than date

Tothe principalKendriya Vidyalaya DelhiSir,sir we want to visit a historical place as for example the Taj Mahal the Lal Kila and other historical places nearby us sofa please we are kindly requested you for gifted of a historical place for enjoyment and there anything this is the ways of all hole my class and we want we learn our culture and our immunity please sir we hope you will do that what we want thank you sir...your obedientlyclass 11th A Kendriya Vidyalaya delhi class captain dash Gaurav nagayach

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thank friend

Let speed be 's ', time be 't ' and distance be 'd 't, t = d / s t = 300 / s,d = (s + 5)(t -2)=> d = (s+5)(300/s-2) {t=300/s}=> 300s = (s+5)s(300/s-2) [multiplyning both sides with s]we get = > 300s = (s+5)(300 - 2s)=> 300s = 300s - 2s2+1500 - 10s=>300s =300s-2s2+ 1500 - 10s=> 0 = -2s2+1500 - 10s=> 0 = 2s2+10s - 1500 [multiplying both sides -1 ]=> 0 = 2s2+60s - 50s - 1500=> 0 = 2s(s + 30) -50(s + 30)=> 0 = (2s - 50)(s + 30)=> s = 25, -30 Since, speed cannot be negative, we take speed 25Km/h.

we have to find the ap

sides of rhombus are equallet side be athen perimeter 4a=80a=20mdiagonal,d=24mas diagonals of rhombus bisect each other at 90°by using pythagoras theorem we getother diagonal(l)as(l/2)^2 =20^2 -12^2l^2 /4=400-144l=32marea of rhombus=l×d /2=24×32/2a=12×32=384m^2

55 is the right answer of the following

Let ABCD be quadrilateral with ab||cdJoin bcIn triangle abd and CBD,Angle abd=angle cdb(alternate angles) Anglecbd=angle adb(alternate angles) Bd=bd(common) Abd=~CBD by asa testAd=BC by cpctSince ad =bc ABCD is a isosceles trapeziumAngle d= angle c - - - - - - (1)Join ac. In triangle adc and bcdCd=cd(common) Angle d =angle c(from 1) Ad =bc(proved above) Triangle adc=~bcd by sas testAc=bd by cpcth. p..

a=2d=8Sn=90Sn=(n/2)[2a+(n-1)d]90*2=n[2*2+(n-1)8]180=n[4+8n-8]180=n[8n-4]180=4n[2n-1] 180/4=n[2n-1]45=n[2n-1]45=2n²-n2n²-n-45=02n²-10n+9n-45=02n(n-5)+9(n-5)=0(n-5)(2n+9)=0

n=5,-9/2

n must be positive integerSo, n=5

an=a+(n-1)dan=2+(n-1)8an=2+8n-8=8n-6

5-V3 × 5+V3/5+V3=(5-V3)^2/5+V3=5+V3

Given cosec theta - sin theta = m, sec theta - cos theta = nGiven that cosec theta - sin theta = m→ !/sin theta - sin theta = m⇒(1-sin² theta)/sin theta = m→ cos² theta/sin theta = mand sec theta - cos theta = n⇒1/costheta - cos theta =n→ (1-cos² theta)/cos theta = nsin² theta/cos theta = nNow (m²n)²/³ + (mn²)²/³⇒(cos⁴ theta/sin² theta × sin² theta/cos theta)²/³ + (cos² theta/sin theta× sin⁴ theta/cos² theta)²/³⇒ (cos³ theta)²/³ + (sin³ theta)²/³⇒cos² theta + sin² theta= 1 Hence proved

Hum jante hai

A + B + C = 180°

(B + C)/2 = 90 - A/2

So, tan(B+C)/2

= tan( 90° - A/2)

= cot (A/2)

Curved Surface Area = 2* (22/7) * r * h= 2*(22/7)*0.5*4.2= 13.2 sq mCost = Area*rate= 13.2 * 15= Rs. 198

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LHS:=(secA − tanA) / (secA + tanA)= (1/cosA − sinA/cosA) / (1/cosA + sinA/cosA)= (1 − sinA) / (1 + sinA)= {(1 − sinA)(1 − sinA)} / {(1 + sinA)(1 − sinA)}= (1 − 2sinA + sin²A) / (1 − sin²A)= (cos²A + sin²A − 2sinA + sin²A) / cos²A= (cos²A − 2sinA + 2sin²A) / cos²A= cos²A/cos²A − 2 (1/cosA)( sinA/cosA) + 2(sin²A/cos²A)= 1 − 2secAtanA + 2tan²A= RHS

sec⁴x - tan⁴x (1+tan²x)² - tan⁴x1+tan⁴x+2tan²x -tan⁴x1+2tan²x

For function to be defined 1+x^2 should not be equal to zero

which is true for all the values of x

hence

option d is the correct answer..

let a/r , a , ar are in GP

A/C to question, (a/r ) × a × ar = 216a³ = 216a³ = (6)³ a = 6

again, sum of their products in pair = 156 (a/r ) × a + a × ar + ar × (a/r) = 156 a²/r + a²r + a² = 15636/r + 36r + 36 = 156 36( 1/r + r + 1) = 156 3(r² + r + 1) = 13r 3r² + 3r + 3 - 13r = 03r² -10r + 3 = 03r² - 9r - r + 3 = 0r = 3 and 1/3

so, numbers are 6/3, 6 , 6×3 => 2, 6, 18 numbers are 6×3 , 6 , 6/3 => 18, 6, 2

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12ab is the product.

The answer is 12ab as 2a*6b=12ab

3/5=0.6 kya ma na jo keya ha vo hi shi ha I am not sour but I am sure this answers

first make a line and write with start point 123 after 123 you write 0 after0 you write 12345

1:48.662:2903:64.28

100 paise = 1 rupee. 25 paise = 0.25 rupees.75 rupees 25 paise = 75.25 rupees.

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