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Posted by **Preston** on Monday, October 14, 2013 at 12:21am.

2) Let x represent the length of the shorter leg in a right triangle. If the longer leg is 1 more than the shorter leg and the hypotenuse has length 7, set up but do not solve an equation to find the two side lengths.

3) The product of two consecutive negative odd integers is 575. Find the integers

4)A rectangular canvas picture measures 13 inches by 7 inches. The canvas is mounted inside a frame of uniform width, increasing the total area covered by both canvas and frame to 160 square inches. Find the uniform width of the frame.

5) A rectangle has perimeter 30 feet and an area of 54 square feet. Find the dimensions of the rectangle assuming the width is smaller than the length.

- Algebra Homework, HELP! -
**Kuai k**, Monday, October 14, 2013 at 2:22am1. A = lw

w =4l-3

4l-3(l) = 297

4l^2 -3l = 297

4l^2 -3l - 297 =0

(4l+33)(l-9)= 0

l = 9

w =33

2. a^2 + b^2 = c^2

short leg = x

Long leg = x+1

x^2 + (x+1)^2 = 7^2

x^2 +x^2 + x + x + 1 = 49

2x^2 +2x + 1 = 49

3. x, x +2

x (x+2) = 575

x^2+ 2x = 575

x^2 + 2x -575 =0

(x -23)(x + 25) = 0

x =23

23,25 and -23, -25

- Algebra Homework, HELP! -
**Bosnian**, Monday, October 14, 2013 at 2:26am1 )

W = 4 L - 3

A = L * W = 297

L * ( 4 L - 3 ) = 297

4 L ^ 2 - 13 L = 297 Subtract 297 to both sides

4 L ^ 2 - 3 L - 297 = 297 - 297

4 L ^ 2 - 3 L - 297 = 0

Solutions :

L = - 33 / 4

and

L = 9

The length can't be negative so L = 9 in

W = 4 L - 3

W = 4 * 9 - 3

W = 36 - 3

W = 33 in

2 )

b = the longer leg = x + 1

c = hypotenuse = sqrt [ x ^ 2 + b ^ 2 ] = 7

c = hypotenuse = sqrt [ x ^ 2 + ( x + 1 ) ^ 2 ] = 7

c = sqrt ( x ^ 2 + x ^ 2 + 2 x + 1 )

c = sqrt ( 2 x ^ 2 + 2 x + 1 ) = 7 Square both sides

2 x ^ 2 + 2 x + 1 = 49 Subtract 49 to both sides

2 x ^ 2 + 2 x + 1 - 49 = 49 - 49

2 x ^ 2 + 2 x - 48 = 0

2 ( x ^ 2 + x - 24 ) = 0 Divide both sides by 2

x ^ 2 + x - 24 = 0

3 )

x = 1st number

x + 2 = 2nd number

x ( x + 2 ) = 575

x ^ 2 + 2 x = 575 Subtract 575 to both sides

x ^ 2 + 2 x - 575 = 0

Solutions :

x = - 25

and

x = 23

Numbers must be negative so :

1st number

x = - 25

2nd number

x + 2 = - 25 + 2 = - 23

4 )

c = width of the frame

New width = 7 + 2 c

New length = 13 + 2 c

( 7 + 2 c ) ( 13 + 2 c ) = 160

13 * 7 + 13 * 2 c + 2c * 7 + 2 c * 2 c = 160

91 + 26 c + 14 c + 4 c ^ 2 = 160

4 c ^ 2 + 40 c + 91 = 160 Subtract 160 to both sides

4 c ^ 2 + 40 c + 91 - 160 = 160 - 160

4 c ^ 2 + 40 c - 69 = 0

Solutions :

c = - 23 / 2

and

c = 3 / 2

Width of the frame can't be negative so :

c = 3 / 2 in

( 7 + 2 * 3 / 2 ) ( 13 + 2 * 3 / 2 ) =

( 7 + 3 ) ( 13 + 3 ) = 10 * 16 = 160 in ^ 2

5 )

P = 2 W + 2 L = 2 ( W + L ) = 30

A = W * L = 54

W * L = 54 Divide both sides by W

L = 54 / W

W ^ 2 - 15 W + 54 = 0 Divide both sides by 2

W + 54 / W = 15 Multiply both sides by W

W * W + ( 54 / W ) * W = 15 * W

W ^ 2 + 54 = 15 W Subtract 15 w to both sides

W ^ 2 + 54 - 15 W = 15 W -15 W

W ^ 2 + 54 - 15 W = 15 W -15 W

Solutions :

W = 6 in

and

W = 9 in

L = 54 / W

When W = 6 in then L = 54 / 6 = 9 in

When W = 9 in then L = 54 / 9 = 6 in

Assuming the width is smaller than the length the dimensions are :

W = 6 in , L = 9 in

- Algebra Homework, HELP! -
**Bosnian**, Monday, October 14, 2013 at 2:31amIn Question 5 delete

W ^ 2 - 15 W + 54 = 0 Divide both sides by 2