Arectangle has perimeter 64cm and area 23cmsquared. Solve the following system of equations to find the rectangle's dimensions.
the equations are:
l= (23/w)
l+w=32
so sub l= (23/w) into l+w = 32
23/w + w = 32
23 + w^2 = 32w
w^2 - 32w + 23 = 0
solve for w, then plug into l=23/w
I got w = .7357
l = 31.2643
check: .7357x31.2643 = 23.001
.7357 + 31.2643 = 32
thankyou !
the second part to the question is.:
solve the system of equations:
x^2 + y^2 = 1
xy = 0.5
do it the same way
from xy = .5 = 1/2
y = 1/(2x)
sub into the first
x^2 + 1/(4x^2) = 1
4x^4 + 1 = 4x^2
4x^4 - 4x^2 + 1 = 0
let x^2 = p
then your equation becomes
4p^2 - 4p + 1 = 0
(2p-1)(2p-1) = 0
p = 1/2
so x^2 = 1/2
x = +/- 1/√2
sub back into xy=1/2 to get the y
To solve this system of equations, let's define the dimensions of the rectangle as length (L) and width (W).
We can set up the following equations based on the given information:
1) Perimeter equation: 2(L + W) = 64
2) Area equation: L * W = 23
To solve this system, we can use substitution or elimination method. Let's use the substitution method.
We can rearrange the first equation to solve for L:
L + W = 32
L = 32 - W
Now, substitute the value of L in equation 2:
(32 - W) * W = 23
Expanding the equation:
32W - W^2 = 23
Rearrange the equation to get it in standard quadratic form:
W^2 - 32W + 23 = 0
Now, we can solve this quadratic equation to find the dimensions of the rectangle.
Using factoring, completing the square, or quadratic formula, we find that the solutions for W are W = 23 and W = 1.
If W = 23,
L = 32 - W = 32 - 23 = 9
If W = 1,
L = 32 - W = 32 - 1 = 31
So, the dimensions of the rectangle are 9 cm by 23 cm or 31 cm by 1 cm.