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last 4th line is algebraic equation

Like fractions are two or more fractions which have the same value.

Let the three consecutive odd numbers be 'x', 'x+2' and 'x+4' respectively.

A/Q---->x + x + 2 + x + 4 = 693x + 6 = 693x = 63x = 21

The numbers are --->x = 21x + 2 = 21 + 2 = 23x + 4 = 21 + 4 = 25

Among 21,23 and 25 it is found that 23 is a prime number.

Inner radius of the bowl= 4cm

Height of the cylindrical bowl= 6-2 = 4 cm

Volume of the cylinder = pi*r^2*h

= 22/7 *4*4*4

= 200.96cm^3

Now.,,, The total soup that can be available for 70 bowls= 200.96*70 =14067.2cm^3

Givensin−1(x)+sin−1(1−x)=cos−1(x)sin−1(1−x)=cos−1(x)−sin−1(x)⇒(1−x)=sin[cos−1(x)−sin−1(x)]Usingsin−1(x)+cos−1(x)=π2So(1−x)=sin[π2−2sin−1(x)]=cos(2sin−1(x))=cos(cos−1(1−2x2))Using2sin−1(x)=cos−1(1−2x2)So(1−x)=(1−2x2)⇒2x2−x=0So2x2−x=0⇒x=0,x=12Now Put into Original Equation we get x=12 and

x=0 satisfy the Given equation.

Given : 5x² - 3x + 2

a + b = 3/5

ab = 2/5

(a + b)--------- ab

3/5------2/5

3/2

336laddos will be there in 12 kg 21 boxes are needed to pack all laddos

12×28=336 336÷16=21

12 × 28 = 336336 ÷ 16 = 21

a) Ratio = 1:8

Let the length of the edge of the cube be 'x' cm

(i) Total surface area = 6x² cm²

Increased length of the edge = 2x

Total surface area = 6(2x)²cm²= 24x²cm²

If the edge of the cube is doubled surface area of the cube increases by 4 times.

Ratio = 1:4

fractions in ascending order -3/4 , -5/8 , -2/7 , -1/5 , -1/8 so middle fraction is -2/7

Is it c(a,2)?

yea

sum =3p+4q-9r+3r+13q-8p=-5p+17q-6rhence to get the resultxp+yq+zr-(-5p+17q-6r)=2p+2q+2rhenceX+5=2,X=-3y-17=2,y=19z-6=2,z=8hencethe expression will be -3p+19q+8r

i want another method

Side of the cube=x metresIncrease in side=1%=0.01×x=0.01xVolume of a cube =l×b×h=x×x×x=x3Step 2:Approximate change in volume =ΔV=dvdx×ΔxV=x3dvdx=3x2[Differentiating with respect to x]ΔV=dvdx×Δx=3x2×0.01x=0.03x3m3

so area of these triangle is =(4\9)²=16\81=16:81 is the right answer

16:81 is the correct answer

area of 1st ∆/area of 2nd ∆= (1st ∆ side)^2/(2nd ∆ side)^2let sides are 4x and 9x = (4x)^2/(9x)^2 =16x^2/81x^2 = 16/811st ∆ area : 2nd ∆ area = 16 : 81

5 root(2) + 2 root(8) - 3 root(32) + 4 root(128)

= 5 root(2) + 2 root(4*2) - 3 root(16*2) + 4 root(64*2)

= 5 root(2) + 2*2 root(2) - 3*4 root(2) + 4*8 root(2)

= (5 + 4 - 12 + 32) root(2)

= 29 root(2)

a^2 - b^2 - a - b

(a-b)(a+b) -(a+b)(a+b)((a-b) - 1)(a+b)(a-b-1)

in Triangle ABC = tan30° = h/511/√3= h/5151/√3 = hor 88.34 m

It is given that the angle of depression is30°30°and after six seconds, the angle of depression is60°60°.



Let AB be the height of the tower,C is the initial position of the car and D is the position after six seconds.

InΔADBΔADB,

ABDB=tan60°ABDB=√3DB=AB√3ABDB=tan60°ABDB=3DB=AB3

InΔABCΔABC,

ABBC=tan30°ABBD+DC=1√3AB√3=BD+DCABBC=tan30°ABBD+DC=13AB3=BD+DC

Simplify further,

AB√3−AB√3=DCDC=3AB−AB√3=2AB√3AB3−AB3=DCDC=3AB−AB3=2AB3

The speed of the car is,

Speed=DistanceTime=2AB√36=AB3√3m/sSpeed=DistanceTime=2AB36=AB33 m/s

The time taken to travel CD that is,2AB√32AB3distance is66seconds.

The time taken to travel BD that is,√33distance is,

Time=62AB√3×AB√3=62=3secondsTime=62AB3×AB3=62=3 seconds

Therefore, car will take3seconds3 secondsto reach the foot of the car.

516 m is to a distance between the bus 🚌 and the boy

44/100 × 170 is the answer please accept it sir

96 % is the correct answer of the given question

length will be - 1.3Lwidth- 0.7 Wnew are will be = 0.91 LBpercentage change in area= 0.91LB- LB/100-9decrease by 9 Percentage

but how will be the length 1.3

l+ ( l*30/100)so it will be 1.3l

why 1+ 1× 30/100

percentage increase is written there that's why

Properties of Similar Triangles:1)Corresponding angles are congruent (same measure) So in the figure above, the angle P=P', Q=Q', and R=R'.2)Corresponding sides are all in the same proportion. Above, PQ is twice the length of P'Q'. Therefore, the other pairs of sides are also in that proportion.

3 m cloth cost = 79.50no. of cloth =79.50no.of cloth in meter=3 79.50/5 =26.5

Now 15m cloth =26.5×15 =397.5 Ans.

answer of your question is 397.5

Let the other no. be xSo, - 14/5+x=-2=> x=-2+14/5=4/5

let other be x, A/c to quex+(-14/5) =-2x=4/5

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AD= AE = lengths of tangents from point A to the ex-circle of triangle ABC.BF = BD = lengths tangents from B onto the excircle.CE = CF = lengths of tangents from C onto the excircle. AD = AB + BD = AB + BF AE = AD = AC + CE = AC + CFAD+ AE = 2 AD= AB + BF+ FC + CA = perimeter of triangle ABC So perimeter = 2 * 5 cm = 10 cm

Answer is not matching from my book

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Area of shaded region = area of semicircle of diameter BC - {area of quadrant of radius AB /AC - area of ∆ABC }

So, area of semicircle of diameter BC = 1/2 πr² = 1/2 × 22/7 × 7√2 × 7√2 [ ∵ BC is hypotenuse of right angle ∆ABC , here AB = BC = 14 so, BC = 14√2 = 2 × radius ⇒ radius = 7√2 ]= 11 × 7 × 2 = 154 cm²

area of quadrant of radius AB/AC = 1/4 πr² = 1/4 × 22/7 × 14 × 14 = 22 × 7 = 154 cm²

area of ∆ABC = 1/2 height × base = 1/2 × 14 × 14 = 98 cm²

Now, area of shaded region = 154cm² - { 154cm² - 98cm²} = 98cm² Hence, area of shaded region = 98 cm²

Time taken in 1 round=3*60+15=195 secondsTime taken in 60 rounds=60*195=11700seconds=195 minutes