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Arpitas present age is 3x +9

Hi Tekram,

Here is the answer to your question.

Three yeats ago Shilpa's age was x.

so present age of Shilpa is (x+3) years

Therefore the age of Arpita =3(x+3) years.

let arpita's present age= yshilpa's present age= x+3according to statement, y= 3(x+3) 3x+9

Arpit present age is 3x +9

Let Atul's age be x yrs. and Parul's age be y yrs.According to the question, x-3=4(y-3)x-3=4y-12x-4y+9=0x=4y-9And, x+5=2(y+5)x+5=2y+10x-2y=54y-9-2y=52y=14y=7 x=4y-9=4(7)-9=28-9=19Therefore, Atul's present age= 19 yrs. and Parul's present age= 7 yrs.

Let present age of Rohit be xLet present age of Shikhar be y

As per given conditions(x-7) = 5(y - 7)^2...... (1)

(y + 3) = 2/5(x + 3)5y + 15 = 2x + 6y = (2x - 9)/5......(2)

Put value of y in eq(1), we get(x - 7) = (2x - 9 - 35)^2 /55x - 35 = 4x^2 + 1936 - 176x4x^2 - 181x + 1971 = 04x^2 - 108x - 73x + 1971 = 04x(x - 27) - 73(x - 27) = 0x = 73/4, 27

Age must be integer value. So, x = 27

y = (2*27 - 9)/5 = 45/5 = 15

Therefore, Present age of Rohit = 27Present age of Shikhar = 15

Let the present age of Varun and Swati are x and y years respectively.

Given – 7 years ago

x – 7 = 5(y-7)^2 ----------------1

Also given – 3 years hence

y + 3 = 2/5 (x + 3)

5y + 15 = 2x + 6

2x = 5y + 9

x = (5y + 9)/2 ------------------2

Substitute the value of x in equation 1

(5y + 9)/2 – 7 = 5 (y-7)^2

5y + 9 – 14 = 10 (y^2 – 14y + 49)

5y – 5 = 10y^2 – 140y + 490

10y^2 – 145y + 495 = 0

Dividing the equation by 5

2y^2 – 29y + 99 = 0

Solving above quadratic equation to find y

y = (-b + sqrt (b^2 – 4ac))/4ac or y = (-b - sqrt (b^2 – 4ac))/4ac

get y = 9 and y = 11/2

consider age as a whole number

Therefore, y = 9 years

Substitute the value of y in equation 2

Therefore, x = (5 * 9 + 9)/ 2 = 27 yearsAnswer – The present age of Varun is 27 years and presentage of Swati is 9 years.

Let Vijay's present age be 'x'.

Then, Sushma's present age will be X+15

Sushma's age after 3 years will be [(X+15) + 3] = X+18

Vijay's age 3 years ago was X-3

Given, Sushma's age 3 years from now will be equal to 8 times of what Vijay's age was 3 years back.

Therefore, X+18 = 8 (X-3)

X + 18 = 8X - 24

8X - X = 18 + 24

7X = 42

X = 6

Therefore, their present ages are as follows:

Vijay is 6 years old.

Sushma is 21 years old.

let Rohans age is X.His mother's age is X+26.as per the details after 3 yearstheir respective age are X+3 & X+26+3

(X+3) (X+26+3)=360》(X+3) (X+29) = 360》X(X+29) +3(X+29) =360》 X" + 29X+ 3X+87= 360》 X"+32X=360-87》X"+32X=273》49+224=273

The answer is 7

S=V+15(S+3)=8(V-3)S+3=8v-24V+15+3=8v-24so 7V=42so V=6and S=21

you are in ,DAV public school in kusmunda

Mary is twice older than her sister. In 5 years time, she will be 2 years older than her sister find how old are they both now.

solution : Let present age of Mary's sister = x therefore Present age of Mary = y = 2x After 5 years , Mary's Sisters age = x+5 Mary's age = 2x+5 According to Question , Mary's age = Mary's Sisters age + two years

2x+5 = (x+5 ) + 2 2x+5 = x+5+2 2x = x+7 -5 2x-x = 7-5 x=2

hence Mary's Sister is 2 years old.and Mary is (2x =2 *2 = 4 )4 years Old !!

thank you soo much

Let the length of the edge of the cube be 'x' cmTotal surface area = 6x2cm2

Increased length of the edge = 2xTotal surface area = 6(2x)2cm2= 24x2cm2

If the edge of the cube is doubled surface area of the cube increases by 4 times

Let the width be x metres. Then, height = 6x. length =7x. Volume = 16128=> lbh = 16128=> 42* x^3 = 16128=> x^3 = 384=> x = 8 metres.

Please leave a like

(2x+1)/(3x-2)=(3x-4)/(x+8)(2x+1)(x+8)=(3x-4)(3x-2)2x^2+16x+x+8=9x^2-6x-12x+87x^2-35x=0x(x-5)=0so x=0 or x=5if x=o then fraction=-1/2if x=5 then fraction=11/13

thanks

let the number be x so8x -10 = 6x +48x-6x = 4+102x = 14x = 14/2 = 7x = 7 number is 7

OK thanks

understood

Suppose number is x6x= 55-136x= 42x= 7please like the solution 👍 ✔️

Let Viru 's present age = x yrs then Sudhir's present age = (5 + 3x) yrs Anil's present age = (5 + 3x)/2 yrs

ratio of the sum of Sudhir's and Viru's age to three times Anil's age is 5 : 6 e.g., (5 + 3x + x)/3(5 + 3x)/2 = 5/6 => 2(5 + 4x)/3(5 + 3x) = 5/6 => 12(5 + 4x) = 15(5 + 3x)=> 4(5 + 4x) = 5(5 + 3x) => 20 + 16x = 25 + 15x => x = 5

therefore virus age is 5yr

Let Viru 's present age = x yrs then Sudhir's present age = (5 + 3x) yrs Anil's present age = (5 + 3x)/2 yrs

ratio of the sum of Sudhir's and Viru's age to three times Anil's age is 5 : 6 e.g., (5 + 3x + x)/3(5 + 3x)/2 = 5/6 => 2(5 + 4x)/3(5 + 3x) = 5/6 => 12(5 + 4x) = 15(5 + 3x)=> 4(5 + 4x) = 5(5 + 3x) => 20 + 16x = 25 + 15x => x = 5

thanks for answer

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Correct

Let Viru 's present age = x yrs then Sudhir's present age = (5 + 3x) yrs Anil's present age = (5 + 3x)/2 yrs

ratio of the sum of Sudhir's and Viru's age to three times Anil's age is 5 : 6 e.g., (5 + 3x + x)/3(5 + 3x)/2 = 5/6 => 2(5 + 4x)/3(5 + 3x) = 5/6 => 12(5 + 4x) = 15(5 + 3x)=> 4(5 + 4x) = 5(5 + 3x) => 20 + 16x = 25 + 15x => x = 5

hence viru's present age is 5yr

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Red line is 2x = 9

2x²+px-15=0x=-5 is root and it will satisfy the equation.2(-5)²+p(-5)-15=02(25)-5p-15=050-5p-15=035=5pp=7

Now the equation p(x²+x)+k=0 has equal roots.px²+px+k=0For equal root, b²=4ac

(-p)²=4(p)(k)p=7

(-7)²=4(7)(k)49=4*7*kk=7/4

let A = ( 3 , y) & B = ( 2 , 7 )

C = ( 1 , -4 ) & D = ( 0 , 6 )

so,

slope of AB = slope of CD { as the point are parallel }

(7 - y) / ( 2-3) = (6 + 4) / (0 - 1) {slope = (y2 - y1) / (x2 - x1) }

(7 - y) / (-1) = (10) / (-1)

7 - y = 10

- y = 10 - 7

y = -3

Sorry I think answer is y=9

Given,Mid point (1, p/3) of line segment joining the points(2,0) and (0, 2/9)

Then,p/3 = (0 + 2/9)/2p = 2*3/9*2 = 1/3

If line 5x + 3y + 2 = 0 pass through point (-1, 1)

Then, 5*-1 + 3*1 + 2= -5 + 5 = 0

The ratio of the density of a substance to the density of water at 4°C is known as the relative density of the substance.

Relative density = Density of the substance/Density of water at 4°C.

As both the numerator and the denominator have the same unit (i.e. kg/m3) they get cancelled. Hence relative density has no unit.

the coefficient of x2in 2+x2+is

this is not my question

X=0,2x-y+3=0hence put X=0hence-y+3=0y=3hence intersection point is (0,3)

(sin5x-2sin3x+sinx)/(cos5x-cosx)

=[{2sin(5x+x)/2cos(5x-x)/2}-2sin3x]/{2sin(5x+x)/2sin(x-5x)/2}

=(2sin3xcos2x-2sin3x)/{2sin3xsin(-2x)}

={2sin3x(cos2x-1)}/{-2sin3xsin2x}

=-(cos2x-1)/sin2x

=(1-cos2x)/sin2x

=2sin²x/2sinxcosx

=sinx/cosx

=tanx (Proved)

x+6+6+6=0x+18=0x= -18

x+6+6+6=0x+18=0x= -18

x+18=0x= -18is the correct answer

part 1

part 2

Thankyu for the ans

Vaise aapki math itni aachi kaise hai. Mujhe to kuch bhi samaj nhi aata. Only 87% hi math m num laa paaya. High school. Aisa lagta hai aap to 95 laogi

Distance between two points=√(x2-x1)^2+(y2-y1)^2=√(2+3)^2+(-6-6)^2=√5^2+12^2=√25+144=√169=13