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nikdorinn [45]
3 years ago
13

Im confused on this one !

Mathematics
1 answer:
ludmilkaskok [199]3 years ago
5 0

I think that since the x values, as they increase become farther and farther from the x-axis, the correct point is (1.5, -3). Because on both sides of the y-axis they are 1.5 units away, since one is positive and one is negative.


Hope this helps! :)

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I need help with this somebody help
Mariulka [41]

Answer:

Width: x + 3

(Length, width):

(5,4) (6,5) (7,6)

Step-by-step explanation:

x² + 7x + 12

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x(x + 4) + 3(x + 4)

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The length of a swimming pool is 3 feet longer than it’s width. The swimming pool is surrounded by a deck that is 2 feet wide an
Firlakuza [10]

Answer:

The Length of swimming pool is 8.35 feet

The width of swimming pool is 5.35 feet

Step-by-step explanation:

Given as :

The length of swimming pool is 3 feet longer that its width

Let The width of swimming pool = w  feet

So, The Length of swimming pool = L = (w + 3) feet

Now, The swimming pool is surrounded by deck of 2 feet wide

The width of deck = w' = w + 2 + 2 = (w + 4) feet

The area of deck = 116 feet²

The length of deck = L' = L + 2 + 2 = (L + 4) feet

So, L' = (w + 3 + 4) feet

I.e L' = (w + 7) feet

∵ The area of deck = 116 feet²

So , L' × w' = 116 feet²

(w + 7)× (w + 4) = 116

Or, w² + 4 w + 7 w + 28 = 116

Or, w² + 11 w - 88 = 0

Solving the quadratic equation as

ax² + bx + c = 0

So, w = \dfrac{-b\pm \sqrt{b^{2}-4\times a\times c}}{2\times a}

Or, w = \dfrac{-11\pm \sqrt{11^{2}-4\times 1\times (-88)}}{2\times 1}

Or, w = \dfrac{-11\pm \sqrt{473}}{2}

or, w = \dfrac{-11\pm 21.7}{2}

or, w = 5.35 , -16.35

The width of swimming pool = w = 5.35 feet

The Length of swimming pool = L = (5.35 + 3) feet = 8.35 feet

Hence, The Length of swimming pool is 8.35 feet

And The width of swimming pool is 5.35 feet

Answer

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4 years ago
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