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Let AB = x, BQ = y, PQ = z

tan30∘=zx+ytan⁡30∘=zx+y

=>z=x+y3–√z=x+y3

tan60∘=zytan⁡60∘=zy

=>z=3–√yz=3y

∴x+y3–√∴x+y3=3–√y=>x=2y=3y=>x=2y

∴y=x2∴y=x2

To go x, it takes 10 minutes.

To go y, it takes 5 minutes

x,it takes 10 minutesy,it takes 5 minutes.answer

After walking 5 km towards south, when he turns right, He is facing west. After walking 3 km towards west, when he turns left, He is facing south Direction.

Hence, he is now walking in south direction

A----------------------------B55km= ABTotal time= 1 hour 20 minTotal distance travelled= 30kmS1= 36km/hrT1= distance travelled/S1=30/36=5/6 hr=50minremaining time= 40 min = 4/6 hrremaining distance= 25kmrequired speed= 25/(4/6) km/hr=25*6/4 km/hr=150/4 km/hr= 37.5 km/hr

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Solution :- cost of one toys = 192/8 = 24 cost of 14 toys = 24 × 14 = 336

10000 invested at interest of 5% p.aso R = 5 and P= 10000given at end of one year so T = 1

now I = PTR/100

so I = (10000×1×5)/100=> I = 500

glt hai

55km= ABTotal time= 1 hour 20 minTotal distance travelled= 30kmS1= 36km/hrT1= distance travelled/S1=30/36=5/6 hr=50minremaining time= 40 min = 4/6 hrremaining distance= 25kmrequired speed= 25/(4/6) km/hr=25*6/4 km/hr=150/4 km/hr= 37.5 km/hr

option B is the correct answer

option b

this is write answer?

d); is the correct answer

In parallel resistance the voltage drops is same across each resistor as current gets divided.

1. interest = 10000*5*1/100

= 500 rupees

2. interest = 3500*7*2/100

= 490 rupees

Let the speed of Joseph = xkm/h

Let the speed of Hamid = ykm/h

When approaching one another, combined speed = (x+y)km/h

Time taken to meet = 30/(x+y) = 1/3 (20 minutes = 1/3 hours)

x+y = 90 ------ (i)

When moving in opposite direction, combined speed = (x-y)m/h

Time taken for Hamid to catch up = 30/(x-y) = 3

x-y = 10 ---- (ii)

Solving (i) and (ii) simultaneously

x+y = 90x-y = 10_______2x = 80x = 40

y = 50

∴ Speed of Joseph =40km/h

Speed of Hamid =50km/h

Let speed of Hamid be vspeed of Joseph be u

Distance covered by Hamid in 20 min= v*20/60 = v/3

Distance covered by Joseph in 20 min= u*20/60 = u/3

v/3 + u/3 = 30v + u = 90.....(1)

Distance covered by Hamid in 3 hrs= 3v

Distance covered by Joseph in 3 hrs= 3u

3v = 3u + 30v - u = 10.......(2)

Add eq(1) and eq(2)

2v = 120v = 60 km/hr

Put value of v = 60 in eq (1)60 - u = 90u = - 30 km/hr

Speed of Hamid = 60 km/hrSpeed of Joseph = 30 km/hr in opposite direction of Hamid

SI = 876*4*8/12*100

=23.36 rupees

I my book your answer is wrong

I had also done like this but answer was wrong so I asked in this app

interest after 8 months=(876×4×8)/(12×100)=23.36

let cot inverse (-√3) = A(-√3) = cotAA = - 30°

120 is the right answer

120 is the correct answer.

120 is the correct answer

120 is right answer of this question. please like my answer

120 is the answer to your question

120 is the correct answer of the following question

Theorem: There is one and only one circle passing through three given non-collinear points.

Given: Three non collinear points P, Q and R

To prove: There is one and only one circle passing through the points P, Q and R.

Construction: Join PQ and QR.

Draw perpendicular bisectors AB of PQ and CD of QR. Let the perpendicular bisectors intersect at the point O.

Now join OP, OQ and OR.

A circle is obtained passing through the points P, Q and R.

Proof: We know that, each and every point on the perpendicular bisector of a line segment is equidistant from its ends points.

Thus, OP = OQ [Since, O lies on the perpendicular bisector of PQ]

and OQ = OR. [Since, O lies on the perpendicular bisector of QR]

So, OP = OQ = OR.

Let OP = OQ = OR =r.

Now, draw a circle C(O,r) with O as centre andras radius.

Then, circle C(O,r) passes through the points P, Q and R.

Next, we prove this circle is the only circle passing through the points P, Q and R.

If possible, suppose there is a another circle C(O′,t) which passes through the points P, Q, R.

Then, O′ will lie on the perpendicular bisectors AB and CD.

But O was the intersection point of the perpendicular bisectors AB and CD.

So, O ′ must coincide with the point O.[Since, two lines can not intersect at more than one point]

As, O′P =tand OP =r; and O ′ coincides with O, we gett=r.

Therefore, C(O,r) and C(O,t) are congruent.

Thus, there is one and only one circle passing through three the given non-collinear points.

this was an alternative of this throrem.

please give me short answer

3 points are known to be collinear if they are in a straight line. But if we want a circle so two points will be collinear and can make a circle. But three collinear points cannot make a circle . Hence it is proved that any three points on a circle cannot be collinear.

Three points can't be collinear coz if they collide then there is a line and no circle is formed.

The integral converges ifs is less than 1 and diverges fors>1

Thales stated this theorem as ratio of any two corresponding sides of an equiangular triangle is always same irrespective of their sizes. And this is known as BPT (Basic Proportionality Theorem).

2t=1. How??

Answer-600please like the solution 👍 ✔️Time= 8 months= 8/12 year

hiii plz like my comments

160 is the correct answer me

B)420 is the right answer as cube root of 1512 is 11.47and cube root of 49000 is 36.59

Option B 420 is correct answer.

Answer is B. 420.Is this correct

Right answer is 420 because 1512 cubevroot 11.47

160 Correct answer hai iska

option b is correct answer.

Given : m||n

To prove : l || n

So, we have alternate angles are equal to 60°. So, using theorem that if twolinesare cut by a transversal and thealternateinteriorangles are congruent,thenthelines are parallel.

Therefore, l || n.

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answer is 56 cm long