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209/247 = 11/13.

initial mean of all 5 numbers = (5+2+7+6+5)/5 = 25/5 = 5now add 2 in all numbers so resultant mean= (7+4+9+8+7)/5 = 35/5 = 7so mean is also increased by 2

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As(45/15)(45/135)=1

3/1)(1/3)= 1

Ans:- 45.

(-3/10)^8-11

= (-3/10)^-3

= -(10/3)³

= -1000/27

The answer is 10/-3 power of 3

_47/4 is thr answer of this question

In algebra, the polynomial remainder theorem or little Bézout's theorem is an application of Euclidean division of polynomials. It states that the remainder of the division of a polynomial by a linear polynomial is equal to In particular, is a divisor of if and only if a property known as the factor theorem

2/9 is the right answer

2^33/9=2^11/3=2^9+2/3= 2^3+2==2^5

2/9 is the best answer

954437176.8888888is correct answer

95444 is the correct answer

4/7=0.57 and 9/11=0.818 then 9/11 is greater than 4/7

let x be common multiplehence 5x and 9x84=5x+9x84=14xX=6hence the greater part is 9xthat is 9*6=54

GivenLM//ONandPQ //RS// TU

if PQ//RS ,thenangle PMN=angle MNS=40° (alternate interior angle)

then, SNO+NOU=180° (co- int. angles)

SNO +75°=180°SNO=105°

if LM//ONthenangle ONM=angle NML (alternate interior angle) SNO+SNM=NML105°+40°=NMLNML=145°

d = 70° (vertically opposite angle)b = 70° ( corresponding angle)c = 70° ( corresponding angle)a = 70° ( corresponding angle)

One ream = 500 sheets

500X6= 3000 sheets

3000X4= 12000 photos

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Here let the number be N.

So, N=121*x+25 where x is quotient.

it can be written as N=11(11*x+2)+3

When N is divided by 11 the remainder is3.

LCM of 8, 12 and 16 = 48

So The nos. which when divided by 12,16 and 8 and leave remainder 3 will be a multiple of 48 + 3These nos. are 51 ( 48×1 + 3), 99 (48×2 + 3), 147 ( 48×3 +3)... and so on. The least of them which leaves no remainder when divided by 7 is 147. So the ans. is 147.

answer is 147 for this question

lcm of 8, 12 and 16=48, remainder 3 multiples 49+3

multiples 49+3=, those numbers of 51, (48+1,+3), 99(48×2+3), 147(48×3+3), which leaves no remainder divided by 7 is 147.

Hi sir can you give me your no pls

fast

Given : In triangle ABC , AD is perpendicular to BC and AD² = BD.DC

To prove : BAC = 90°

Proof : in right triangles ∆ADB and ∆ADCSo, Pythagoras theorem should be apply , Then we have , AB² = AD² + BD² ----------(1)AC²= AD²+ DC² ---------(2)

AB² + AC² = 2AD² + BD²+ DC²= 2BD . CD + BD² + CD² [ ∵ given AD² = BD.CD ] = (BD + CD )² = BC²Thus in triangle ABC we have , AB² + AC²= BC²

hence triangle ABC is a right triangle right angled at A

∠ BAC = 90°

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In triangle ABE and triangle ACF

BE=CF (Given)

∠A=∠A (Common)

∠AEB=∠AFC=90°

By AAS congruent rule ABE ≅ ACF

AB=AC (CPCT)

Similarly

BCF ≅ ABD

AB=BC

AC=BC

AB=AC

AB = BC = AC..... Proved

The triangle is a equaliteral triangle .

In triangle ABC, we apply angle sum property,therefore, x+y+z=180°=>y+x=180°-x. --------(1)Now, exterior angle EBC is sum of interior opposite angles x and z.so angle EBC= x+z=> angleEBC/2=(x+z)/2=>angleOBC=(x+z)/2. --------(2){OBC is half angle EBC as BO is angle bisector}

Similarly, angle DCB=x+yso angleDCB/2=(x+y)/2so angle OCB= (x+y)/2. --------(3)

now in triangle OBC we apply angle sum property.angle OBC + angle OCB + angle BOC=180°

using (2) and (3)(x+z)/2+(x+y)/2+angle BOC=180°=>x+(y+z)/2+angle BOC = 180°

Using (1)x+(180°-x)/2 + angle BOC= 180°=>x/2 +90° +angle BOC=180°=>angle BOC= 90°-x/2hence proved.