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LHS-(sinA + cosA)(sinA/cosA + cosA/sinA)

​(sinA + cosA)(sin^ + cos^2A/sinAcosA)

sinA + cosA/sinAcosA

​RHS-

1/cosA + 1/sinA=sinA+cosA/sinAcosA=secA+cosecA

Let the radius of the cone be r and height be h.

Now the height is double to 2h.

So, percentage change in volumenew volume- initialvolume/initial volume1/3πr^2(2h-h)/1/3πr^2h*(100)100 percentage

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By euclid's division algorithm225=135×1+90135=90×1+4590=45×2+0hence the hcf is 45

1cm -> 4m6.5cm -> 4 × 6.5 = 26m

Surface charge density (σ) is the quantity of charge per unit area, measured in coulombs persquaremeter (C. m−2), at any point on a surface charge distribution on a two dimensional surface. Linear charge density (λ) is the quantity of charge permeter length, measured in coulombs per meter. Option D

it will be not loss and not profit

it will not be a loss nor profit

Find the buying price i.e

100%=Buying price

100+5=105% i.e selling price105%=63 rs what about 100% to get the buying price100/105 *63 =60 i.e the Buying priceProfit/loss = Selling price -Buying priceSo ;52.50-60= -7.5When the answer is negative it mean it is a lossLos of rs 7.5

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12 RS per metre selling price

Let the angles are(a - d), a, (a + d)

a - d = 50°........ (1)

Sum of all angles of triangle is 180°(a - d) + a + (a + d) = 1803a = 180a = 180/3 = 60

Put value of a in eq(1)60 - d = 50d = 60 - 50 = 10

Therefore, Angles are:a - d = 50°a = 60°a + d = 60 + 10 = 70°

thanks a lot

Thanks its great

=-5/4+(-1/4)=-5-1/4=-6/4=-3/2

tqsister

141.75 is the correct answer of the given question

Translate D by vector AB, and let its image be E. Quadrilateral ABED is a parallelogram, so EB is equal in length to DA. Now consider triangle EBC. All three sides are known.

EB = 13 cmBC = 15 cmCE = 14 cm

Use Heron's formula. For triangles that is. There is no Heron's formula for quadrilaterals.

(13 + 15 + 14)/2 = 21

area(EBC) = √[21(21 - 13)(21 - 15)(21 - 14)]= √[21(8)(6)(7)]= √(7056)= 84

Let EC be the base of that triangle. Find its corresponding height.

(base)(height)/2 = area14h/2 = 84h = 12

The height of triangle EBC is 12 cm. That is also the height of the trapezium.

area(ABCD) = (AB + DC)h/2= (30 + 44)(12)/2= 444 cm²

Ar(ABC)/ar(PQR)=K²

K IS CONSTANT

K=3/2

SINCE TRIANGLES ARE SIMILAR

BC/PR=K

PR=BC/KPR=15x2/3

PR=10

Area of ∆ ABC/Area of ∆QRP = 9/4

AB = 18 cm , BC = 15 cm So PQ = ?

We know when two triangles are similar then " The areas of two similar triangles are proportional to the squares of their corresponding sides.

Area of ∆ ABC/Area of ∆ QRP = AB2/QR2 = BC2/PR2 = AC2/QP2

Now substitute all given values and get

9/4 = 152/PQ2

Taking square root on both hand side , we get

3/2 = 15/PQ

PQ = 10 cm

ar(ABC)/ar(PQR)=K²

K IS CONSTANT

K=3/2

SINCE TRIANGLES ARE SIMILAR

BC/PR=K

PR=BC/KPR=15x2/3

PR=10

in ABD D=90So DC=root(225-144)=root(81)=9in ABD D=90so BD=root(169-144)=root(25)=5so BC=BD+DC=9+5=14

We know 1km=1000m =1000×100cm =1000×100×10mm

1 kilometre=

100000 centimetres

1 kilometre

100000 Centimeters.

1 kilometre=100000 centimetres

100000 Centimeters.

Time=4hoursDistance=140kmSpeed=Distance/time=140/4=35km/hr

cost of 35 apple =961apple is=96/32=3then cost of 65 apple is =65×3195is right answer

195 is the correct answer of the given question

x=65*36/32x=65*3x=195

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