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The train is running at a speed of 42kmph. So the distance covered by train in meters per second is:

42*(5/18)=35/3 m/sec

Now the time required to the train to cross a man of is the same amount of time that train will take to cross itself that is 280 m.

So 280*(3/35) = 24 seconds.

an = - 4n + 15 also 15 - 4nan = a + ( n - 1 ) da1 = 15 - 4 × 1 = 15 - 4 = 11 a2 = 15 - 4 × 2 = 15 - 8 = 7 a3 = 15 - 4 × 3 = 15 - 12 = 3

AP : 11 , 7 , 3

sn = n/2 { 2a + (n-1) d } sn = 3 / 2 { 2 × 11 + (3-1) -4 sn = 3 /2 { 22 + 2 × - 4 } sn = 3/2 { 22 + (-8) }sn = 3/2 { 22 - 8 }sn = 3/2 × 14 sn = 3 × 7 sn = 21

Hence Sn = 21

Question is same just nomenclature is changed

Given X and Y are two circles touch each other externally at P. AB is the common tangent to the circles X and Y at point A and B respectively.

To find : ∠APB

Proof: let ∠CAP = α and ∠CBP = β.

CA = CP [lengths of the tangents from an external point C]

In a triangle PAC, ∠CAP = ∠APC = α

similarly CB = CP and ∠CPB = ∠PBC = β

now in the triangle APB,

∠PAB + ∠PBA + ∠APB = 180° [sum of the interior angles in a triangle]

α + β + (α + β) = 180°

2α + 2β = 180°

α + β = 90°

∴ ∠APB = α + β = 90°

i) √27 = 3√3 is irrationalReason : √3 is irrational number.

ii) √441 = 21 is rational

Reason : 21 is natural number.

iii) Not rational

iv) 7.4848..... is rational

Reason : can be written in p/q form

v) 11.213243..... irrational Reason : Decimal expansion not follow any pattern

vi) not rational

Let x= 3.2 bar on 2 mark it equation 1Multiply equation by 10So, 10*x= 3.2 bar on 2*1010x= 32.2 bar on 2Subtract eq 1 from 210x-x= 32.2 bar on 2 - 3.2 bar on 29x= 29X= 29\3

The train will cover a distance of 350 km in = 350/75 = 14/3= 4 hrs 40 mins .

350/75=14/3=4 hrs 40mins.

find the mean of first ten whole number

Sum of the numbers 1 through 10 = 55 Average = 55/10 = 5.5

1) a=10, d= 10Tn = a +(n-1)dSo T1=10T2=20T3=30T4=40AP: 10,20,30,40

2) AP: -2,-2,-2,-23) AP: 4,1,-2,-54)AP: -1,-1/2,0,1/2

LHS=1/(cosecA-cotA) -1/sinA

=(cosecA+cotA)/cosec^2A-cot^2A) -cosecA

=cosecA+cotA-cosecA

=cotA

RHS=1/sinA-1/(cosecA+cotA)

=cosecA-(cosecA-cotA)/(cosec^2A-cot^2A)

=cosecA-(cosecA-cotA)

=cotA

LHS=RHS

|0 a -b||-a o -c||b c 0|= 0(0×0 - c×-c) + (-a )(-a×0 - b×-c) + (-b)(b×0 - -ac)= 0+ (-a)(bc) + (-b)(-ac)= -abc + abc=. 0

Let the first term of AP = acommon difference = dWe have to show that (m+n)th term is zero ora + (m+n-1)d = 0

mth term= a + (m-1)dnth term = a + (n-1) d

Given thatm{a +(m-1)d} = n{a + (n -1)d}⇒am +m²d -md = an + n²d - nd⇒am - an +m²d - n²d -md + nd = 0⇒a(m-n) + (m²-n²)d-(m-n)d = 0⇒a(m-n) + {(m-n)(m+n)}d -(m-n)d = 0⇒ a(m-n) + {(m-n)(m+n) -(m-n)} d = 0⇒a(m-n) +(m-n)(m+n -1) d = 0⇒ (m-n){a+ (m+n-1)d} = 0⇒a + (m+n -1)d = 0/(m-n)⇒a + (m+n -1)d = 0

Km

Light answer bro I have solve this question

Sorry light nahi right

9/14

0

9/14 is greater than 00 is greater than -3/11

the

I can help you bro 😎 be cool

tq bro

tq bro☺️

The first four terms of the sequence (n-1)(2-n)(3+n) are 1 ) put n = 1, we get = 02) put n=2 , we get = 03) put n=3 , we get =2*(-1)*(6) = -124) put n=4 , we get = 3*-2*7 = -42.

the sequence is 0,0,-12,-42

him can you give me your no

hi

the line drawn through the midpoint of one side of a triangle parallel to another side half of the third side.

1.Non Terminating2. Non Terminating3.Terminating4.Non Terminating5.Terminating

1.Non Terminating2.Non Terminating 3.Terminating 4.Non Terminating 5.Terminating

v) is the correct answer

1.Non terminating2. Non terminating3.Terminating4.Non terminating5.Terminating

2 is the correct answer

1.non terminating2.non terminating3.terminating 4. non terminating5.terminating

if You would be copying from brainly.in , I'd directly go their instead of asking here. Go off!!

(2x+1)(2x+1) + (x+1)(x+1) = 6x+47 4x*x+4x+1+x*x+2x+1 = 6x+47 5x*x = 45 x*x = 9 x = -3,3

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