algebra word problem..pls help me...?

Question

a rectangular building was constructed so that its depth is twice as frontage. the building is divided into two parts by a position that is 30ft from parallel to the front wall. if the rar position of the building contains 3500ft^2, find the dimensions of the building(pls give a answer with a quadratic eq'n)

Answer 1

You are beyond help, leo. When in doubt, just add together all the numbers in a problem and then multiply that by 2. You will often be surprisingly close to the right answer.

Answer 2

Frontage = x
Depth = 2x

Dimensions of rear part are x and 2x - 30, and its area is

x(2x - 30) = 3500
2x^2 - 30x - 3500 = 0

Solve this quadratic equation for x and you can determine the overall dimensions of the building

A little thought gives the factors of the equation as

(2x + 70)(x - 50)

So the 2 roots are - 70 and 50

Discard, of course, the negative root

So the dimensions of the building are 50 and 100

If it is not easy for you to factor the quadratic equation, see the hard case on the Purplemath lesson.

Answer 3

let the frontage be 'x'.so the depth will be 2x
area of the rear portion=(2x-30)x is given as 3500
=>2x^2-30x-3500=0=>x^2-15x-1750=0
=>(x-50)(x+35)=0 therefore x=50
ignoring the negative value of -35
so the dimensions are frontage=50'
and the depth 100'

Answer 4

let x be frontage. 2x be depth.

then, i have rear area:

x*(2x-30) = 3500

x^2 -15 x - 1750 =0.

x=50.

so dimensions : 50 ft x 100 ft.

Answer 5

If the scholars modern grade is a ninety% then we’re of route assuming it’s out of one hundred% ninety/one hundred Now because the scholars grade more advantageous 20% from the only accurate grading era then you honestly upload that to the previous one hundred% providing you with ninety/ 100 and twenty = .seventy 5 and for the reason that grades are in opportunities you multiply .seventy 5 by using 1 hundred providing you with seventy 5% because the scholars previous grade

Answer 6

Plz don't ruin people's weekends.

Answer 7

entire building dimensions = x * 2x

back portion: 3500 = x * (2x-30)
3500 = 2x^2 - 30x

x
---------- ... front
|...........| ... 30 ........... }
---------- ... partition..... } 2x
|...........| .......................}
|..3500.| ... back
|..sqft...|
----------