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According to Euclid's division lemmaa=bq+r, 0<=r<b85=51*1+3451=34*1+1734=17*2+017=[51-34*1]17=[51-{85-(51*1)}*1]17=[51-85*1+51*1]17=[51*2-85*1]17=[(-85*1)+51*2]17=85x+51yx = (-1), y = 2

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answer of the question is option number- c

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c is answer because cosecQ is opposite of sacaQ

this question is option number answer is -c

Bbbbbbbbbbbbbbbbb (√10/3)

b options is correct

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By calculating the value of sec thita is eqaul to option B. so answer is B

h²=p²+b²(√10)²=1²+b²b²=9b=3secA=√10/3

answer is 10√3 bro.thank,follow meplzzzzz

option c is correct as sec = 1/cosec

answer option c is the correct answer

option cccccccccccccccccccccccccccccccccccccc

cccccccccccccccccccccccccccccccccccccc

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cosec¢= h/p= √10/1sec¢= h/b = √10/3 then correct answer is option B

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17y-3=4817y=48-3y=45/17

17y -3=48 ,17y=48+3, 17y=51, y=51/17, y=3

17y=48+3y=51/17y=3

17y=48+317y=51y=51/17y=3

17y-3=48:- 17y-3=48 17y=48+3 17y=51 y=51/17 y=3

17y-3=4817y =48+317y =51y. =51/17y. =3

17y-3=4817y=48+317y=51y=51/17y=3

17y-3=4817y=51y=51/17

17y-3=48or 17y=48+3or 17y=51 y=51÷17 =3

17y-3=4817y=48+317y =51y. =51\17y. =3

17y-3=48 , 17y=48+3 ,17y=51 , y=51÷17 , y=3

17y-3=4817y=51y=51/17y=3

17y-3=48,17y=48+3=51,y=51/17=3 = y=3

17y=48+3then 17y=51now y=51/17and ans is 3y=3

17y-3=4817y=48+3y=51/17y=3

17y-3=48= 17y=48+3= 17y=51= y=51/17= y=3 ans...

y=3 this is the write answer

y=3 is correct answer

17y-3=4817y=48+317y=51y=51/17y=3 ans

17y-3=4817y=48+317y =51y=51/17y=3

17y —3=4817y=48+3y=51 /17y=3

17y -3 = 48

17y= 51

y=51/17so' y=3

17y-3=4817y. =48+317y. =51y. =51/17y. =3 Ans=3

17y=48+3=5117y=51y=3

17y-3=4817y=48+317y=51y=51/17 y=3

17y - 3 = 4817y = 48+317y = 51y = 51/17y = 3

17y-3=48,17y=48+3,17y=51,y=51÷17,y=3

17y=48-3y=41÷3y=17

17y-3=48...17y=48+317y=51y=51/17=3 so.... answer =3/1=3..

when we solve this we found answer is 3

17y-3=48 17y=48+3 17y=51y=51/17y=3

17y-3=48 , 17y=48+3 , 17y=51, y=51/17, y=3

17y-3=48; 17y=48+3; 17y=51; y=51/17

the value of y is 3 ,17y=48+3 ,17y = 51 , y=3

17y-3=4817y=48+3y=51÷17y=3

17y-3=48

17y=48+317y=51y=51/17y=3

17y - 3 =4817y=48+ 317y=51y=3

17-3=4817y=48+317y=51y=51/17y=3

17y=48+317y=51y=51/17=3

17y-3=48,17y=48+3,17y=51,y=51/17,y=3

17y=48+3y=51/17y=3 ans

17y - 3=48 +3

17y = 51

y = 51 ÷17

y = 3

y=3 is right answer of this question

17y=48+3y=51/17y=3 is correct answer

17y-3=48 17y=48+3 17y=51 y=51÷17 y=3

17y-3=4817y=48+317y=51y=51/17y=3is the best answer

17y-3+3=48+3, 17y=51,y=3

17y - 3 = 48 17y = 48+3 y. = 51/17 y. =. 3

17y-3=48:-17y=48+3:-17y=51:-y=51/17:-y=3

17y-3=48:- 17y-3=48 17y=48+3 17y=51 y=51/17 y=3

17y-3=48,17y=48+3,17y=51,y=51/17,y=3

17 ×3-3=48. So the value of y is 3

Answer is y=3 okay if you like my answer please like the button

15x + 17y = 85.....(1)17x + 15y = 75.....(2)

Multiply eq(1) by 17 and eq(2) by 15

15*17x + 17*17y = 85*1717*15x + 15*15y = 75*15

Subtract these equations

(17*17 - 15*15)x = 1445 - 1125(289 - 225)x = 32064x = 320x = 320/64 = 5

Put value of x = 5 in eq(1)

15*5 + 17y = 8517y = 85-75y = 10/17

X + Y = 5 + 10/17 = 95/17

First Subtract Both the equations.17x + 12y =68-12x - 17y =-48___________5x -5y =205(x-y)=20x-y=20/5x-y=4

Solution

2^(x) = 4^(y) =8^(z)convert all base in same variablefrom this equation, we get 2^(x) = 2^(2x) = 2^(3x)y =x/2 and z=x/3put this value in second equation, we get

1/2x + 1/(4*x/2) + 1/(4*x/3)=4

1/2x + 1/2x + 3/4x = 4

by solving this we get, 7 = 16xx = 7/16

Let the provision day to be last now xstudents at beginning 200and provision 30 daysbut 50 students left so obviously provision increase so its inverse variation.200×30=150×x6000=150xx=40 days

There is provision for 120 men for 30 days

After 5 days five men more joined, let for x more days they can sustain on provision

Hence,120*30 = 120*5 + 125*x3600 = 600 + 125x125x = 3000x = 3000/125x = 24

Therefore for 24 more days they can sustain on provision

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15x + 17y = 21........(1)17x + 15y = 11........(2)

Multiply eq (1) by 17 and eq (2) by 15255x + 289y = 357255x + 225y = 165

Subtract eq(2) from eq(1)64y = 192y = 3

Put value of y in eq(1)

15x + 17*3 = 2115x = 21-5115x = - 30x = - 2

X=-2, Y = 3

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tell me the answer

Area= 1/2 × base × altitudeSo, area of given ∆ is:- 1/2 × 7/2 × 4/7=> 1...

Given,Base=7/2Altitude=4/7Therefore, area of the triangle=1/2×7/2×4/7 =2/2 =1

Letthe fixed chargefor taxi= Rs x

Andvariable cost per Km= Rs y

Total cost= fixed charge + variable charge

Given that for a distance of 10 km, the charge paid is Rs 105

x + 10 y= 105… (1)

x= 105 – 10 y

Given that for a journey of 15 km, the charge paid is Rs 155

x + 15 y= 155

Plug the value of x we get

105 – 10 y + 15 y= 155

5y= 155 – 105

5y= 50

Divide by 5we get

y= 50 / 5= 10

Plug this value in equation (1) we get

x= 105 – 10 * 10

x= 5

People have to pay for traveling a distance of 25 km

= x + 25 y

= 5 + 25*10

= 5 + 250

=255

Answer pay for 25 Km is Rs 255

Speed of slower train=xSpeed of faster train = x + 156{x +(x+15)}=570 speed units km/h 6 is hours, so distance will be km.divide by 6 and collect terms(2x+15)=95subtract 15 from both sides2x=80divide by 2x=40 km/h (slower)x+15= 55 km/h (faster)6 hoursslower train 6*40=240 kmfaster train 5+55=330Total is sum, or 570.

(17×11×2) + (17×11×5)

= (17×11)(2+5)

=(17×11×7) is the number.

So, the number is divisible by 17, 11 and 7, and thus, obviously, it's composite.

thanks

this is not my answer

73 km 155 m + 24 km 19 m + 31 km 90 m= ( 73 + 24 + 31) km (155 + 19 + 90) m= 128 km 264 m

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u r rt

ection A:

Number of students = 25

Mean marks = 47

We know that,Sum of observations = Mean × Number of observations

∴ Sum of marks obtained by students of section A = 47 ×25= 1175

Section B:

Number of students = 35

Mean marks = 51

∴ Sum of marks obtained by students of section B = 51 ×35 = 1785

Section C:

Number of students = 30

Mean marks = 53

∴ Sum of marks obtained by students of section C = 53 ×30 = 1590

Now, sum of marks obtained by all students of three sections

= Sum of marks obtained by students of section A + Sum of marks obtained by studes of section B + Sum of marks obtained by students of section C

= 1175 + 1785 + 1590

= 4550

Total number of students = 25 + 35 + 30 = 90

Mean marks scored by 90 students

=sum of marks scored by 90students ÷90

=4550/90

=50.6(approx)

Given:AD is the median of triangle ABC.

E is the mid point of AD.BE produced meet ad at F.

to prove:AF=1/3AC.

Construction:From point D draw DG parallel to BF.

Proof: so by Convere of mid point theorum we get F as the mid point of AG⇒AF=AG(1)

similarly we have G as the mid point of CF⇒FG=GC(2)

from 1 and 2 we get

AF=FG=GC (3)

now AF + FG + GC =AC

From (3) we getAF+AF+AF=AC

3(AF)=AC

AF=1/3AC

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Given ΔPMS andΔQMR are similar. So the ratios of corresponding sides are: PM/ MR = QM / MS = PS/ QR

We have thatAr(ΔPSR) = Ar(ΔQSR) , as the altitude and base are equal.

Hence Ar(ΔPMS+ΔMSR) = Ar(ΔQMR +ΔMSR)Hence Ar(ΔPMS ) = Ar(ΔQMR)

The ratio ofareas of similar triangles = square of ratio of corresponding sides.

If the areas of similar triangles are equal, it means the correspondingsides are equal.

Hence, PS = QR

Length of parallel sides=15 cm and 40 cmTo find height...by Pythagoras theorem..c^2= a^2 + b^241^2= a^2 + 40^21681-1600= a^281=a^2a=9so, the height of the trapezium is 9cm.Area of trapezium= 1/2 *h*(a+b)=1/2*9*(15+40)=4.5*55=247.5cm^2

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Given: ABC is a triangle , DE parallel BC and AD= 3 cm, DB=2 cm and DE=6 cm

To find: AE

sol: Since DE parallel BC

angle ADE= angle ABC (corresponding angles)

and angle AED = angle ACB

Triangle ADE ≈ triangle ABC ( by AA similarity)

therefore , AD/AB=DE/BC=AE/AC (1)

From (1)

AD/AB=AE/AC

2/5=x/18

2×18=5x

36=5x

x=36/5

x=7.5cm

∵ AE = 7.5 cm

Thanks

C.P. of two TV = Rs. 8000

S.P. of tv horse = Rs. 4000

Gain = 25%

∴ C.P. of this tv = Rs (100/125) x 4000 = Rs.3200

C.P. of another tv = Rs. (8000 - 3200) = Rs. 4800

S.P. of this tv = Rs. 4000∴ Loss% = (800/4800) x 100 % = 162/3%

AC= 3cm is the correct answer

C.P.=2*4000=8000now he sold one with 25% gain so S.P. of that TV is 5000so S.P. of another TV must be 3000so that will be loseso lose percentage is 1000/4000*100=25%