9 and 12 have a common factor of 3 so 9/12 = 3/4
Answer: Anything between 0 and 10, excluding both endpoints.
In terms of symbols we can say 0 < w < 10 where w is the width.
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Explanation:
You could do this with two variables, but I think it's easier to instead use one variable only. This is because the length is dependent on what you pick for the width.
w = width
2w = twice the width
2w-5 = five less than twice the width = length
So,
which lead to
area = length*width
area = (2w-5)*w
area = 2w^2-5w
area < 150
2w^2 - 5w < 150
2w^2 - 5w - 150 < 0
To solve this inequality, we will solve the equation 2w^2-5w-150 = 0
Use the quadratic formula. Plug in a = 2, b = -5, c = -150
![w = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\\w = \frac{-(-5)\pm\sqrt{(-5)^2-4(2)(-150)}}{2(2)}\\\\w = \frac{5\pm\sqrt{1225}}{4}\\\\w = \frac{5\pm35}{4}\\\\w = \frac{5+35}{4} \ \text{ or } \ w = \frac{5-35}{4}\\\\w = \frac{40}{4} \ \text{ or } \ w = \frac{-30}{4}\\\\w = 10 \ \text{ or } \ w = -7.5\\\\](https://tex.z-dn.net/?f=w%20%3D%20%5Cfrac%7B-b%5Cpm%5Csqrt%7Bb%5E2-4ac%7D%7D%7B2a%7D%5C%5C%5C%5Cw%20%3D%20%5Cfrac%7B-%28-5%29%5Cpm%5Csqrt%7B%28-5%29%5E2-4%282%29%28-150%29%7D%7D%7B2%282%29%7D%5C%5C%5C%5Cw%20%3D%20%5Cfrac%7B5%5Cpm%5Csqrt%7B1225%7D%7D%7B4%7D%5C%5C%5C%5Cw%20%3D%20%5Cfrac%7B5%5Cpm35%7D%7B4%7D%5C%5C%5C%5Cw%20%3D%20%5Cfrac%7B5%2B35%7D%7B4%7D%20%5C%20%5Ctext%7B%20or%20%7D%20%5C%20w%20%3D%20%5Cfrac%7B5-35%7D%7B4%7D%5C%5C%5C%5Cw%20%3D%20%5Cfrac%7B40%7D%7B4%7D%20%5C%20%5Ctext%7B%20or%20%7D%20%5C%20w%20%3D%20%5Cfrac%7B-30%7D%7B4%7D%5C%5C%5C%5Cw%20%3D%2010%20%5C%20%5Ctext%7B%20or%20%7D%20%5C%20w%20%3D%20-7.5%5C%5C%5C%5C)
Ignore the negative solution as it makes no sense to have a negative width.
The only practical root is w = 10.
If w = 10 feet, then the area = 2w^2-5w results in 150 square feet.
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Based on that root, we need to try a sample value that is to the left of it.
Let's say we try w = 5.
2w^2 - 5w < 150
2*5^2 - 5*5 < 150
25 < 150 ... which is true
This shows that if 0 < w < 10, then 2w^2-5w < 150 is true.
Now try something to the right of 10. I'll pick w = 15
2w^2 - 5w < 150
2*15^2 - 5*15 < 150
375 < 150 ... which is false
It means w > 10 leads to 2w^2-5w < 150 being false.
Therefore w > 10 isn't allowed if we want 2w^2-5w < 150 to be true.
Y=3/100x
y= 3/100 (150)
y=4.5
150+4.5=154.50 total cost of CD player.
Therefore that is the answer. Please mark as brainliest and give a thanks.
<span><span>Graph <span>x2<span> = 4</span>y</span><span> and state the vertex, focus, axis of symmetry, and directrix.</span></span><span>This is the same graphing that I've done in the past: </span><span>y = (1/4)x2</span><span>. So I'll do the graph as usual:</span></span><span> </span><span>The vertex is obviously at the origin, but I need to "show" this "algebraically" by rearranging the given equation into the conics form:<span>x2 = 4y</span> Copyright © Elizabeth Stapel 2010-2011 All Rights Reserved<span>
(x – 0)2 = 4(y – 0)</span><span>This rearrangement "shows" that the vertex is at </span><span>(h, k) = (0, 0)</span><span>. The axis of symmetry is the vertical line right through the vertex: </span><span>x = 0</span>. (I can always check my graph, if I'm not sure about this.) The focus is "p" units from the vertex. Since the focus is "inside" the parabola and since this is a "right side up" graph, the focus has to be above the vertex.<span>From the conics form of the equation, shown above, I look at what's multiplied on the unsquaredpart and see that </span><span>4p = 4</span><span>, so </span><span>p = 1</span><span>. Then the focus is one unit above the vertex, at </span>(0, 1)<span>, and the directrix is the horizontal line </span><span>y = –1</span>, one unit below the vertex.<span>vertex: </span>(0, 0)<span>; focus: </span>(0, 1)<span>; axis of symmetry: </span><span>x<span> = 0</span></span><span>; directrix: </span><span>y<span> = –1</span></span></span><span><span><span>Graph </span><span>y2<span> + 10</span>y<span> + </span>x<span> + 25 = 0</span></span>, and state the vertex, focus, axis of symmetry, and directrix.</span><span>Since the </span>y<span> is squared in this equation, rather than the </span>x<span>, then this is a "sideways" parabola. To graph, I'll do my T-chart backwards, picking </span>y<span>-values first and then finding the corresponding </span>x<span>-values for </span><span>x = –y2 – 10y – 25</span>:<span>To convert the equation into conics form and find the exact vertex, etc, I'll need to convert the equation to perfect-square form. In this case, the squared side is already a perfect square, so:</span><span>y2 + 10y + 25 = –x</span> <span>
(y + 5)2 = –1(x – 0)</span><span>This tells me that </span><span>4p = –1</span><span>, so </span><span>p = –1/4</span><span>. Since the parabola opens to the left, then the focus is </span>1/4<span> units to the left of the vertex. I can see from the equation above that the vertex is at </span><span>(h, k) = (0, –5)</span><span>, so then the focus must be at </span>(–1/4, –5)<span>. The parabola is sideways, so the axis of symmetry is, too. The directrix, being perpendicular to the axis of symmetry, is then vertical, and is </span>1/4<span> units to the right of the vertex. Putting this all together, I get:</span><span>vertex: </span>(0, –5)<span>; focus: </span>(–1/4, –5)<span>; axis of symmetry: </span><span>y<span> = –5</span></span><span>; directrix: </span><span>x<span> = 1/4</span></span></span><span><span>Find the vertex and focus of </span><span>y2<span> + 6</span>y<span> + 12</span>x<span> – 15 = 0</span></span></span><span><span>The </span>y<span> part is squared, so this is a sideways parabola. I'll get the </span>y stuff by itself on one side of the equation, and then complete the square to convert this to conics form.<span>y2 + 6y – 15 = –12x</span> <span><span>
y</span>2 + 6y + 9 – 15 = –12x + 9</span> <span>
(y + 3)2 – 15 = –12x + 9</span> <span>
(y + 3)2 = –12x + 9 + 15 = –12x + 24</span> <span>
(y + 3)2 = –12(x – 2)</span> <span>
(y – (–3))2 = 4(–3)(x – 2)</span></span><span><span>Then the vertex is at </span><span>(h, k) = (2, –3)</span><span> and the value of </span>p<span> is </span>–3<span>. Since </span>y<span> is squared and </span>p<span> is negative, then this is a sideways parabola that opens to the left. This puts the focus </span>3 units to the left of the vertex.<span>vertex: </span>(2, –3)<span>; focus: </span><span>(–1, –3)</span><span>
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