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Speed of flowing of water from canal is 10km/h , means length of water flows in 1 hour = 5 km , so length of water flows in 30 minutes = 5km

Now, volume of water flowing from canal = length of water flows in 30 minutes × breadth of canal × deep of canal = 5000m × 6m × 1.5m = 45000 m³

Let area of field = x m² Then, volume of water irrigates into the field = area of field × height of water during irrigation = x m² × 8× 10⁻² m = 0.08x m³

Now, volume of water flowing from canal = volume of water in field 45000 m³ = 0.08x m³ x = 562500 m²

We know, 1 hectare = 10000 m² So, area of field in hectare = 562500/10000 = 56.25 hectares

Y is the mean proportional between x and z

so, y/z = x/y

y^2 =xz

now,

LHS =xyz(x + y + z)^3

=y(xz){x + y + z }^3

=y^3(x + y + z)^3

=(xy + yz + zx)^3 = RHS

Every year except feb all have 29 in the monthso 11×400=4400now every 4 year feb has 29 days except 100th,200th,300th year so that's 97.so total 4400+97=4497 times 29th day of a month occur in 400.consecutive years.

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Interior angles on the same side of thetransversal:

The pair of interior angles on the same side ofthe transversal are called consecutive interior angles or allied angles or cointerior angles.

If a transversal intersects two Parallel Linesthen each pair of interior angles on the same side of the transversal is supplementary.

If a transversal intersects two lines such thata pair of alternate interior angles is equal then the two lines are parallel.

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Given,

PQ || ST, ∠PQR = 110° and ∠RST = 130°

Construction,

A line XY parallel to PQ and ST is drawn.

∠PQR+∠QRX= 180° (Angles on the same side of transversal.)⇒110° + ∠QRX= 180°⇒∠QRX= 70°

Also,∠RST+ ∠SRY= 180° (Angles on the same side of transversal.)⇒130° + ∠SRY= 180°⇒∠SRY= 50°

Now,∠QRX+∠SRY+∠QRS = 180°⇒70°+ 50°+ ∠QRS= 180°⇒∠QRS= 60°

sin + cos = 1sqauring both sidessin² + cos² + 2sinsin = 1so sincos = 0/2 = 0option (c) is correct

0.1/30 hrs =L/V

0.1*3600/3012sec

but hoe

60degrees

Length of cross section of pipe = 2 dm = 0.2 mBreadth of cross section of pipe = 1.5 dm = 0.15 mArea of cross section = 0.2 × 0.15 = 0.03 m^2Speed of water = 15 km/h = 15000 m/hVolume of water flown in 1 h = 0.03 × 15000 = 450 m^3Dimensions of the water tank = 150 × 100Volume of water for a depth of 3 m = 150 × 100 × 3 = 45000 m^3Time taken to fill the tank to a depth of 3 m = (45000) / (450) = 100 hrs.

Given :PA and PB are tangent to the circlewith centre O.

CD is a tangent to the circle at E which intersects PA and PB in C and Drespectively and PA = 12 cm.

Lengths oftangents drawn from as external point to a circle are equal.

∴ PA = PB = 12 cmCA = CEDB = DENowperimeter of ΔPCD = PC + CD + PD

= PC + (CE+ ED) + PD

= PC + (CA+ DB) + PD

= (PC +CA) + (DB + PD)

= PA + PB

= 12 cm +12 cm= 24 cm.

A line XY parallel to PQ and ST is drawn.

∠PQR+∠QRX= 180° (Angles on the same side of transversal.)⇒110° + ∠QRX= 180°⇒∠QRX= 70°

Also,∠RST+ ∠SRY= 180° (Angles on the same side of transversal.)⇒130° + ∠SRY= 180°⇒∠SRY= 50°

Now,∠QRX+∠SRY+∠QRS = 180°⇒70°+ 50°+ ∠QRS= 180°⇒∠QRS= 60°

Thank you

in the provided solution consider points B and D as vice versa

ΔABD andΔPQD are similar, as the corresponding sides are parallel. x/z = BD / QD => 1/x = QD /(z * BD)

ΔCDB andΔPQB are similar, as corresponding sides are parallel. y/z = BD/ BQ => 1/y = BQ / (z * BD)

Add the two equations: 1/x + 1/y = (BQ + QD) / (z * BD) = 1 /z

ΔABD andΔPQD are similar, as the corresponding sides are parallel. x/z = BD / QD => 1/x = QD /(z * BD)

ΔCDB andΔPQB are similar, as corresponding sides are parallel. y/z = BD/ BQ => 1/y = BQ / (z * BD)

Add the two equations: 1/x + 1/y = (BQ + QD) / (z * BD) = 1 /z

4356121................

4356121 is the right answer

435612 is the correct answer

435612 is the correct answer of the given question

1,00,00,000-5,643,879=4,356,121 is the answer

4356121....is right ans .

4356121 is the best answer

the correct answer is 4356121 of the following question

4356121 is the correct answer

A compiler is a translator which transforms source language (high-level language) into object language (machine language). In contrast with a compiler, an interpreter is a program which imitates the execution of programs written in a source language. Another difference between Compiler and interpreter is that Compiler converts the whole program in one go on the other hand Interpreter converts the program by taking a single line at a time

10000000-4793821=5206179

Required number = 1 crore ‒ 564387 = 10000000 ‒ 5643879 = 4356121

∴ 5643879 is smaller than one crore by 4356121.

10000000 – 5643879Answer = 4356121

Required number = 1 crore ‒ 564387 = 10000000 ‒ 5643879 = 4356121

∴ 5643879 is smaller than one crore by 4356121.

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The mark price of an article is RS 285.6

15.1725 is the correct answer

15 .1725 is the correct answer of the given question

15. 1725 is the correct answer of the following question

Given,

PQ || ST, ∠PQR = 110° and ∠RST = 130°

Construction,

A line XY parallel to PQ and ST is drawn.

∠PQR+∠QRX= 180° (Angles on the same side of transversal.)⇒110° + ∠QRX= 180°⇒∠QRX= 70°

Also,∠RST+ ∠SRY= 180° (Angles on the same side of transversal.)⇒130° + ∠SRY= 180°⇒∠SRY= 50°

Now,∠QRX+∠SRY+∠QRS = 180°⇒70°+ 50°+ ∠QRS= 180°⇒∠QRS= 60°

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y=130 (vertically opposite angle) x=130(corresponding angle)

Volume of heap=l*b*h=5*9*3=135dm^3

Stride length = 9 dm

Distance covered = 18 dm

Number of strides required= 18/9= 2