# The total area of the rectangle is 10 ft^2 What is the width and length of the rectangle, given that the width is 3 feet less than the length?

x is 5 feet because the length is 5 ft and the width is 2 ft.

I found it by trial and error. You can try quadratic formula to solve the problem.

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Let ( w ) represent the width and ( l ) represent the length of the rectangle. Since the width is 3 feet less than the length, we can express this relationship as ( w = l - 3 ).

The formula for the area of a rectangle is ( \text{Area} = \text{width} \times \text{length} ).

Given that the total area of the rectangle is ( 10 , \text{ft}^2 ), we can write the equation:

[ 10 = w \times l ]

Substituting ( w = l - 3 ), we get:

[ 10 = (l - 3) \times l ]

Expanding and rearranging the equation:

[ 10 = l^2 - 3l ]

This equation is a quadratic equation. To solve for ( l ), we can rearrange it into the standard form:

[ l^2 - 3l - 10 = 0 ]

Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Once we find the value(s) of ( l ), we can substitute it back into ( w = l - 3 ) to find the corresponding value(s) of ( w ).

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Let the length of the rectangle be ( L ) feet and the width be ( W ) feet. Given that the width is 3 feet less than the length, we can express this relationship as:

[ W = L - 3 ]

The formula for the area ( A ) of a rectangle is:

[ A = \text{Length} \times \text{Width} ]

Given that the total area of the rectangle is 10 ft(^2), we can set up the equation:

[ 10 = L \times (L - 3) ]

Expanding and rearranging:

[ 10 = L^2 - 3L ]

This is a quadratic equation. To solve for ( L ), we set the equation equal to zero:

[ L^2 - 3L - 10 = 0 ]

Now, we can factor or use the quadratic formula to solve for ( L ). Let's use the quadratic formula:

[ L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]

where ( a = 1 ), ( b = -3 ), and ( c = -10 ).

[ L = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-10)}}{2(1)} ] [ L = \frac{3 \pm \sqrt{9 + 40}}{2} ] [ L = \frac{3 \pm \sqrt{49}}{2} ] [ L = \frac{3 \pm 7}{2} ]

So, ( L = \frac{3 + 7}{2} = 5 ) feet or ( L = \frac{3 - 7}{2} = -2 ) feet.

Since the length cannot be negative, we discard the negative solution. Therefore, the length of the rectangle is 5 feet.

Now, we can find the width using ( W = L - 3 ):

[ W = 5 - 3 = 2 ] feet.

Therefore, the width of the rectangle is 2 feet and the length is 5 feet.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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