1) A rectangle has an area of 297 square inches and a width which is three inches less than four times the length. Find the length and the width of the rectangle.

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.

1. 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

1 )

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

In Question 5 delete

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

The length of the base measures

65ft less than twice the width. The perimeter of this base is 770
ft. What are the dimensions of the​ base?

1) To find the length and width of the rectangle, we need to set up a system of equations based on the given information.

Let's assume the length of the rectangle is "L" and the width is "W".

From the problem, we know that the area of the rectangle is 297 square inches, so we can write the equation:
L * W = 297

We are also given that the width is three inches less than four times the length, which can be written as:
W = 4L - 3

Now we have a system of two equations with two variables. To solve, we can substitute the second equation into the first equation:

L * (4L - 3) = 297

Simplify the equation and solve for L:
4L^2 - 3L - 297 = 0

Using factoring or the quadratic formula, we can find the values of L. Once we have the value of L, we can substitute it back into the second equation to find the value of W.

2) To set up the equation for the side lengths of the right triangle, we can use the Pythagorean theorem. The theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides.

Let's assume x represents the length of the shorter leg.

According to the given information, the longer leg is 1 more than the shorter leg, so we can represent it as (x + 1).

The hypotenuse has a length of 7, so we can represent it as 7.

Using the Pythagorean theorem, we can set up the equation:
x^2 + (x + 1)^2 = 7^2

Simplifying and expanding the equation will give us the equation needed to solve for the side lengths.

3) To find the consecutive negative odd integers, we can set up an equation based on the given information.

Let's assume the first consecutive negative odd integer is "n". Since the next consecutive odd integer follows n, we can represent it as (n + 2).

The problem states that the product of these two integers is 575, so we can set up the equation:
n * (n + 2) = 575

Solving this equation will give us the values of the consecutive negative odd integers.

4) To find the uniform width of the frame, we need to set up an equation based on the given information.

The canvas picture measures 13 inches by 7 inches, so the area of the canvas is 13 * 7 = 91 square inches.

Let's assume the uniform width of the frame is "w".

The area covered by both the canvas and the frame is 160 square inches, so we can set up the equation:
(13 + 2w)(7 + 2w) = 160

After expanding and simplifying the equation, you can solve for the value of the uniform width "w".

5) To find the dimensions of the rectangle, we can set up equations based on the given information.

Let's assume the length of the rectangle is "L" and the width is "W".

According to the problem, the perimeter of the rectangle is 30 feet, so we can write the equation:
2L + 2W = 30

The area of the rectangle is given as 54 square feet, so we can also write the equation:
L * W = 54

Now we have a system of two equations with two variables. Solving this system will give us the dimensions of the rectangle, assuming the width is smaller than the length.