algebra question.....?

Hey evey,

Here's your answer...

L = length

W = width

A = area = 104 m^2

The problem says that L = W + 5

The area of a rectangle is:

A = L x W

or

A = (W + 5) x W = W^2 + 5W

or rearranging terms:

W^2 + 5W - A = 0

This is a quadratic equation in W. It takes the typical quadratic form of:

ax^2 + bx + c = 0

Here, a = 1, b = 5, c = -A = -104

The solution for a quadratic equation is:

x = { -b +/- sqrt(b^2 - 4ac) } / 2a

So in this case x = W, so:

W = { -5 +/- sqrt(25 - 4(1)(-104)) } / 2(1)

= 8 m

(we neglect the negative solution as there cannot be a negative distance)

Since W = 8 m, that means that

L = W + 5 = 8 + 5 = 13 m

So the dimensions of the rectangle are:

W= 8 m

L = 13 m

That's your answer!

anAGAH.

A=l*w

w=x

l=x+5

A=104m^2

104m^2=x*(x+5)

x^2+5x-104=0

then use the quadratic equation

x=[-5+/-sqrt(5^2-4(1)(-104))]/2

x=(-5+21)/2 x=(-5-21)/2

x=8 x=-13

it must be positive therefore x=8

therefore the rectangle is 8m wide and 13m long

L = W + 5

A = L x W

104 = (W + 5) x W

104 = W^2 + 5W

0 = W^2 + 5W - 104

0 = (W - 8)(W + 13)

W = 8, -13

Since it can only be a positive number W = 8m

Therefore L = 8 + 5 = 13m

The dimensions are 13m by 8m

P = 104

a = b + 5

P = a * b

P = (b + 5) * b

P = b^2 + 5b

104 = b^2 + 5b

b^2 + 5b - 104 = 0

b = [-5 +/- sqrt(5^2 + 4* 104)] / 2

b = [-5 +/- 21] / 2

b = 8 or b=-13 (it can't be -13 cause it's a width)

a = b + 5

a = 8 + 5 = 13

so a = 13, b = 8

Let the width be x, then the length would be x + 5 (all in metres).

x(x + 5) = 104 ;

x^2 + 5x - 104 = 0 ;

x^2 + 13x - 8x - 104 = 0 ;

(x - 8)(x + 13) = 0 ;

x = 8 (omitting negative value).

Dimensions-----> 13m by 8m.

Legnth=13

Width=8

(5+x)*x=104

/Edit:

2nd person's question: m= 3, b=-4; You'd continue to go 3 up and one across to the right for each point. Or you could go 3 down and one to the left.

Hey I know your asking for a question ..but if help me..I'll be glad to help you.

Here is my problem..

Graph Using y=mx+b

A. y=3x-4

m= ?

b= ?

And how would you graph it?