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Total distance will be 2*5km=10km

Menaka travelled 5 km from house to school and back againTotal distance she traveled is = 5+5=10 km

Area of Trapezium=1/2×sum of parallel sides× perpendicular distanceSo,34=1/2×(10+x)×4Or,34=2(10+x)Or,17=10+xOr,x=17−10=7cm

AD/DB= AE/ECsohereAD/7.2= 1.8/5.4AD= 2.4cm

m=1 is the right answer

so that the answer is m = 1

m= 1 is the correct answer

m=1 is the correct answer

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sol.(-3)m+1+5=(-3)7=(-3)m+6=(-3)7=m+6=7=m=7-6=m=1 ans.

(-3)^m+1*(-3)^5=(-3)^7(-3)^m+1+5=(-3)^7m+6=7m=1

m= 1is the correct answer

m+5+1=7m=7-6m=11 is the answer

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.....m+1+5=7then m=1

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ఆన్సర్ ఇస్ వన్ (-3)m+⁵☆ (-3)⁵ = (-3)⁷(-3)m+¹+⁵ =(-3)⁷(-3)m+⁵ = (-3)⁷

therefore m+6 =7m=7-6m = 1

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m+5+1=7m=7-6Therefore m=1

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( -3)m+1+5=(-3)7, m+6= 7,m= 7-6=1 right answer is m=1

here bases are equal then powers should be added so m+1+5=7;m+6=7; m=7-6;m=1 is the answer

5+7 is12 then 12-1is11 then m is 11

m=1 is the right answer.

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construct a right triangle right angled at B.

construction:construct a perpendicular BD on side AC.

GIVEN:angleB=90

TO PROVE:AB^2+BC^2=AC^2

PROOF: In triangle ABD and triangle ABC,

angle A=angle A

angle ADB=angle BDC=90°

therefore, triangle ADB is similar to triangle ABC.

=>AD/AB=AB/AC (sides are in proportion)

=>AD×AC=AB^2-------1

Similarly, triangle BDC is similar to triangle ABC

BC/DC=AB/BC (sides are in proportion)

which implies AC×DC=BC^2-----2

adding 1 and 2

AD.AC=AB^2

AD.AC+AC.DC=AB^2+BC^2

=AC(AD+DC)=AB^2+BC^2

=AC^2=AB^2+BC²

Hence proved.

Let the lengths of parallel sides = 10cm and b cm

height = 4cm

area = 1/2 × (10+b) × 4

34 = 2 × (10+b)

10+b = 34/2 = 17

b = 17-10 = 7cm

the length of the other parallel side is 7cm

117 letters in one day so in 28 days 28×117 = 3276 letters.

Let the policeman catch the thief in n minutes.

Given that uniform speed of the thief = 100m/min.

Given that After 1 minute a policeman runs to catch him at the speed of 100 in first minutes.

(n + 1) minutes = 100(n + 1) minutes.

Given that speed of policeman increase by 10m/min.

The speed of police in 1 minute = 100m/min.

The speed of police in 2 minutes = 110m/min.

The speed of police in 3 minutes = 120m/min.

The speed of police in 4 minutes = 130m/min.

Hence 100,110,120,130 are in AP.

Let a be the first term and d be the common difference.

a = 100,d = 110 - 100 = 10.

We know that sum of n terms of an ap = n/2(2a + (n - 1) * d)

= n/2(2(100) + (n - 1) * 10) ------ (1)

The distance traveled by the thief = Distance traveled by the police

100(n + 1) = n/2(2 * 100+ (n - 1) * 10)

100n + 100 = n/2(200 + 10n - 10)

200n + 200 = 10n^2 + 190n

10n^2 + 190n - 200 = 200n

10n^2 - 10n - 200 = 0

n^2 - n - 20 = 0

n^2 -5n + 4n - 20 = 0

n(n - 5) + 4(n - 5) = 0

(n -5)(n + 4) = 0

n = 5 (or) n = -4.

n cannot be negative.

Therefore the time is taken by the policeman to catch the thief = 5minutes.

Let the police catch the thief in n min

As the thief ran 1 min before the police...time taken by the thief before being caught = (n+1) min

Distance travelled by the thief in (n+1) min = 100(n+1)m

Speed of police in 1st min=100m/min

Speed of police in 2nd min=110m/min

...3rd min = 120m/min.........so on

100,110,120,............ this forms an AP

Total distance travelled by the police in n min = n/2(2 x 100 +(n-1)10)

On catching the thief by police,distance traveled by thief= distance travelled by the police

100(n+1)= n/2(2 x 100 + (n-1)10)

100n + 100= 100n +n/2(n-1)10

100=n(n-1)5

n2-n-20 = 0

(n-5)(n+4) = 0

n-5 = 0

n= 5 OR n= -4...(but this is not possible)

so, n= 5

Time taken by the policeman to catch the thief = 5min

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Thief speed=100m/mintime taken by police before catching the thief= xthen the time of thief being catched= x+1as the police increases his speed 10 m/min every succeeding minute,his time taken forms an ap- 100,110,120......

distance of thief=time*speed=100*(x+1)-----------------1

distance covered by police=sum of x terms of the apsn=x/2 (2(100) + (x-1) 10)sn=x(100+5x-5)sn=95x-5x square---------------2as the distance covered by police and thief are same,1=295x-5x square=100*(x+1)95x-5x square=100x + 1005x square-5x-100=0xsquare-x-20=0

if we solve this equation,we will get x= -4 and x=5as time cannot be negative, 5 is the right answer.thus the time taken by the police to catch the thief=5 mins.

maan lo ki anupat x hai to sankhya 3x aur 8x ho jaegi therefore , dono ka antar hai 116 so, 8x - 3x = 116 5x = 116 x = 116/ 5badi sankhya 7x = 7 × 116/5= 162.4 hai

Forpolygons(includingtriangles),similaritymeans that the corresponding angles are same. The converse is also true, that is, if the corresponding angles of twopolygonsareequal , then the twopolygonsaresimilar. ... If all the sides are scaled by a common factor, the two polygonsare not necessarilysimilar.

The degree to which a fluid resists flow under an applied force, measured by the tangential friction force per unit area divided by the velocity gradient under conditions of streamline flow.

The degree to which a fluid resists flow under an applied force, measured by the tangential friction force per unit area divided by the velocity gradient under conditions of streamline flow, is the right answer

coefficients of viscosity. The degree to which a fluid resists flow under an applied force, measured by the tangential friction force per unit area divided by the velocity gradient under conditions of streamline flow.

the ratio of applied stress upon strain. (n). or the degree to which a fluid resist flow under applied pressure is knows as viscosity

the degree to which a fluid resists flow under an applied force is called velocity

the degree to which a fluid reststs flow under an applied force, measured by the tangential friction force per unit area divided by the velocity gradient under conditions of streamline flow.

coefficient of viscosity. noun. pl.coefficients of viscosity. The degree to which a fluid resists flow under an applied force, measured by the tangential friction force per unit area divided by the velocity gradient under conditions of streamline flow.

The degree to which a fluid resists flow under an applied force, measured by the tangential friction force per unit area divided by the velocity gradient under conditions of streamline flow.

the drgree yo which a fluid resists

coefficient of viscosity. noun. pl.coefficients of viscosit tangential friction force per unit area divided by the velocity gradient under conditions of streamline flow.

Viscosity is defined as thedegree up to which a fluid resists the flow under an applied force; it is measured by the tangential friction force acting per unit area divided by the velocity gradient under conditions of streamline flow.

coefficient of viscosity is defined as the viscous dragging force which maintain a unit velocity gradient between two parallel layers each of unit area. it decreases with temperature its unit poise

the degree to which a fluid resist Flow under an applied force is called applied

the degree to which a fluid resists flow under an applied force measured by the tengential friction force per unit area divided by the velocity gradient under conditions of streamline

The degree to which a fluid resists flow under an applied force, measured by the tangential friction force per unit area divided by the velocity gradient under conditions of streamline flow.......,.

the slow flowing property of liquid is called viscocity

the measure of the viscosity of a fluid, equal to the force per unit area required to maintain a difference of velocity of one unit distance per unit time between two parallel planes in the fluid that lie in the direction of flow and are separated by one unit distance: usually expressed in poise or centipoise.

The degree yo which a fluid resists

the coefficient of viscosity. noun. pl.coefficients of viscosity. The degree to which a fluid resists flow under an applied force, measured by the tangential friction force per unit area divided by the velocity gradient under conditions of streamline flow.

The degree to which a fluid resists flow under an applied force, measured by the tangential friction force per unit area divided by the velocity gradient under conditions of streamline flow

It may be defined as The degree to which a fluid resists flow under an applied force, measured by the tangential friction force per unit area divided by the velocity gradient under conditions of streamline flow

Coefficient of viscosityis the degree to which a fluid resists flow under an applied force. It is expressed as the ratio of the shearing stress to the velocity gradient. As the temperature increases thecoefficient of viscosityof liquids decreases because the bonds between molecules are weakened

coefficient of viscosity- a measure of the resistance to flow of a fluid under an applied force

Given In ΔABC, PQ || AB and PR || AC and RQ || BC.To show BC = 1/2 QRProof In quadrilateral BCAR, BR || CA and BC|| RASo, quadrilateral, BCAR is a parallelogram.BC = AR …(i)Now, in quadrilateral BCQA, BC || AQand AB||QCSo, quadrilateral BCQA is a parallelogram,BC = AQ …(ii)On adding Eqs. (i) and (ii), we get2 BC = AR+ AQ⇒ 2 BC = RQ⇒ BC = 1/2 QRNow, BEDF is a quadrilateral, in which ∠BED = ∠BFD = 90°∠FSE = 360° – (∠FDE + ∠BED + ∠BFD) = 360° – (60° + 90° + 90°)= 360°-240° =120°

13/30 is the answer .total chances is 30 So probability is 13/30

Total students = 30Girls = 13So probability = 13/30.

thanks to all my friends for answering my question

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12cm is the correct answer

12 cm is the following question answer.

when the value of x is rational

then, f(x) = x , f(f(x)) = x ...

when x is irrational..then , f(x) = 1-x , f(f(x))= 1-(f(x)) = 1-(1-x) = x...

so both the functions are same..

and rational and irrational makes up real set.

therefore option 1 is correct..

Solution:

Given: ΔABC

Prove that sinb/sinc = c- acosb/b-acosc

Use cosine formulas:

cosb = (a²+c² - b²)/2ac and cosc = (a² - c² + b² )/ 2ab

RHS:

c- acosb/b-acosc = (c- a (a²+c² - b²)/2ac)/ (b - a (a² - c² + b² )/ 2ab)

= (b/c) × ((2c² - a² + c² - b²)/ (2b² - a² - b² +c²))

= (b/c) × ((c² - a² + b²)/ (b² - a² +c²))

= b/c ... (1)

Now, use the sine rule,

sina/a = sinb/b = sinc/c = i/k

So, a = asina, b = ksinb and c = ksinc

Substitute the value of b and c in equation (1).

b/c = ksinb / ksinc

= sinb / sinc

Hence proved.

32√3