the length of a photograph is 1 cm less than twice the width. the area is 28 cm squared. find the dimensions of the photograph.

So the formula for the area of this photograph would be the following:

area = length x width
(let's fill in what we know from the question into this formula)

28 =(2w-1) * w
(Now let's use the distribution rule and multiply everything in the brackets by 'w')
28 = 2w^2-w
(Now let's set this equation to zero)
0=2w^2-w-28
(If you noticed, this equation in the form of y=ax^2+bx+c)

Now at this point you could use the quadratic question to solve for 'w'.

What I got for my values were -3.5 and 4
Since -3.5 does not make sense realistically, you would go with 4 cm.

Let's plug this back into the question to see if we're correct:

28 =(2w-1) * w
28= (2[4]-1)*4
28 = (8-1)*4
28 = 7*4
28 = 28

Therefore the width is 4 cm and length is 7 cm

Oh, photography, picture perfect! Let's solve this riddle together.

Let's say the width of the photograph is "x" cm. According to the information given, the length would then be 2x - 1 cm (1 cm less than twice the width).

Now, to find the dimensions, we have to multiply the length by the width and set it equal to the given area. In this case, that's 28 cm².

So, we have the equation:

(x)(2x - 1) = 28

Expanding the equation, we get:

2x² - x = 28

Rearranging and simplifying, we have:

2x² - x - 28 = 0

Now, let me put on my clown hat and solve this quadratic equation for you:

(x - 4)(2x + 7) = 0

Now, we have two possible solutions:

x - 4 = 0 or 2x + 7 = 0

If we solve these separately, we find:

x = 4 or x = -7/2

Hmm, negative dimensions don't make sense in this context, so we'll discard x = -7/2.

Therefore, the width is x = 4 cm. Now we can find the length:

Length = 2x - 1
Length = 2(4) - 1
Length = 8 - 1
Length = 7 cm

Ta-da! The dimensions of the photograph are 4 cm by 7 cm. Happy framing! 📸

To find the dimensions of the photograph, we can set up two equations based on the given information.

Let's assume the width of the photograph is x cm.
According to the given information, the length of the photograph is 1 cm less than twice the width, which can be represented as (2x - 1) cm.

The area of a rectangle is given by the formula: Area = length × width.
So, we can set up the equation:

Area = length × width
28 = (2x - 1) × x

Now, we can solve this quadratic equation for x.

28 = (2x^2 - x)
Rearranging the equation:
2x^2 - x - 28 = 0

To solve a quadratic equation, we can either factorize or use the quadratic formula. Let's use the quadratic formula.

x = (-b ± √(b^2 - 4ac))/(2a)

In our equation, a = 2, b = -1, and c = -28.

x = (-(-1) ± √((-1)^2 - 4(2)(-28)))/(2(2))
x = (1 ± √(1 + 224))/4
x = (1 ± √225)/4
x = (1 ± 15)/4

This gives us two possible values for x:

x = (1 + 15)/4 = 16/4 = 4
or
x = (1 - 15)/4 = -14/4 = -7/2 (rejected since length and width cannot be negative)

Therefore, the width of the photograph is 4 cm.

Now, we can find the length using the equation length = 2x - 1:

length = 2 × 4 - 1 = 8 - 1 = 7 cm

So, the dimensions of the photograph are:
Width = 4 cm
Length = 7 cm

To find the dimensions of the photograph, let's assign variables to the width and length. Let's say the width is 'w' cm.

According to the given information, the length of the photograph is 1 cm less than twice the width. So, the length can be expressed as (2w - 1) cm.

We also know that the area of a rectangle is given by the product of its length and width. In this case, the area is given as 28 cm². So, we can set up an equation:

w * (2w - 1) = 28

Now, let's solve this equation to find the value of 'w' (width).