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(12.1-9.8)/9.8 and multiply it by 100this would be the answerin this case it is 23.469 % increment

B. 55.0percentage decrease

B. 55.0percentage decrease

71-51round off70-60=10 that is the estimated differenceand exact difference is 71-59=12

2)400-300=100 est. diff.425-315=110 exact diff.

3)5000-4000=1000 est. diff5180-3992=1188 exact diff.

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Area of field=length * breadth as breadth is 2/3 of length i.e300so breadth =2/3*300=200m so area=300*200 =60000m² now when 10 m road is built so length=300-20=280m and breadth=200-20=180m so area=length *breadth=180*280=50400m²so area of road=area of rectangle outer-area of rectangle inner =60000-50400 =9600m²

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1)i 90ii 410iii 3,950iv 4,4102)i 690ii36,150iii13,650iv 9,620

i. 90 ii. 410. iii.3950. iv. 4410

i. 700. ii.36200iii.13600. iv.93600

Smallest number is 5995 and largest number is 6004

photo is not complete but largest number is 6004

smallest number is 5995 and largest number is 6004

HCF of 404 and 96404 = 2*2*10196 = 2*2*2*2*2*3Common Factors = 2*2HCF of 404 and 96 = 4

LCM of 404 and 96404 = 2*2*10196 = 2*2*2*2*2*3LCM = 2*2*2*2*2*3*101 = 9696LCM of 404 and 96 is 9696

HCF× LCM = Product of the two numbers⇒ 4*9696 = 404*96⇒ 38784 = 38784

LHS = RHS

Hence, it is verified

into prime factor s

404 = 2 × 2 × 101 = 2² × 101

96 = 2×2×2×2×2×3 = 2^5×3

HCF(404,96) = 2² = 4

[ Product of the smallest power

of each common prime factors

of the numbers ]

LCM( 404,96) = 2^5 × 101 × 3

= 9696

************************************

We know that ,

If h, l are HCF and LCM of

two numbers a and b then

l × h = a × b

************************************

Verification :

l = 9696 ,

h = 4 ,

a = 404 ,

b = 96

l × h = 9696 × 4 = 38784 --(1)

a × b = 404 × 96 = 38784---(2)

Therefore ,

From (1) and (2) , we conclude

that

l × h = a × b

404= 2*2*10196=2*2*2*2*2*3HCF (404,96) = 2*2=4LCM (404,96) =2*2*101*2*2*2*3=9696HCF(404,96)*LCM(404*96)=4*9696=38784product of two numbers =38784Hence, HCF *LCM = product of two numbers .

HCF of 404 and 96

404 = 2*2*101

96 = 2*2*2*2*2*3

Common Factors = 2*2

HCF of 404 and 96 = 4

LCM of 404 and 96

404 = 2*2*101

96 = 2*2*2*2*2*3

LCM = 2*2*2*2*2*3*101 = 9696

LCM of 404 and 96 is 9696

HCF× LCM = Product of the two numbers

⇒ 4*9696 = 404*96

⇒ 38784 = 38784

LHS = RHS

Hence, it is verified.

Thanx bro

The main feature of flowers are following :-

Flowers arise from apical meristems similar to vegetative shoots but, unlike them, have determinate growth. The floral primordia develop into four different kinds of specialized leaves that are borne in whorls at the tip of the stem (see Figure). The two outer whorls are sterile, the inner two fertile. The first formed outer whorl—thecalyx— is the most leaflike and its individual parts, thesepals, often are green. Thepetalsof the next whorl, thecorolla, frequently are brightly colored and in a majority of flowers retain some semblance to leaves. (Together the calyx and the corolla are called theperianth.) The next two whorls, theandroeciumand thegynoecium, are composed of highly modified.

Flowers arise from apical meristems similar to vegetative shoots but, unlike them, have determinate growth. The floral primordia develop into four different kinds of specialized leaves that are borne in whorls at the tip of the stem (see Figure). The two outer whorls are sterile, the inner two fertile. The first formed outer whorl—thecalyx— is the most leaflike and its individual parts, thesepals, often are green. Thepetalsof the next whorl, thecorolla, frequently are brightly colored and in a majority of flowers retain some semblance to leaves. (Together the calyx and the corolla are called theperianth.) The next two whorls, theandroeciumand thegynoecium, are composed of highly modified reproductive structures that have lost their leaf‐like appearance. The androecium is composed ofstamensand the gynoecium ofcarpels. (Pistilis sometimes used as the term for a single carpel or a group of fused carpels.) The stamens are microsporophylls and have a stalk, thefilament, at the top of which thepollen‐bearinganthersare located. A carpel is a megasporophyll and has as its base an enlargedovaryfrom which thestylebearing astigmaarises. The whorls are attached to thereceptaclearea at the end of the flower stalk orpedicel. Some flowers arise singly, but more are produced and arranged in groups calledinflorescences. The stalk of an inflorescence is thepeduncleand the extension of the axis in the inflorescence is therachis, to which the pedicels of the individual flowers are attached.

flowers have smell in them

Plants are majorly classifiedon basis of presence or absence of flower into flowering and non- flowering plants. A flower is a characteristic feature of flowering plants and is actually an extension of the shoot meant forreproduction. Flowers are attractive and appear in different colours and shapes to attract pollinators who help inpollen transfer.

Parts of a Flower



(Source: anmh.org)

Most flowers have four main parts:sepals, petals, stamens, and carpels. The stamens are the male part whereas the carpels are the female part of the flower. Most flowers are hermaphrodite where they contain both male and female parts. Others may contain one of the two parts and may be male or female.

Flowers arise from apical meristems similar to vegetative shoots but, unlike them, have determinate growth. The floral primordia develop into four different kinds of specialized leaves that are borne in whorls at the tip of the stem (see Figure). The two outer whorls are sterile, the inner two fertile. The first formed outer whorl—thecalyx— is the most leaflike and its individual parts, thesepals, often are green. Thepetalsof the next whorl, thecorolla, frequently are brightly colored and in a majority of flowers retain some semblance to leaves. (Together the calyx and the corolla are called theperianth.) The next two whorls, theandroeciumand thegynoecium, are composed of highly modified reproductive structures that have lost their leaf‐like appearance. The androecium is composed ofstamensand the gynoecium ofcarpels. (Pistilis sometimes used as the term for a single carpel or a group of fused carpels.) The stamens are microsporophylls and have a stalk, thefilament, at the top of which thepollen‐bearinganthersare located. A carpel is a megasporophyll and has as its base an enlargedovaryfrom which thestylebearing astigmaarises. The whorls are attached to thereceptaclearea at the end of the flower stalk orpedicel. Some flowers arise singly, but more are produced and arranged in groups calledinflorescences. The stalk of an inflorescence is thepeduncleand the extension of the axis in the inflorescence is therachis, to which the pedicels of the individual flowers are attached.

1 dozen = 12 pieces 12 pens are of 180 so 1 pen is of 180/12 = 15 8 ball pen is of 56 so 1 ball pen is of 56/8 = 7 Cost of a pen to the cost of a ball pen = 15 : 7

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a=22d=20-22=-2an=0n=?Henceformula is an=a+(n-1)dhence0=22+(n-1)(-2)-22=-2(n-1)11=n-1n=12hence 12th term is 0

wrong answer

😡wrong answer😡

Sn=0a=18d=16-18=-2n=?Sn=n/2 (2a+(n-1)d)0=n/2 (36+(n-1)(-2))36-2n+2=038-2n=038=2nn=19

A.T.QGivenSn=0T1=a1=18d=16-18 =-2n=?then, we know thatSn=n/2[2a+(n-1)d]0=n/2[2(18)+(n-1)(-2)]0=n[36-2n+2]0=n[38-2n]0/n=38-2n0=38-2n38=2nn=38/2 =19 Ans

Given:a=27,d=-3, sn=0Formula :S=n/2(2a+(n-1) d) 0=n/2(2*27+(n-1) (-3))0=54-3n^23n^2=57n=19therefore no of terms =19

he drives 52.5 km in 1 hour

Cost price of 8 = 8xCost price of 2 = 2xLoss = (2x/8x)*100 = 25%

If he sells 8 pens,=8xhe is left with 2 pens=2xTherefore,Loss=(8x/2x)×100=(1/4×100)%=25%

eight is a the answer

what is your question?

tysm😊

cubic millimeter to cubic centimeterV cm cube=V mm cube *0.001

between 50 and 250 are4³=645³=1256³=216

answer is 3.

side of a cube is 7.5 cmvolume of a cube side *side*side7.5*7.5*7.5= 421.88cm^3

this answer is incorrect

Please mention which step you found error?

its answer is 2868cm3

and thank you for asking

Cubic has 3 zeorsthanks

3 zero the answer is correct

a cubic polynomial has 3 zeroes

A cubic polynomial has 3 zeroes

yep three is the right answer

A cubic polynomial has 3 zero

a cubic polynomial has 3 zeros

1) It is impossible that the entire class is on leave on the same day.2) It is likely that the maths teacher will teach maths.3) It is It is certain that you will improve your concepts in various subjects.4) It is possible that you will start liking the subjects that you dislike in a weeks time.

a)unlike B)possible c)impossible d)equal

Time taken to travel 340 km at speed of 85km/hr = 340/85=4 hrsHence the bus started at 9.15 A.M

Sin(60+45) + cos(60+45)

=[sin60cos45 +cos60sin45] + [cos60 cos45 - sin60sin45]

=cos 60 sin 45 +cos60 cos45

=1/2 * 1/√2 + 1/2 *1/√2

=1/√2

cos45

LHS = sin105° + cos105°

= sin(60 + 45 ) + cos ( 60 + 45 )

= ( sin 60 cos 45 + cos 60 sin 45 )

+ ( cos 60 cos 45 - sin60 din 45)

= { ( √3/2 × 1/√2 ) + ( 1/2 × 1 √2)} + { ( 1/2 × 1√2 ) - ( √3/2 × 1/√2)

= ( √3 /2√2 + 1 /2√2 + 1 2√2 + √3 /2√2 ) = 1/√2 = RHS

LHS = sin105° + cos105°

= sin(60 + 45 ) + cos ( 60 + 45 )

= ( sin 60 cos 45 + cos 60 sin 45 )

+ ( cos 60 cos 45 - sin60 din 45)

= { ( √3/2 × 1/√2 ) + ( 1/2 × 1 √2)} + { ( 1/2 × 1√2 ) - ( √3/2 × 1/√2)

= ( √3 /2√2 + 1 /2√2 + 1 2√2 + √3 /2√2 ) = 1/√2 = RHS=cos45°