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4/5 × -3/8-3/8×1/4+19/20= 4(8)+3(5)/40 - 3(4)+1(8)/32+19/20= 32+15/40 -32+8/32+19/20;+=47/40-26/32+19/20

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Let there be x students.Each student donates ₹x.Total amount = x² = 5625∴ Number of students in the class = √5625 = 75

bro no of student is equal to the money each has given so x^2=5625 and therefore x=75

Vol. of cylinder = πr^2h= 22/7*(4.5/2)^2*10= 22/7*5.0625*10= 159.107143 cm^3vol.of plate = π*r^2*h= 22/7*(1.5/2)^2*0.2= 22/7*0.5625*0.2=0.35357143 cm^3No.of plates = vol.of cylinder/vol.of plates= 159.107143/0.35357143= 450 plates.

2 sin2θ = √3

sin2θ = √3/2

2θ = 60°

θ = 30°

So, cos 30° = √3/2

InΔQMP andΔQPR <QMP =<QPR (EACH = 90 degree) <Q =<Q (COMMON)⇒ ΔQMP SIMILARΔQPR ....(1) (AA similarity) Again inΔPMR andΔQPR, <PMR = <QPR (EACH = 90 degree) <R = < R (COMMON)⇒ ΔRMP similarΔQPR .......(2) (AA similarity)From (1) and (2) we get ΔQMP similarΔPMRTherefore, The corresponding sides are proportional QM/PM = PM/RM⇒ QM.RM= PM. PM (BY CROSS-MULTIPLICATION)⇒ PM² = QM.RM

LetQS = aThusQT = 2aand QR = 3a

In triangle PQRPR^2 = PQ^2 + QR^2PR^2 = PQ^2 + 3a^2PR^2 = PQ^2 + 9a^2 Eqn 1

In triangle PQTPT^2 = PQ^2 + 4a^2 Eqn 2

In Triangle PQSPS^2 = PQ^2 + a^2 Eqn 3

Thus3 PR^2 + 5 PS^2 = 3 PQ^2 + 27a^2 + 5 PQ^2 + 5a^2 = 8 [ PQ^2 + 4a^2 ] = 8 PT^2Hence proved

identity is right answer

Distributive propertyis the best answer

distributive is right because 6 is multiply to both b and c

distributive property is a right answer

distributive property is the best answer

distributive property is the best answer

Distributive is correct answer.

distributive property

commutative is the correct answer

Distance covered, 2S = 8/2 => S=2. Hence, S²=4 square units

A = { 1,{a,b} }

n(A) = 2

P(A) = 2ⁿ

Here n is 2

P(A) = 2² = 4

Solution :-

Given length of pieces - 45 cm, 75 cm and 81 cm

To find the length of the rope we have to find the L.C.M. of 45, 75 and 81. __________________3 | 45, 75, 81 |________________ 3 | 15, 25, 27 |________________3 | 5, 25, 9 |________________3 | 5, 25, 3 |________________5 | 5, 25, 1 |________________5 | 1, 5, 1 |________________ | 1, 1, 1 |

L.C.M. = 3*3*3*3*5*5= 2025 cm

So, the least length of the rope should be 2025 cm which can be cut into whole number of pieces of length 45 cm, 75 cm and 81 cm.Answer.

2025 cm is the answer of the following

length pieces = 45cm, 75 cm, 81 cm, lcm of 45, 75, 81 3 |45, 75 , 81 |_________________ 3| 9, 25, 27 | ________________ 3| 3, 25, 9 |_________________ 3| 1, 25, 3 |__________________ 3| 1, 25, 1 |_________________ 5| 1, 5, 1 |__________________ 5| 1, 1, 1 |___________________ lcm 3x3x3x3x3x5x5=9×9×3×25 =81×3×25=243×25= 6075 cm

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let no of boys be Xhence girls will be 2/5xhence X+2/5x=497x/5=49X=49*5/7=35.. boyshence girls=2/5x=2/5*35=14 girls

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Secondary School

Math

5 points

A triangle PQR is right angeled at Q and points S and T trisect side QR.prove that 8PT^2= 3PR^2+5PS^2.

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byAhadiabdul27.08.2016



from where did you find this question



frm my book



which class and whose publication



class IX ML AGARWAL APC PUBLISHERS



I am from I.C.S.E board

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Let,QS=xQS=ST=TR=xQT=2xQR=3xInΔPQT,∠Q=90°PT²=PQ²+QT² =PQ²+4x² -------------- 1InΔPQR,∠Q=90°PR²=PQ²+QR² =PQ²+9x²InΔPQS,∠Q=90°PS²=PQ²+QS² =PQ²+x²From eq.18PT²=8PQ²+32x²--------------------- eq.2[by multiplying eq.1 by 8]3PR²+5PS²=3(PQ²+9x²)+5(PQ²+x²) =3PQ²+27x²+5PQ²+5x² =8PQ²+32x² =8(PQ²+4x²) =8PT² [from eq.2]

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cost of one heater=7599.60\8 =949.95 is the right answer

8 heaters cost=7599.601 heater cost = 7599.60/8 =949.95

949.95 rupyacorrect answer for this question

8 heaters =7599.601 heater =7599.60 /8949.95

7599.60÷8=949.95is the correct answer

949.95 is the best answer

The cost of one heater=7599/.60/8 = 949.95

The cost of 8 heaters is Rs 7599.60. find the cost of one heater so 7599.60 / 8 = 949.95

the cost of 1 heater=7599.60/8=949.95

The cost of 8 heaters =₹7599.60Cost of one heater =₹7599.60/8=₹949.95

the cost of 1 heater = 7599.60/8=949.95

949.95 is the correct answer

949.95 is correct answer.

Volume of cubes = (15 cm)³

Volume of box = 1.5 m × 90cm × 75 cm= 150 × 90 × 75 cm³

Numbers of cubes fitted = Volume of box/ volume of cubes= 150 × 90 × 75 / 15³= 300

As the difference between them is 36-28=8cm

By Pythagoras theorem,

15²+8²=(distance bet. Their tops) ²

225+64

√289 =distance between tops

17m=distance between Their tops

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mean height = (152+170+156+164+158)/5 = 800/5 = 160 cm

Length of plastic box (l) = 1.5 cm= 0.015mWidth of plastic box (b) = 1.25 mDepth of plastic box (h) = 0.65 m(i) The area of sheet required to make the box is equal to the surface area of the box excluding the top.Surface area of the box = Lateral surface area + Area of the base = 2(l+b)×h+ (l×b) = 2[(0.015+ 1.25)×0.65]+ (0.015× 1.25)m2 2.5675The sheet required required to make the box is 2.5675m^2

Describe plz

this is the formula..it can't be described more

Asscending order= 682,600,720,735

perimeter of square garden = 48 m1 side length = 48/4 m = 12 mArea of garden = 144m²Area of flower bed = 18m²Area not covered by flowers = (144-18)m²= 126m​²Fractional part covered by flower bed = 18/144 = 1/8Ratio = 18 : 126 = 1 : 7

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140 cm is the correct answer which is got by multiplying 160 &7/8