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for only 1 root D = 0

=> (-k)²-(4*1*49) = 0=> k²-196 = 0=> k² = 196=> k = ±14

it is wrong

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1

48 × 4 × 27 / 18 × 27 × 16 2 / 3

(4(root(3)+root(2)))/3(root(3)-root(2))=4/3 * (root(3)+root(2))^2/1=4/3(3+2+2root(6))=20/3+8root(6)/3

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27:x::x:4827/x=x/4827*48=x^21296=x^2X=36

value of x=36 .should be right answer

GIVEN - ABCD is a quadrilateral and it hascircumscribing a circle Which has centreO.

CONSTRUCTION - Join -AO, BO, CO, DO.

TO PROVE :-Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

PROOF - In the given figure , we can see that∠DAO = ∠BAO [Because, AB and AD are tangents in the circe]

So , we take this angls as 1 , that is , ∠DAO = ∠BAO = 1

Also in quad. ABCD , we get,∠ABO = ∠CBO { Because, BA and BC are tangents }

⇒Also , let us take this angles as 2. that is , ∠ABO = ∠CBO= 2

⇒ As same as , we can take for vertices C and as well as D.

⇒ Sum. of angles of quadrilateral ABCD = 360° { Sum of angles of quad is360°}

Therfore ,

2(1 + 2 + 3 + 4) = 360° { Sum. of angles of quad is -360° } 1 + 2 + 3 + 4 = 180°

Now , in Triangle AOB, ∠BOA= 180 –(a + b ) { Equation 1 }

Also , In triangle COD,∠COD = 180 – (c + d) { Equation 2 }

⇒From Eq. 1 and 2 we get , Angle BOA + Angle COD = 360 – (a + b + c + d)

= 360° – 180°

= 180°

⇒So , we conclude that the line AB and CD subtend supplementary angles at the centreO.

GIVEN ;-

⇒ ABCD is a quadrilateral and it hascircumscribing a circle Which has centreO.

CONSTRUCTION ;-

⇒ Join -AO, BO, CO, DO.

TO PROVE :-

⇒Opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

PROOF ;-

⇒ In the given figure , we can see that ⇒∠DAO = ∠BAO [Because, AB and AD are tangents in the circe]

So , we take this angls as 1 , that is , ⇒ ∠DAO = ∠BAO = 1

Also in quad. ABCD , we get, ⇒∠ABO = ∠CBO { Because, BA and BC are tangents }

⇒Also , let us take this angles as 2. that is , ⇒∠ABO = ∠CBO= 2

⇒ As same as , we can take for vertices C and as well as D.

⇒ Sum. of angles of quadrilateral ABCD = 360° { Sum of angles of quad is360°}

Therfore ,

⇒2(1 + 2 + 3 + 4) = 360° { Sum. of angles of quad is -360° } ⇒1 + 2 + 3 + 4 = 180°

Now , in Triangle AOB, ⇒∠BOA= 180 –(a + b ) ⇒ { Equation 1 }Also , In triangle COD, ⇒∠COD = 180 – (c + d) ⇒ { Equation 2 }

⇒From Eq. 1 and 2 we get , ⇒Angle BOA + Angle COD

= 360 – (a + b + c + d)

= 360° – 180°

= 180°

⇒So , we conclude that the line AB and CD subtend supplementary angles at the centreO

⇒Hence it is proved that -opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.

Formula: 4/3 pi* r^3i) r=7cmvolume= 4/3*(22/7)*7*7*7=1437.3 cm cubeii)r=0.63 mvolume=4/3 *pi* 0.63^3=1.05 m cube

7- (-8) - 6 ÷ (-2) - (10/100*(-2)-(7-17)= (7+8 - 6)÷(-2 - (10*(-2 + 10))= (9) ÷ (-2 - (10*8))= 9 ÷ (-2 - 80)= 9/82

7 1/2 *8 1/2=15/2 * 17/2=255/4=63 3/4

As we know15/2*17/2= 15*17/4255/4thanks

the answer is 56 power 1/2

Hi Raman,

Here is the answer to your question.

iii) 3 41---- + 5------ 5 7

8/5 + 39/7

56 + 195--------------- 35

251------ 35

1/7 +3/9=1/7+1/3=3+7/3×7=10/21

8/5+39/756+195/35=251/35

Let the larger number = xThen the square of thesmaller number= 8 times the larger number = 8xand the square of thelarger number = x²According to the question, x² - 8x = 180 => x²- 8x - 180 = 0 => x²- 18x + 10x - 180 = 0 => x(x - 18) + 10(x - 18) = 0 => (x - 18) (x + 10) = 0 => x - 18 = 0orx + 10 = 0 => x = 18orx = -10

Thus, the larger number = 18 or -10Then, the square of the smaller number= 8(18)or 8(-10)= 144 or -80

The square of a number can't benegative, so, thesquare of smaller number =144Hence, thesmaller number =√(144) =12The numbers are12and18

thank you

Let the larger number = xThen the square of the smaller number = 8 times the larger number = 8xand the square of the larger numbe r = xAccording to the question,x - 8x = 180=> x - 8x - 180 = 0=> x - 18x + 10x - 180 = 0=> x(x - 18) + 10(x - 18) = 0=> (x - 18) (x + 10) = 0=> x - 18 = 0 or x + 10 = 0=> x = 18 or x = -10Thus, the larger number = 18 or -10Then, the square of the smaller number = 8(18) or 8(-10)= 144 or -80The square of a number can't be negative, so, the square of smaller number = 144Hence, the smaller number = sqrt(144) = 12The numbers are 12 and 18

Let the larger number = xThen the square of the smaller number = 8 times the larger number = 8xand the square of the larger numbe r = xAccording to the question,x - 8x = 180=> x - 8x - 180 = 0=> x - 18x + 10x - 180 = 0=> x(x - 18) + 10(x - 18) = 0=> (x - 18) (x + 10) = 0=> x - 18 = 0 or x + 10 = 0=> x = 18 or x = -10Thus, the larger number = 18 or -10Then, the square of the smaller number = 8(18) or 8(-10)= 144 or -80The square of a number can't be negative, so, the square of smaller number = 144Hence, the smaller number = sqrt(144) = 12The numbers are 12 and 18 .

12 and 18 is the right answer

total 12 triangle are there

let larger number is x, smaller number is ythen A/Q-x^2-y^2=180..........(1)y^2=8x........ (2)from equation (1)and (2)x^2-8x=180x^2-8x-180=0x^2+10x-18x-180=0x(x+10)-18(x+10)=0(x+10)(x-18)=0x+10=0 and x-18=0x=-10 and x=18thereforelarger number is 18And smaller number is y=√8×18=√144=12

Markup Price = MP CP cost price SP = sale Price Total Cost of buying = n * CP, if n items are in the shop.Total Sales amount = n * SP , n items are sold

MP = 1.40 CPSP = 0.95 MP = 1.4 * 0.95 CPTotal sales amount = n SP = 1.4 * 0.95 * n * CP = 1.4 * 0.95 * Total Cost => 1064 = Total Cost * 1.4 * 0.95 Total cost = Rs 800 Profit = 1064 - 800 = Rs 264

profit =264 ok bro 😎😎😎😎😎😎😎😎😎😎😎😎😎😎😎😎😎

thank you

let the other no be xproduct of no = 155/6one no be = 20/3155×3÷6×20=x31/8=x

We know in triangle exterior angle is equal to sum of opposite two interior angles

Given,Exterior angle = 130One interior angle = 60Let other interior angle = x

Then, x + 60 = 130 x = 130 - 60 x = 70

Other opposite interior angle is 70°

aadha solutions kat ja Raha hi

thanks

2^(2/3)*2^(1/5)

=2^(2/3+1/5)

=2^(2(5)+1(3)/15)

=2^(10+3/15)

=2^(13/15)

-38-(-34)= -38+34= -4 answer

= -(38) - (-34)= -38+34= -4

-34-(-38)-34+384 ans

-(38) -(-34) _38+34-4 ans..

-38-(-34)=-38+34=-4

-38-(-34)=-34+38=-4

We know that, The subraction of two negative integer results in form of sum of the negative integers...So the answer is -72

-38-(-34)=-38+34=-4 answer

-38-(-34)=-38+34=-4 answer

-4 is the write answer

-38-(-34)=-38+34-4 answer

A : B = 2 : 3 B : C = 3 : 4 A : B : C = 2 : 3 : 4

-(266)-(364)-(81)-630-81-711

(288)/54x = 4/3

=> x = 288*3/4*54 = 4

similarly y = 14

so, x/y = 4/14 = 2/7

C.P. = Rs. 1200S.P. = 125% of Rs. 1200 = Rs.(125×1200)/100= Rs. 1500

Let marked price bex Then 80% ofx= 1500x=(1500×80)÷100= 1875so Marked price = 1875.

cost of 2 dozen = 40or 24 oranges in 40 rupees1 orange = 40/241.66 rupee

area of 4 quadrants= 4x1/4xπr^2 =4x1/4xx22/7x1x1=22/7 sq. CM. diameter of circle= 2 CMradius=2/2=1 CMarea of circle=πr^2 22/7x1x1 =22/7sq.cm.area of saare =4x4=16 sq. cm. area of shaded region=16-(22/7+22/7) 16-44/7 =112-44/7 =68/7 sq cm.