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7. in the given figure PQ is a line pQ.OS is another ray lying between rays 0P and OR.prove that Ros =1/2 QOS- POS

Draw circle O and mark points A & B on it (I suggest such that central angle AOB is around 30 to 60 degrees.

Elsewhere, marks points C & D on the circle with central angle COD equaling angle COD. The arrangement of points on the circle will be such that you go from A to B to C to D by moving clockwise around the circle.

Then equal chords AB & CD have equal arcs AB & CD.

Note that arc ABC will equal arc BCD, because arc AB + arc BC = arc BC + arc CD.

The chords of arc ABC & arc BCD will therefore be equal ("equal arcs have equal chords, on a given circle").

Therefore AC = BD

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degree of differential equation is 2.order of differential equation is 2.

A is skew symmetric

A = -A^{T}

|A| = - |A| { ∴ |A| = |A^{T}| }

2|A| = 0

|A| = 0

Therefore, det A = 0

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solution kay hoga

5.30 P. M. [(D)] is the correct answer

(b)4.30 pakad lega sahi he na please

In cylinder,r=7cm=0.7ml=15km=15000m

In tank,l=50mb=44mh=0.21m

Vol.of water in tank=lbh =50*44*0.21 =462m³

Height of cylindrical pipe=Vol. /πr² =462/(0.07)²(22/7) =462/0.0154 =30000m

Time = 30000/15000 = 2 hours

This is the sridharacharya formula .There are other ways to solve the quadraticequationinstead of using the quadraticformula, such as factoring, completing the square, or graphing. Using the quadraticformulais often the most convenient way.

Yes the answer is yes

41.19 percentage

Ans :- Let no.of oranges bought=100

then percentage of oranges rotten=13%

then remaining oranges=100-13%of 100 i.e.=87percentage given to charity out of left oranges=75% of 87 i.e.=21.75

by unitary method 100 - 21.75 then if 522 are left then no.of oranges bought = 522*100/21.75 i.e.=2400

I don't know this answer

Cost price of 11 oranges CP = Rs 10

Selling price of 10 oranges = Rs 11Selling price of 11 oranges SP= 11*11/10 = 121/10 = 12.1

Gain Percentage = [(SP - CP) /CP]*100= [(12.1 - 10)/10]*100= 2.1*10= 21%

when 3 lemons are bought for Rs 4 and 4 lemons are sold for Rs 3 what will be loss

58(51-58)

58(-7)

-406

1 is the difference between the last number and the first number

1 is the difference between the last number and first number

∠FOB + ∠BOE = 180° ...(1)| Linear Pair Axiom

∠FOA + ∠AOE = 180° ...(2)| Linear Pair AxiomFrom (1) and (2),∠FOB + ∠BOE = ∠FOA + ∠AOE ...(3)But ∠BOE = ∠AOE| ∵ Ray OE bisects ∠AOB∴ From (3),⇒ ∠FOB = ∠FOA.

a=2infinite sum=a/(1-r)=6so 1-r=a/6=2/6=1/3so r=1-1/3=2/3

Let Original price per Egg was X.

Increased Price = X + 30% of X = 1.3XNow,

(7.80/X ) - (7.80 /1.3X) = 3

On solving, we get,

X = 0.6Price per egg was Rs. 0.6 .Present price = 1.3X = 1.3 * 0.6Present price for 12 eggs = 1.3 * 0.6 * 12 = Rs. 9.36.

As CA = BC (given)ABC = BAC.

DBC = ECF ( corresponding angles)So,20° + ABC = 70° ABC = 50°

So,GAC = 180° - ABC (linear pair)GAC = 180° - 70° GAC = 150° (Ans)

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DBC=ECF=70° (CORRESPONDING ANGLES)

70°-20°=50°

Therefore, ANGLE GNC=50 °

hi every one I am ludo king

First you should make a rough sketch of triangle.

∠ABD +∠ABC = 180° ….(i) (Angle on the same line)

∠ACE +∠ACB =180° ……….(ii)

Adding (i) and (ii), we get

∠ABD +∠ABC +∠ACE +∠ACB =360°

∠ABD +∠ACE + (∠ABC +∠ACB) =360° …………..(iii)

Now, In ΔABC,

∠ABC +∠ACB +∠BAC = 180°

∠ABC +∠ACB = 180°-∠BAC

Putting the value of∠ABC +∠ACB in eq.(iii), we get

∠ABD +∠ACE +180°-∠BAC =360°

∠ABD +∠ACE = 360°- 180°+∠BAC

∠ABD +∠ACE = 180°+∠BAC

Which proves that sum of the two exterior angles formed is greater than angle A by180°

also figure

2L 750mL=2+750/1000 L=2+0.75 L=2.75 L

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