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Alexus [3.1K]
4 years ago
5

At the park there is a pool shaped like a circle. A ring-shaped path goes around the pool. Its inner diameter is 22yd and its ou

ter diameter is 32yd . 22yd 32yd. We are going to give a new layer of coating to the path. If one gallon of coating can cover 6yd2 , how many gallons of coating do we need?

Mathematics
1 answer:
jeka944 years ago
6 0
First, we need to think about what we're actually solving for: the area of the path. To find this, we can find the area of the whole thing, minus the area of the pool.

The diagram would look something like the picture I attached. We know that to find the area of a circle, you do πr^2. The radius is half of the diameter, so halve the diameters:

22/2 = 11 yards
32/2 = 16 yards

Then solve for the area of both circles.
π11^{2}= 121π ≈ <span>380.13 (area of the pool)</span>
π16^{2}= 256π ≈ <span>804.25 (area of the whole thing)
</span>
Next, subtract the area of the pool from the area of the whole thing to find the area of the walkway:
804.25 - 380.13 = 424.12 (area of the walkway)

Finally, divide the area of the walkway by 6 to see how many gallons of paint you'll need:

424.12 ÷ 6 ≈ 70.69

So you'll need 71 gallons of paint!

Hope I helped and let me know if you have any questions :)


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Any quadratic function can be rewritten in standard form by completing the square. (See the section on solving equations algebraically to review completing the square.) The steps that we use in this section for completing the square will look a little different, because our chief goal here is not solving an equation.

Note that when a quadratic function is in standard form it is also easy to find its zeros by the square root principle.

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= (x2 - 6x )+ 7.        Group the x2 and x terms and then complete the square on these terms.

= (x2 - 6x + 9 - 9) + 7.

We need to add 9 because it is the square of one half the coefficient of x, (-6/2)2 = 9. When we were solving an equation we simply added 9 to both sides of the equation. In this setting we add and subtract 9 so that we do not change the function.

= (x2 - 6x + 9) - 9 + 7. We see that x2 - 6x + 9 is a perfect square, namely (x - 3)2.

f(x) = (x - 3)2 - 2. This is standard form.

From this result, one easily finds the vertex of the graph of f is (3, -2).

To find the zeros of f, we set f equal to 0 and solve for x.

(x - 3)2 - 2 = 0.

(x - 3)2 = 2.

(x - 3) = ± sqrt(2).

x = 3 ± sqrt(2).

To sketch the graph of f we shift the graph of y = x2 three units to the right and two units down.

If the coefficient of x2 is not 1, then we must factor this coefficient from the x2 and x terms before proceeding.

Example 4.

Write f(x) = -2x2 + 2x + 3 in standard form and find the vertex of the graph of f.

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We add and subtract 1/4, because (-1/2)2 = 1/4, and -1 is the coefficient of x.

= -2(x2 - x + 1/4) -2(-1/4) + 3.

Note that everything in the parentheses is multiplied by -2, so when we remove -1/4 from the parentheses, we must multiply it by -2.

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In some cases completing the square is not the easiest way to find the vertex of a parabola. If the graph of a quadratic function has two x-intercepts, then the line of symmetry is the vertical line through the midpoint of the x-intercepts.

The x-intercepts of the graph above are at -5 and 3. The line of symmetry goes through -1, which is the average of -5 and 3. (-5 + 3)/2 = -2/2 = -1. Once we know that the line of symmetry is x = -1, then we know the first coordinate of the vertex is -1. The second coordinate of the vertex can be found by evaluating the function at x = -1.

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The second coordinate of the vertex is f(-2) = (-2 + 9)(-2 - 5) = 7*(-7) = -49.

Therefore, the vertex of the graph of f is (-2, -49).

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