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kvv77 [185]
3 years ago
13

A sand box has an area of 45 ft. The length is 4 feet longer than the width. What are the dimensions of the sand box? Solve by c

ompleting the square
Mathematics
1 answer:
erica [24]3 years ago
6 0
So,

The sand box's area is 45 ft.

The length is 4 feet longer than the width.

We can solve by translating these sentences into mathematical form.

Let l represent length and w represent width.

First equation: lw = 45

Second equation: l = 4 + w

Substitute 4 + w for l in the first equation.
(4 + w)w = 45

Distribute
w^{2} + 4w = 45

Subtract 45 from both sides
w^{2} + 4w - 45

Factor
(w + 9)(w -5) =  w^{2} + 4w - 45

Set both factors equal to zero.
w + 9 = 0
w - 5 = 0

Subtract 9 from both sides
w = -9

Add 5 to both sides
w = 5

We have to cross out -9 for the width, because, logically, it is impossible to have a negative width.

Substitute 5 for w in the second original equation.
l = 4 + (5)
l = 9

Check
5 * 9 = 45
45 = 45 This checks.

9 = 4 + 5
9 = 9 This also checks.

The length is 9 ft. and the width is 4 ft.
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a road follows the shape of a parabola f(x)=3x2– 24x + 39. A road that follows the function g(x) = 3x – 15 must cross the stream
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the coordinates where the bridges must be built is (3,-6) and (6,3) .

<u>Step-by-step explanation:</u>

Here we have , a road follows the shape of a parabola f(x)=3x2– 24x + 39. A road that follows the function g(x) = 3x – 15 must cross the stream at point A and then again at point B. Bridges must be built at those points.We need to find Identify the coordinates where the bridges must be built. Let's find out:

Basically we need to find values of x for which f(x) = g(x) :

⇒ f(x)-g(x)=0

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⇒ 3x^2- 24x + 39-3x + 15  =0

⇒ 3x^2- 27x + 54  =0

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Value of g(x) at x = 3 : y=3x -15 = 3(3)-15 = -6

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Therefore , the coordinates where the bridges must be built is (3,-6) and (6,3) .

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