Math Problem Hellp!!!!?

Question

Image there is a Big rectangle. Imagine there is a smaller rectangle inside.
Big Rectangle: W= x+3 L= x+5
Small Rectangle: W= x L= 2x-1


If the outer area is 120cm², then what is the inner area?

Answer 1

Hi,


Area of bigger Rectangle A = LW
(x+3)(x+5) = 120

x² + 8x + 15 = 120
x² + 8x - 105 = 0

(x + 15)(x - 7) = 0

x = -15 or x = 7

For a length x is positive, so x= 7.


Area of smaller rectangle since x = 7
x(2x - 1) = 7(13) = 91 cm²

I hope that helps!! :-)

Answer 2

Area = L x W

Therefore for outer rectangle

120 = (x+3)(x+5)
x^2 +8x + 15 -120 =0
x^2 +8x - 105 = 0
(x -7)(x + 15) = 0
x = 7 or x =-15
since L> 0 x = 7

so area of inner rectangle
A=x(2x-1)
A=7(2*7-1)
A=91 cm^2

Answer 3

this really means that the area of the big rectangle is x^2+8x+15 and the area of the smaller rectangle is 2x^2-x

idk what the "outer" area is but the inner is the area of the smaller rectangle

Answer 4

(x+3)(x+5) = 120
x^2 +8x +15 =120
x^2+8x-105 = 0
x = [-8 +/- sqrt(64+4*105)]/2
x = (-8 +/-22)/2
x = 14/2 = 7 = W
2x-1 = 13=L
So A inner = 7*13= 91cm^2

Answer 5

(x+3)(x+5) = 120
x^2 +8x +15 =120
x^2+8x-105 = 0
x = [-8 +/- sqrt(64+4*105)]/2
x = (-8 +/-22)/2
x = 14/2 = 7 = W
2x-1 = 13=L
7*13= 91cm^2

Answer 6

I'm not sure how you get the answer but i'm pretty sure it wouldn't be 192 divided by 2 because if it's 192 DIGITS then that would only work if every number had 1 digit, which is obviously incorrect. You would have to work it out by doing something like this: pages 1 -> 9 = 9 digits pages 10 -> 40 = 62 digits right up until 192 .. that's how I would figure it out anyway =)