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8/100× x=2x=(2×100)/8 = 25

the number whose 8% is 2 is 25(a(

We have to find the inside distance and outside distance.for inside distance , radius = 140/2 =70 mlength of circumference =2π×radius =2×(22/7)×70 =440 mnow, outside distanceradius = 70 +7 = 77 mlength of circumference = 2π× radius = 2×(22/7)×77 = 484 mtotal length = 440 +484 =924 mcost = Rs 7× 924 = Rs 6468

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z=1/4+3/4i is the right answer

Let PQ be the chord of a circle with center OLet AP and AQ be the tangents at points P and Q respectively.Let us assume that both the tangents meet at point A.Join points O, P. Let OA meets PQ at RHere we have to prove that ∠APR = ∠AQRConsider, ΔAPR and ΔAQRAP = AQ [Tangents drawn from an internal point to a circle are equal]∠PAR = ∠QARAR = AR [Common side]∴ ΔAPR ≅ ΔAQR [SAS congruence criterion]Hence ∠APR = ∠AQR [CPCT]

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TheHimalayasstretch across the northeasternportionof India. They cover approximately 1,500 mi (2,400 km) and pass through the nations of India, Pakistan, Afghanistan, China, Bhutan and Nepal.

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3x + 2y - 4 = 0

9x + 6y - 12 = 0

Ratio are

3/9 = 1/3

2/6 = 1/3

(-4)/(-12) = 1/3

So, this case of infinite solution

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James Millhas divided Indian history into three periods. According to Mill before the Britishcame to India, Hindu and Muslim dictators ruled the country.

FIRST LAW OF THERMODYNAMICS

The quantity of heat energy of heat energy absorbed by the system is equal to the sum of increases in the internal energy of the system and the external work done by it.

dQ = du + dw

But, dw = Pdv

∴ dQ = du + Pdv

{ SIGN CONVENTION }

(1) Heat is absorbed by the system dQ = +ve

(2) Heat is released by the system dQ = +ve

(3) Work done by the system dw = +ve

(4) Work done on the system dw = -ve

(5) Increase on internal Energy du = +ve

(6) Decrease in Internal Energy du = -ve

Now, Let us consider a gas with a massless and frictionless pistion. Let the gas expands by a very small volume dV.

Let p is pressure and V is volume of the gas. A is the area of cross-section of the piston, where dx is the distance moved by the piston. Let the volume expansion is dV.

So, the work done by the gas

dW = force× dz dW = pA× dx = p( A dx) = pdV [∴ Adx = dV]Now, dW = pdV

For small displacement dx, the pressure is assumed to be constant.

Thus,from first lawΔQ =ΔU + pdV

where, ΔQ = heat transition between system and surrounding ΔU = change in internal energy of the system.

{ RELATION BETWEEN Cp and Cv }

→ We called this relation as Mayer's Formula.

We can establish the relation between specific heat capacity at constant volume (Cv) and specific heat capacity at constant pressure (Cp) of a gas.

For an ideal gas, the relation between Cp and Cv is

Cp - Cv = R -------------------> (1)

∴ This relation is known as Mayer's Formula.

Now, To establish the relation, we need to begin from the first law of thermodynamics for 1 mole of gas.

ΔQ =ΔU + pΔV

If heatΔQ is absorbed at constant volume,

∴ pΔV = 0 andΔQ = CvΔT for one mole of a gas

Now,ΔV = 0

Then, Cv = (ΔQ/ΔT)v = (ΔU/ΔT)v = (ΔU/ΔT)-----------------> (2)

where the V is dropped in the last step, since U of an ideal gas depends only on the temperature, not on the volume.

Now, heatΔQ is absorbed at constant pressure, then

ΔQ = CpΔT

Cp = (ΔQ/ΔT)p = (ΔU/ΔT)p = (ΔU/ΔT)p

Now, p can be the dropped from the first term since U of an ideal gas depends only on T, not on pressure.

Now, by using Eq. (2)

or Cp = Cv +p(ΔV/Δp)p --------------------> (3)

Now, for 1 mole of an ideal gas, PV = RTIf the pressure is kept constant

p(ΔV/ΔT)p = R -----------------------> (4)

From Eq. (2) , (3) and (4)

{ Cp - Cv = R }

Here, Cp and Cv are molar specific heat capacities of an ideal gas at constant pressure and volume and R is the universal gas constant.

The ratio of Cp and Cv is notified asγ

γ = Cp/Cv

And it is also known as heat capacity ratio,

Cv = R/r-1 and Cp =γ R/r-1

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4x+6y-9x-6y=8-33-5x=-25so x=510+3y=4so y=-2

Total degree in a circle is 36040% of 360°=40*360/100144°option D is correct

given me your number

According to rule3x+2+5x+6= 1808x= 172x= 172/8= x= 21.5°

Given,Cosec theta = 2/root(3) = H/P

Using pythagoras theoram

H^2 = P^2 + B^2(2*2) = (root(3)*root(3)) + B^24 = 3 + B^2B^2 = 4 - 3 = 1B = 1

Therefore,Other trigonometric functionSin theta = P/H = root(3)/2Cos theta = B/H = 1/2Sec theta = 1/Cos theta = 2Tan theta = P/B = root(3)Cot theta = 1/Tan theta = 1/root(3)

The ratio is 1:24 is answer

400:(6×1000)= 400:60000= 1: 150

1m : 1000km400m : 400000km

400000 : 6=200000 : 3

Prime Factors of 197197^1

The Least Common Multiple is 197

Length of diagnol of floor=dd²=10²+10²=100+100=200d=10√2m

Length of largest pole=pp²=(10√2)²+5²=200+25=225p=15m

Or

Length of diagonal of room=√(l²+b²+h²)=√(10²+10²+5²)=√(100+100+25)=√225=15m

Linesareparallelif they are always the same distance apart (called"equidistant"), and will never meet.

6/7=0.853/4=0.75hence6/7 >3/4

7/8 and 5/12 7×12. 5×884. 40 so ,7/8 and 5/127/8. > 5/12

7/8 and 5/12 7×12. 5×8 84. 40 so, 7/8 and 5/12 7/8.>5/12

5/12<7/8 is the correct answer of the given question

-8/9 is greater than 4/-5

-4/5 is greater than -8/9

-8/9 is greater than 4/-5