Homework Help: Math 11

Posted by Need Help on Wednesday, January 11, 2012 at 10:06pm.

I need help on this math problem.

The width of a certain painting is 5cm less than twice the length. The length of the diagonal distance across the painting is 107cm. Find the length and width. Round your answer to 2 decimal places.

Substitute:
diagonal distance: 107cm
length: L
width: 2L-5

Is the substitutes correct?
Should I use the pythagoras method?
Can someone put me in the right track please so I can work out the problem?

please help and thank you

Math 11 - Steve, Thursday, January 12, 2012 at 6:08am
yes
yes

L^2 + (2L-5)^2 = 107^2
L^2 + 4L^2 - 20L + 25 = 11449
L = 49.84
Math 11 - HelpPlease, Thursday, January 12, 2012 at 9:38am
the next step is:

5L^2-20L=11424

right?
Math 11 - Steve, Thursday, January 12, 2012 at 10:44am
Almost. Set the equation to zero to solve:

5L^2-20L-11424 = 0

then use the quadratic formula to get L.

Answer this Question

Related Questions

Math 11 - I need help on this math problem. The width of a certain painting is ...
Math - I need help on this math problem. The width of a certain painting is 5cm...
Math - A square mirror has sides measuring 2 ft. less than the sides of a square...
Math - a square painting is surrounded by a 2.5cm frame. if the total are of the...
Help please? Math! - Let W repersent the width of a rectangle, the length if 7cm...
math - the length of a rectangle is 1 cm less than twice its width. the area is ...
Math - a square painting is surrounded by a 2.5cm frame. if the total are of ...
Math - Another painting has length : width ratio approximately equal to the ...
math - the length of a rectangular field is 24 meters. this is 3 meters less ...
math - the length of a rectangle is 1 inch less than twice the width, and the ...

For Further Reading

First Name:
School Subject:
Answer: