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26 = 2 × 13

39 = 3 × 13

LCM = 2 × 3 × 13 = 78

23*23*23=12167hence 23 is the cube root

Diameters are 112 and 70 cm.So radius=56 and 35 cm

Area of parapet=π(56²-35²)=(22/7)(56+35)(56-35)=(22/7)(91)(21)=22*91*3=22*273=6006 cm²

Avg rainfall= 9+11+8+20+10+16+12/7= 86/7= 12.2

avg rainfall=(9+11+8+20+10+12)/7=86/7=12.2

P(x)=kx3+9x2+4x-8

g(x)= x+3

x= -3

p(-3)= -27K+81-12-8=10-10k

= -17k+51=0

k= -51/ -17

k=3

A leap year is a year which has 366 days. The extra day is the 29th February. There is a leap year every four years.

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Here's a geometric explanation. Considering this is the triangle you're trying to construct (provided it exists).

How'd you construct this triangle? You would draw triangle QRS first, which is pretty basic because you know the length of two sides and the included angle. Now, you draw the perpendicular bisector of QS and P is where RS extended meets the bisector.

Except that SR extended on this side of QS will never meet the perpendicular bisector. That's because angle QSP is obtuse.

To understand why, look at this diagram below, where QS' = RS' by construction:

It's easy to see that RS' = 72√≈ 4.94 cm.

diagram

The answer is D because there's 36 inches in a yard, and 36 x 8 yards = 288 inches.

The prime factors are: 2 x 2 x 2 x 3 x 5 x 7

or also written as { 2, 2, 2, 3, 5, 7 }

Written in exponential form: 23x 31x 51x 71

thanks

total length2m70cm=200+70=270cm2m60=200+60=260cmhence total270+260=530cm5m30cm

1. (a) x + 5 (b) y - 3 (c) 2z

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Please before posting questions make sure the whole question is captured in the image. Also post questions one by one.

can anyone solve this

area of ∆ = 1/2*|(x1(y2-y3) +x2(y3-y1) +x3(y1-y2)| = 1/2*|(2*(0+4) + -1(-4-3) + 2(3-0)| = 1/2*|8+7+6| = 21/2 sq units.

A leap year has 366 days.

Now 364 is divisible by 7 and therefore there will be two excess week days in a leap year.

The two excess week days can be (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday).

So, the sample space S has 7 pairs of excess week days. i.e. n(S) = 7.

Now we want the desired event E to have 53 Sundays and 53 Mondays. E consists of only one pair in S which is (Sunday, Monday). So n(E) = 1

Hence, the probability that a leap year will contain 53 Sundays and 53 Mondays = n(E)/n(S) = 1/7

normally year consists of 365 days .but a leap year has 365+1days. 365 days divided into weekly days has 365/7 i.e 52days+2extra days.

For solving this question we should know a basic fact about a year i.e. it has 52 weeks.

52 weeks means 52×7 = 364 days.

If the year in an ordinary year, the one odd day could be Monday,Tuesday, Wednesday…… Sunday. Thus the probability of getting a sunday is1/7.

If the year is a leap year, the two odd days could be

Mon Tue, Tue Wed, Wed Thu , Thu Fri , Fri Sat , SatSun,SunMon.

Thus the probability of getting a sunday is2/7

Area of triangle = 1/2*Base*Height

For given triangleB = 7 cm, Height = 5 cm

Area = 1/2*7*5 cm^2

Thus Ramu is correct

he spent 50 rupees in total

total money spent by Shantanu= ₹30+ ₹ 20 =₹50

No. Of seconds spent = 8401 second = 1/60 minutesThen 840 seconds = 840/60 min = 14 minutes



1

Secondary SchoolMath6 points

A matchbox is 4 cm long, broad 2.5cm and 1.5 cm in height.Its outer sides are to be covered exactly with craft paper. How much paper will required to do so??Explain in steps . Don't want rubbish answers

Ask for detailsFollowReportbyVenicca200509.01.2018

Answers



rishiraj5Helping Hand

It will take around 4 centimetres of paper

3.7

3 votes

THANKS

8

CommentsReport



YASH3100Genius

To calculate the amount of paper required to cover the matchbox,We need to calculate the surface area of that matchbox with,Length = 4cmBreadth = 2.5cmHeight = 1.5cmTherefore, the surface area of matchbox = 2(lb + bh +hl)Surface area = 2(4*2.5 + 2.5*1.5 + 1.5*4)=> 2( 10 + 3.75 + 6 )=> 2( 19.75 )=> 39.5 cm^2Therefore area of paper required to cover the matchbox = 39.5 cm^2

Hope it helps you.Thank you.Pls do add it as brainliest if u liked it.

39.5cm is the answer of this question

Good

√9/5=3/5 which is less than 1and hyperbola can never have eccentricity less than 1

Heterotrophs are referred to those organisms which cannot prepare their own food. They are dependent on green plants or animals for their food. Their mode of nutrition is known as the heterotrophic mode of nutrition. All the non-green plants and animals, inclusive of human beings, are called heterotrophs.

The non-green plants lack chlorophyll which is necessary to carry out the process of food referred to as photosynthesis. Therefore, they depend on other organisms i.e. plants and animals in order to obtain food.

The non-green plants, for example. Fungi, yeast, mushroom, bread mold, are called heterotrophs. They have a heterotrophic mode of nutrition. Therefore, all animals like dogs, cats, cow, buffalo, lion, tiger, deer as well as human beings are called heterotrophs. Their mode of nutrition called ‘ heterotrophic mode of nutrition ‘.

The probability of a year being a leap year is1/4and being non-leap is3/4

A leap year has 366 days or 52 weeks and 2 odd days. The two odd days can be {Sunday,Monday},{Monday,Tuesday},{Tuesday,Wednesday},{Wednesday,Thursday},{Thursday,Friday},{Friday,Saturday},{Saturday,Sunday}.

So there are 7 possibiliyies out of which 2 have a Sunday. Sothe probability of 53 Sundays in a leap year is2/7.

A non-leap year has 365 days or 52 weeks and 1 odd day. The odd day can be Sunday,Monday, Tuesday,Wednesday,Thursday,Friday or Saturday.

So there are 7 possibilities out of which 1 is favorable. Sothe probabilityof53 Sundays in non-leap year is1/7.

cotx=7/8:; (hypo)^2=7^2+8^2=64+49=113; (hypo)=V113; we see height of (1+sinx)(1- sinx)/(1+ cosx)(1- cosx)=(1- sinx^2)/(1-cosx^2)= (1-64/V113)(1-49/V113)=(V113-64)/V113/(V113-49)/V113=64/49=8/7

total 1236 33.33% on books so money spent on books are 1236*33.33/100 = 412

1140 is the correct answer

suppose average is x8 have spent 12 so 12*8=96ninth have spent 8 more than xso 96+x+8=9x104=9xso total money spent is 104

pp68 4pt612i4i3i37427i2

l7324143y626273374837

a) and c) are not polynomial as the power of variable is negitive and fraction respectively.

b,d where is solve it

c is polynomial as variable power is Integral.

there is no d) question.

sorry

c ko copy me Kaise karenge reason do

Given:

Dimension of matchbox = 4cm × 2.5cm × 1.5cm

l = 4 cm, b = 2.5 cm & h = 1.5 cm

Volume of cuboid = (l × b × h)

Volume of one matchbox = Volume of cuboid

Volume of one matchbox =4 × 2.5× 1.5

= 15 cm³

Volume of a packet containing 12 such boxes = 12× volume of one match box

(12 ×15) cm3= 180cm³

Hence, the volume of a packet containing 12such boxes= 180 cm³

total men and women watched match = 7,12,956total women watched the match = 2,65,567total men watched the match= total members-women =7,12,956-2,65,567 =447,389

√2 is the eccentricity of rectangular hyperbola because the length of transverse axis is equal to the length of the conjugate axis.

X+y(√5-√3)^2+(√5+√3)^2/(√5+√3)(√5-√3)=5+3-2√15+5+3+2√15/5-3=16/2=8

x^2+y^2=(X+y)^2-2xy=(8)^2-2(5-3/5-3)=16-2=14

first π/6= 30° because π means 180 so isi trha sab me π ke jagah 180 rakho then sabka man rakh ke jaise sin30°= 1/2 isi trha sabka man rakh ke aap is 1.question ko proof Kar sakte ho good luck

isi trha next solve karo aayega