I need help with this math problem!!!! This is for a test tomorrow!!?

No issue my frnd i will tell u
Area of rectangle = lb =33 cm^2
So, let the l = 33/b ~[eq 1]

And perimeter of the given rectangle =23 cm
23= 2* ( l+b )
23/2 = l+b
l = 23/2 - b ~[eq 2]
Now by substituiting the value of l from eq 1 into 2 we get,
33/b = 23/2 - b

33/b + b = 23/2
( 33+ b^2 )/ b = 23/2

33 + b^2 = 11.5b
b^2 - 11.5b +33 =0
Now by splitting the middle term
b^2 - ( 6 + 5.5 )b + 33
b^2 -6b -5.5b + 33
b( b- 6 ) -5.5 (b -6)
( b-6) (b-5.5)
So ,
b-6 = 0 b = 6
b-5.5 = 0 b= 5.5
So we have now two values of b
Put this in eq. No.1
WHen b =6cm then,
l =33/b
= 33/6
= 5.5cm
When b=5.5cm
Then,
l= 33/b
= 33/ 5.5
= 6cm
So we have now two values of both measurements........
1- ( l- 6cm and b - 5.5cm)
2- ( l- 5.5cm and b- 6cm)
Both are applicable
U can check it by putting the values
Best of luck????
JUst rate ......????

L=6, W=5.5

We recommend learning methods because answers only work once.
Lw = 33, with L > w
2L + 2W = 2L + 2*33/L = 23
2L^2 – 23L + 66 = (L – 6)(2L – 11)
L = 6, w = 5.5

P = 23 cm
A = 33 cm^2
L + W = 11.5 cm
LW = 33 cm^2
L(11.5 - L) = 33
2L^2 - 23L + 66 = 0
(L - 6)(2L - 11) = 0
Solutions:
L = 11/2
L = 6
The length equals 6 cm and the width equals 5 1/2 cm.

Put rectangle dimensions (L , W). Now 2(L + W) = 23...[1] and LW = 33...[2], From [1] we have L = 23/2 -W and
substitution in [2] gives W(23/2 - W) = 33, ie., W(23 - 2W) = 66, ie., 2w^2 - 23W + 66 = 0, ie., 4W = 23(+or)D,
where D^2 = (-23)^2 - 4(2)(66) = 529 - 528 = 1 so W = (1/4)[23(+or-)1] = 6 or 11/2 so length is 6, width is 5.5
[length units = cm., area units = cm^2]. we, by convention, usually label the larger of 2 values length, and the
smaller of the values, width. That's why I chose L = 6 and W = 5.5.

2L + 2W = 23

LW = 33 ⇒ W = 33/L

2L + 66/L = 23

2L² - 23L + 66 = 0

We need two divisors of (2)(66) = 132 that sum to -23. They are -11 and -12

2L² - 11L - 12L + 66 =

L(2L - 11) - 6(2L - 11) =

(L - 6)(2L - 11) = 0

L ∈ {11/2, 6}

Dimensions of rectangle are 6 cm x 5.5 cm ☜

A rectangle has perimeter 23 cm. Its area is 33 cm^2. Determine the dimensions of the rectangle.

2x + 2y = 23
xy = 33

x = 33/y

Substitute 33/y for x in the first equation.

You have :
---------------

A = ( L ) ( W ) = 33 cm^2

P = 2L + 2P = 23 cm

Let W = A/L

2L + ( 2 ) ( A/L ) = 23

Multiply both sides by L:

( 2 ) ( L )^2 + ( 2 ) ( 33 ) = 23 L

2L^2 - 23L + 66 = 0

Apply the quadratic formula :

ax^2 + bx + c = 0

x = { - ( b ) +/- SQRT [ ( b )^2 - 4ac ] } / { 2a }

a = 2
b = - 23
c = 66

L = { - ( - 23 ) +/- SQRT[ ( -23 )^2 - ( 4 ) ( 2 ) ( 66 ) ] } / { ( 2 ) ( 2 )}

L = { 23 +/- SQRT[ 529 - 528 ] } / { 4 }

L = { 23 +/- 1 } / { 4 }

L = { 23 + 1 } / { 4 } = 6 cm -----> W = 33 cm^2 / 6 cm = 5.5 cm

L = { 23 - 1 } / { 4 } = 5.6 cm ------> W = 33 cm^2 / 5.5 cm = 6.0 cm

L = 6.0 cm <------------

W = 5.3 cm <-------------

Check the result :
-------------------------

A = ( L ) ( W ) = ( 6.0 cm ) ( 5.5 cm ) = 33 cm^2 ----->[ OK ]

P = 2L + 2W = ( 2 ) ( 6.0 cm ) + ( 2 ) ( 5.5 cm ) = 12 cm + 11 cm = 23 cm --->[OK]

Of rectangles,

::Perimeter = 2(length + width) = 2(length) + 2(width)

::Area = (length) * (width)
-------------------
23 = 2(length + width), and you divide out 2:

11.50 = length + width
Solve for width by subtracting the length:

11.50 - length = width >>> We are going to use this in the next equation with the substitution method.

33 = (length)(11.50 - length)

33 = (length)(11.50) - (length)(length)
I am going to use the L because I hate the lower case l on here, it looks exactly like an I, dude!

33 = -L^2 + 11.50(L)
Set equal to zero:

0 = -L^2 + 11.50(L) - 33 <--------> 0 = a(L)^2 + b(L) + constant

Note: a = -1, b = 11.50, and c = -33

Quadratic formula:

L = { -(b) ± √[(b)^2 - 4(a)(c)] } / (2 * a)

L = { -(11.50) ± √[(11.50)^2 - 4(-1)(-33)]} / (2 * -1)

L = (-11.50 ± √(1 / 4)) / (-2)

L = (-11.50 ± 0.50) / -2
The plus or minus sign gives us two solutions. However, since we consider the dimensions to exist, we only use the positive Length.

L = 6 centimeters.

Find the width from the simplest equation:

11.50 - L = w

11.50 - 6 = w

5.50 = width
---------
Dimensions: width = 5.50 cm, and length = 6.00 cm

W*L = 33
W+L = 23/2
W(23/2 - W ) = 33
W(23-2W) = 66
2W^2--23W+66 = 0
(2W-11)(W-6) = 0
Dimensions are 6 cm by 5.5 cm