Can someone please do this math problem with process shown

here ...

let us suppose breadth of the room be x.

then, length = 2x

now ,the cost of plastering the floor is RS 1080 at rate of RS 15 per ...

hence

the area of floor{length *breadth} =1080/15

x*2x=72

or,2x^2=72

or,x=√(72/2)

hence, the value of x= 6m

(ie. breadth {x}=6. &. length {2x}=12)

again

we have ,

volume of the room (length *breadth*height)is 396m^3

12*6*height=396

72*height=396

height=396/72

hence , height=5.5

we know,

the area of 4 walls = 2( length+breadth)*height

=2(12+6)*5.5

=198

hence the cost of plastering 4 walls at the rate of RS 20 per SQ meter = 198*20=RS 3960

the length of o'p' is .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

the scale factor is 1.5.

multiply both sides by op.

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is .