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Area of the well=2*pi*(5.6)^2 cm2=197.12 m2Area of the rectangular field=150*70 cm2=10500 m2Volume of the mud dug out=Area of the well *depth of the well = 197.12*20 m3=3942.4m3Area of the remaining field=10500-197.12 m2=10302.88 m2let the height by which the field is raised be h m.B.T.P,

10302.88*h=3942.4or, h=0.342 m

Blood, Blood vessels, Heart and lymph are the components of the transport system in human beings.

♦ Blood transports nutrients, respiratory gases (CO2 and O2) waste products, harmones, water, inerganic ions and chemicals.

♦ Blood-vessels (Arteries and Viens)

Arteries carry blood from different organs of the body whereas veins collect the blood from different organs and bring it back to the heart.

♦ Heart - Heart is a pumping organ which keeps blood continuously moving in the blood vessels.

♦ Lymph - Lymph carries digested and absorbed fat from intestine and drains excess fluid from extra cellular space back into the blood.

10 underroot 7 is the answer of question

Total surface area=2(lb+bh+hl)=2(1*.5+.5*1.5+1.5*1)=5.5m^2Two boxes area =5.5*2=11m^2Cost of 1m^2=120RsCost of 11m^2=120*11=1320Rs

Given:1 quintal of onion for 2000rsSelling rate =18rs/kgSolution:1 quintal =100kgCost price of 100kg onions =2000rs1kg =2000/100=20₹100kg onions sp=18x100=1800Loss=2000-1800=200Answer: Therefore the loss is 200rs

2/1.5=3/x so x=(1.5×3)/2=4.5/2=2.250.5/0.005=0.005/x so x=0.005/10=0.0005

Diameter of the well = 7 mRadius of the well = 7/2 mDepth of the well = 20 m

Volume of the mud dug from the well = πr2h= 22/7 × 7/2 × 7/2 × 20= 770 m3

Area of the rectangular plot where the mud is spread = 22 x 14Let the height of the plotfarm be h metres.

Volume of the plot form = Volume of the mud taken from the well22 x 14 x h = 770h = 770 / 22x14h = 2.5 m

it is divisible by x-1 and x-2 so when we put 1,2 for x we will get 0so 1-10+a+b=0so a+b=98-40+2a+b=0so 2a+b=322a+b-a-b=32-9so a=2323+b=9 so b=-14

We have to find the inside distance and outside distance.for inside distance , radius = 140/2 =70 mlength of circumference =2π×radius =2×(22/7)×70 =440 mnow, outside distanceradius = 70 +7 = 77 mlength of circumference = 2π× radius = 2×(22/7)×77 = 484 mtotal length = 440 +484 =924 mcost = Rs 7× 924 = Rs 6468

thank you so much 😃😄😁

x+ 9 = 20=> x = 20 - 9=> x = 11.

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x+ 9 = 20=> x = 20 - 9=> x = 11.

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sin^2pi/18 + sin^2pi/9 + sin^27pi/18 + sin^24pi/9

=sin^2pi/18 + sin^27pi/18 + sin^2(pi/2-7pi/18) + sin^2(pi/2-pi/18)

=sin^2pi/18+ cos^2pi/18 + sin^27pi/18 + cos^27pi/18

1 +1

= 2

Tell me plc

Mundyaa j nhi dsna ta na dsss mintaa jiyaan kdai jnda

Width of the the rectangular plot = 50m

Cost of fencing per metre = ₹ 18

Total cost of fencing = ₹ 4680

SOLUTION-----------------

Therefore ,

Distance covered in fencing =

=> 4680/18

=> 260 m

Perimeter of the rectangular plot =

=> Length + Breadth + Length + Breadth

=> 2 ( Length + Breadth )

Perimeter = 260 m ( solved above )

260 = 2 ( Lenght ( L ) + 50 )

=> 260 = 2L + 100

=> 2L = 260 - 100

=> 2L = 160

=> L = 160/2

=> Lenght = 80 m

ANS. The Lenght of the rectangular plot is 80 m .

plz write clean

Surface area of cuboid box = 2(lxb) + 2(bxh) + 2(lxh)= 2(1x.5 + .5x1.2 + 1x1.5)= 2(.5 + .6 + 1.5)= 2(2.6)=5.2 m^2

Surface area for 2 box = 2 x 5. 2 = 10.4Cost of metal sheet = 10.4 x 120= Rs 1248

a correct answer is 2 days

Here, a well is dug and Earth taken out of it is used to form an embankment.

Given:

Inner Diameter of the well= 14 m

Inner Radius of the well (r) = 14/2 m = 7 m

Height of the well(h) = 15 m

Volume of the earth taken out of the well = πr²h

= 22/ 7 ×(7)²×15

= 22× 7×15= 2310 m³

Width= 7m

Outer radius of the embankment R =inner radius + width

Outer radius (R)= 7 + 7 = 14m

The embankment is in the form of cylindrical shell, so area of embankment

Area of embankment = outer area - inner area

= πR² - πr² = π(R²-r²)

= (22/7) ( 14² - 7²)

= 22/7(196-49)

= 22/7 × 147

= 22 × 21

= 462 m²

Volume of embankment= volume of earth taken out on digging the well

Area of embankment × height of embankment= volume of earth dug out

Height of embankment= volume of earth dug out/area of the embankment

Height of the embankment = 2310 / 462

Height of embankment= 5 m

Hence, the height of the embankment so formed is 5 m

thanku so mch

The height of embankment is 60m

60 m is the height of embankment..

The height of the embankment is 60 m .

between every 2 terms.. the common difference is increasing by 1

in between 2 and 5 ... common difference is 3

in between 5 and 9 , the common difference is 4

so between 9 and other.. the common difference would be 5 so the missing term is 9+5 = 14.

option D

1)Three2)63)84,44)9m²5)8:146)107,00

1

2

3

4

Height of the well = 14 mDiameter of the well = 3 mSo, Radius of the well = 3/2 mVolume of the earth taken out of the well=πr²h= 22/7*(3/2)²*14= 99 cu mOuter radius of the embankment = R = (3/2 + 4)m = 11/2 mArea of embankment = outer area - inner area⇒ = πR² -πr²= 22/7*[(11/2)² - (3/2)²]= 22/7*[(121/4) -(9/4)]= 22/7× 112/4= 88 m²Height of the embankment = Volume/Area= 99/88Height of the embankment = 1.125 m

x + y = 19if x = 1then 1 + y = 19y = 19 - 1 = 18

Answer :b) 18Given: x+y=19x=1Hence 1+y=19y=19-1=18

This is your correct answer

Height = 7m and Diameter = 2m

Find Radius:

Radius = Diameter ÷ 2

Radius = 2 ÷ 2 = 1 m

Find volume:

Volume = πr²h

Volume = π(1)²(7) = 22 m³

diagram plz

Total surface area of hemisphere=4πr^2now as diameter is 2cm hence radius is 1cmTotal surface area=4*π*1=4πcm^2