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3 years ago

The focus of a parabola is (0, -3) and the directrix is y = 3. What is the equation of the parabola?

1 answer:

5 0

Focus has coordinates (0, a)
a = -3
Thus, 4a = 4(-3) = -12

Because a is negative, we know it will be a concave down or concave left parabola. Hence, A and D can be eliminated. And we know it has to be B because the directrix is the horizontal line y = 3. Hence, the equation of the parabola becomes:

$x^{2} = -12y$ or $y = -\frac{1}{12}x^{2}$

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3 years ago

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A bus company has contracted with a local high school to carry 450 students on a field trip. The company has 18 large buses whic

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Answer:

The answer is below

Step-by-step explanation:

Let x represent the big buses and y represent small buses. The large buses can carry 30 students and the small buses can carry 15 students. The total number of students are 450, this can be represented by the inequality:

30x + 15y ≤ 450

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x + y ≤ 20

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After plotting the graph, the minimum solution to the graph are at:

A (15,0), B(18,0), C(10, 10), D(18, 2).

The cost function is given as:

The total cost of operating one large bus is $225 a day, and the total cost of operating one small bus is $100 per day.

F(x, y) = 225x + 100y

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F(x, y) = 225(15) + 100(0) = $3375

At point B:

F(x, y) = 225(18) + 100(0) = $4050

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F(x, y) = 225(10) + 100(10) = $3250

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F(x, y) = 225(18) + 100(2) = $4250

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3 years ago

Write an equation that has a graph with the shape of y = x^2, but shifted left 3 units.

svetoff [14.1K]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf % template detailing
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\$

$\bf \bullet \textit{ vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}$

now, with that template in mind, let's see

$\bf y=x^2\implies 
\begin{array}{llll}
y=(&1x&-0)^2\\
&B&C
\end{array}$

so, just change C to +3, thus C/B is 3/1

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3 years ago