Posted by Ariza on Saturday, October 24, 2009 at 11:28pm.
Arectangle has perimeter 64cm and area 23cmsquared. Solve the following system of equations to find the rectangle's dimensions.
- Math - Ariza, Saturday, October 24, 2009 at 11:29pm
the equations are:
- Math - Reiny, Saturday, October 24, 2009 at 11:42pm
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
- Math - Anonymous, Sunday, October 25, 2009 at 12:08am
the second part to the question is.:
solve the system of equations:
x^2 + y^2 = 1
xy = 0.5
- Math - Reiny, Sunday, October 25, 2009 at 12:24am
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
Answer this Question
More Related Questions
- Advanced Functions - A rectangle has perimeter 64cm and area 23cm squared. Solve...
- Math - A) A rectangle has perimeter 64 cm and area 23 cm^2. Solve the following ...
- basic geometric - The width of a rectangle is 6 centimeters less than the lenght...
- Math - Find the Maximum area for the given perimeter of a rectangle. State the ...
- math - the length of a rectangle is equal to triple the width. which system of ...
- Math - The perimeter of a rectangle is 42ft. The length is 7ft. longer than the ...
- PreCalc (still confused) - Hello and Happy Halloween! Trying to get this problem...
- Math - 2. A rectangle whose perimeter is 80 m has an area of 384 m^2. Find the ...
- math ,correction - Directions: Solve each of the following problems. Be sure to ...
- Math - The perimeter of a rectangle is 74 centimeters. The width is doubled and...