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Putting the limits, we will get - arcsin(-1/2) =30 degree or (pi)/6

Given:secA = 17/8

We know, sec A = Hypotenuse/Base = 17/8

So, draw a right angled triangle PQR, right angled at Q such that∠ PQR = ABase QR = 8 and Hypotenuse PR = 17

Using Pythagoras Theorem in ΔPQR

PR² = PQ² + QR²⇒ (17)² = PQ² + (8)²⇒ 289 = PQ² + 64⇒ PQ² = 289 - 64⇒ PQ² = 225⇒PQ = 15

L.H.S. = 3-4sin²A/4cos²A-3

⇒ 3-4(15/17)²/4(8/17)² - 3

⇒ 3-4×(225/289)/4×(64/289)-3

⇒ {(867-900)/289}/{(256-867)/289}

⇒-33/289× 289/-611= 33/611

R.H.S = (3-tan²A)/(1-3tan²A)

⇒ 3-(15/8)²/1-3(15/8)²

⇒ 3- (255/64)/1- (675/64)

⇒-33/64× 64/-611

⇒ 33/611

So, L.H.S. = R.H.S. Hence proved

-6 is the right answer

answer is 6 of that statement

accept me best and like my ans.

(-18)-(-12)=-18+12=-6

point (2,3):x=2, y=3Both x and y coordinates are positive, therefore the given point must lie in the first quadrant

you have taken more time . so ot is not available. bcs my student is boring

(HCF)(LCM) = Number1xNumber2

(7360)(20) = Number1xNumber2

So r = (360*2.2)/(pi) = 252cmAs r is the distance from eyes. So the coin should be kept at a distance of 252cm.

Concentric circles are circle having same center of a circle.

Let O bethe centre of the circle. A, B and C represent the positions of Reshma, Salmaand Mandip.

AB = 6cm and BC = 6cm.

Radius OA = 5cm.(given)

Draw BM ⊥ AC

ABC is an isosceles triangle as AB = BC, M ismid-point of AC.

BM isperpendicular bisector of AC and thus it passes through the centre of thecircle.

Let AM = y and OM = x then BM = (5-x).

In ΔOAM,

OA²=OM²+AM² ( by Pythagorastheorem)

5²=x²+y²—(i)

In ΔAMB,

AB²=BM²+AM²(by Pythagoras theorem)

6²= (5-x)²+y²— (ii)

Subtracting (i) from (ii),

36 – 25 = (5-x)² -x²

11 = (25+x²– 2×5×x) - x²

11= 25+x²-10x - x²

11= 25-10x

10x = 14

x= 7/5

Substituting the value of x in (i), we get

y²+ 49/25 = 25y²=25 – 49/25y²=(625 – 49)/25y²=576/25

y = 24/5

Thus,

AC = 2×AM

AC = 2×y

AC= 2×(24/5) m

= 48/5 m = 9.6 m

Hence, the Distance between Reshma and Mandipis 9.6 m.

answer is wrong

suppose x boys and y girlsso x+y=18060x+45y=180×504x+3y=12×50=6004x+3y-3x-3y=600-540so y=60so x=120so 120 boys and 60 girls

Exaction is the action of demanding and obtaining something from someone, especially a payment.

then RS =6 cm Because both angles are same

Let the volume of the pool = V

Let the time taken by second pipe to fill the pool = x hours

time taken by first pipe to fill the pool = (x+5) hours,

and time taken by third pipe to fill the pool = (x-4) hours.

In One hour,

First pipe can fill parts of volume of pool

Second pipe can fill parts of volume of pool

Third pipe can fill parts of volume of pool

Now, It is given that,

Time taken by first and second pipe simultaneously = Time taken by third pipe alone

Neglect the negative value.

Thus,the time taken by second pipe to fill the pool = x hours=10 hours.

time taken by first pipe to fill the pool = (x+5) hours= 10+5=15 hours.

time taken by third pipe to fill the pool = (x-4) hours=1-4=6 hours.

Please ost one question to be solved.

11.Let the number be x

2x + 7 = 492x = 42 x = 21

To.find this add all the number of bowls470+309+163=942 bowls

Thuesday-470.Wednesday -309.Thursday -163.add all 470+309+163=942

1 bowl costs = 90/6 = 15 rupees

10 bowl costs = 10*15 = 150 rupees

Cost of 6 bowls is 90

So, cost if one bowl is 90/6 = 15

Therefore, cost of 10 bowls

is 15 × 10 = Rs 150

Let x hours be the total time taken by the larger pipe to fill the tank

so in 1 hour it would fill 1/x part of the tank

similarly y hours are needed for the smaller pipe ,then in 1 hour it would fill 1/y part

so, y=10+x -(1)

4/x + 9/y = 1/2

using (1),4/x + 9/10+x =1/2 -(2)

solve (2), to get

x^2 -16x-80=0

(x-20)(x+4)=0since value of x cannot be -vetherefore x=20 and y=30

For volume V of the pool,

the time taken by the pipe 1,2 and 3 respectively are:

(x+5), x and (x-5) hours.

And volume they can fill on the behalf of volume V are:

=> V/(x+5), V and V/(x-5) respectively.

Now given that,

time taken by third pipe is equal to the sum ofthe first two pipes.

Therefore,

V/(x+5) + V/x = V/(x-4)

Solving we get,

x = 10 and -2

Hence,

time required by each pipe to fill the pool is:

10, 15 and 6 hours.

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6soup bowls=1800 ml1 soup bowl=1800/6=300ml

5 bullock carts can hold = 5*525

= 2625 kg

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Using BODMAS rule,

10/9 = 1.1110*1.11 = 11.110+10+11.1 = 31.11

The answer is 31.11

A packet of soup makes 9 bowl of servings

we have to findHow many 3/4 bowl of servings are possible.⇒9÷(3/4)⇒9×4/3⇒12 servings

thanks!

a) 6/12 = 1/2

b) 3/12 = 1/4

c) 3/12 = 1/4

d) 9/12 = 3/4

e) 9/12 = 3/4

f) 9/12 = 3/4

a) 6/12 = 1/2

b) 3/12 = 1/4

c) 3/12 = 1/4

d) 9/12 = 3/4

e) 9/12 = 3/4

f) 9/12 = 3/4

thank you

Soup held by each bowl = Amount of soup/ Number of bowls= 1800/6 = 300 ml