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Bhai substitution method se batao

proof

proof

y=18 x=36 m=18/7 subtract 2nd eq from 1st and you will get value of y put it in 1 you will get x then put them in 3rd you will get m

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since bisector of 180° will make angle of 90° with AC so it will be perpendicular to AC

dear student use pencil to draw angle

Given Equationx + 3y - 4 = 02x + y - 5 = 0

1/2 3/1(-4)/(-5) = 4/5

Unique solution exist

we have given that:curved surface area of cone = 1159 5/7 = 8118 / 7 cm²

andarea of its base = 254 4/7 = 1782 / 7 cm²

we have to find :-volume of cone = ?

solution:-we know that:curved surface area of cone =πrlandarea of base =πr²

andvolume of cone = 1/3×πr²handh =√ (l² - r²)

where,r is radius , l is slant height , h is height of cone .

According to given:

=> area of base = 1782 / 7 cm²

=>πr² = 1782 / 7

=> 22/7× r² = 1782 / 7

=> 22 r² = 1782

=> r² = 81

=> r = 9 cm

And,=> C.S.area = 8118 / 7

=>πrl = 8118 / 7 cm²

=> 22/7× 9× l = 8118 / 7

=> 198 l = 8118

=> l = 41 cm

now,Height of cone;=> h =√(l² - r²)

=> h = √ (41² - 9² )

=√ (1681 - 81 )

=√ 1600 = 40 cm

now,volume of cone = 1/3×π× r²× h

=> 1/3× 22/7× 9²× 40

=> 22×81×40 / 21

=> 3394.28 cm³ Answer

1st

2nd

3rd, so let this answer be f(t):final answer will be:f(2)-f(0)

12.87*4.2

= 1287*42/100*10

= 54054/1000

= 54.054

Option D is correct as both carbon dioxide and Sulphur Dioxide do turn limewater milky.

it's wrong

(x + 4) ( x + 10)

x² + 10x + 4x + 40

x² + 14x + 40

Use the identity(x + a)(x +b) = x² + (a + b)x + ab

(x + 4)(x + 10) = x² + (4 + 10) + (4 × 10) = x² + 14x + 40

Thanks mem

24/5÷3/5 of 5 + 4/5 × 3/10 - 1/5

BODMAS

24/5÷3 + 4/5 × 3/10 - 1/524/5× 1/3+ 4/5 × 3/10 - 1/58/5+ 12/50 - 1/58(10)+12(1)/50 -1/580+12/50 -1/592/50 - 1/592-1(10)/50=92-10/50=82/50

1/4 to 3/4= (1/4)÷(3/4)= (1/4)*(4/3)= 1/3

4^(1/3)×2^(1/3)×3^(1/2)÷9^(1/4)

2^(2/3)×2^(1/3)×3^(1/2)÷3^(2/4)

2^(2/3+1/3)×3^(1/2)÷3^(1/2)

=2^1*1=2

Multiplying 1st equation by 3,9x-3y=99x-3y=9

Both equations are same.Therefore, they are equations of coincident lines.

Hence,Infinitr number of solutions exist.

Aage ka

kha h

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please post a question which is not blurry...

nice answer............

BC²=AC²+AB²

=4²+3²

=16+9

=25

BC=5cm

Area of biggest semicircle= 1/2 * 3.14 * r²

=0.5*3.14*5*5

=0.5*78.5

=39.25 cm²

Area of 2nd Largest Semicircle = 0.5*3.14*r²

=0.5*3.14*3²

=0.5*3.14*9

=14.13 cm²

Area of smallest Semicircle = 0.5*3.14*r²

=0.5*3.14*4²

=0.5*3.14*16

=25.12cm²

Area of Triangle = 1/2 *b*h

= 0.5*3*4

=6 cm²

Area of shaded region = (6+25.12+14.13) - 39.25

= 45.25-39.25

=6 cm²

Eq is2x²-x-1 =0(dividing by 2root3)Now roots by formula, Are (1+- 3)/(2*2) =1, - 1/2Also sumof roots =-1/2 which is same as by eq =-2root3/4root3 =-1/2Product of roots=-1/2 =-2root3/4root3 =-1/2

x-20x/100 = 200

80x = 20000

x = 250

profit = 250 + 250*10/100

= 250+25 = 275 rupees

Let the original separation distance be x. Then distance to be covered starting with relative velocity zero is s = 800 + x (800m being the overtaking distance from front meets rear to rear leaves front).Distance s covered at acceleration a starting from (relative) speed zero = 1/2 at^2 which for t=50 gives:= 0.5 x 50^2 = 1250mSo 800 + x = 1250 Therefore x = 1250 -800 = 450m.