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Nataliya [291]
4 years ago
11

Question 5 (1 point)

Mathematics
1 answer:
soldier1979 [14.2K]4 years ago
7 0

Answer:

C. 12/15 is not equivalent to the rest

Step-by-step explanation:

12/15 simplifies to 4/5, whereas all the others simplify to 2/3

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A simple index of three stocks have opening values on day 1 and day 8 as shown in the table below.
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APEX (Financial Lit): D. 11.4%

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The graph of the function f(x) = (x-3)(x + 1) is shown.
Drupady [299]

Answer:

x < -1

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Draw a parabola (the graph) passing trough two x-intercepts -1 and 3 and open up.

8 0
2 years ago
What is the function of the problem below?
Fed [463]
–2x<span> + 6. (</span>f<span> – </span>g)(x<span>) = </span>f<span> (</span>x<span>) – </span>g(x). = [3x + 2] – [4 – 5x]. = 3x + 2 – 4 + 5x. = 3x + 5x + 2 – 4. = 8x – 2. (f<span> × </span>g)(x) = [f<span> (</span>x)][g(x)]. = (3x + 2)(4 – 5x). = 12x + 8 – 15x2<span> – 10x ... of the </span>functions<span> at </span>x<span> = 2 and then work from there. It's probably simpler in this case to evaluate first, so: </span>f<span> (2) = 2(2) = 4. </span>g(2) = (2) + 4 = 6. h(2) = 5<span> – (2)</span>3<span> = 5 – 8 = –</span><span>3</span>
3 0
3 years ago
Which equation represents a hyperbola with a center at (0, 0), a vertex at (−48, 0), and a focus at (50, 0)?
Lerok [7]

bearing in mind that "a" is the length of the traverse axis, and "c" is the distance from the center to either foci.

we know the center is at (0,0), we know there's a vertex at (-48,0), from the origin to -48, that's 48 units flat, meaning, the hyperbola is a horizontal one running over the x-axis whose a = 48.

we also know there's a focus point at (50,0), that's 50 units from the center, namely c = 50.


\bf \textit{hyperbolas, horizontal traverse axis } \\\\ \cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1 \qquad \begin{cases} center\ ( h, k)\\ vertices\ ( h\pm a, k)\\ c=\textit{distance from}\\ \qquad \textit{center to foci}\\ \qquad \sqrt{ a ^2 + b ^2}\\ \textit{asymptotes}\quad y= k\pm \cfrac{b}{a}(x- h) \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}


\bf \begin{cases} h=0\\ k=0\\ a=48\\ c=50 \end{cases}\implies \cfrac{(x-0)^2}{48^2}-\cfrac{(y-0)^2}{b^2}=1 \\\\\\ c=\sqrt{a^2+b^2}\implies \sqrt{c^2-a^2}=b\implies \sqrt{50^2-48^2}=b \\\\\\ \sqrt{196}=b\implies 14=b~\hspace{3.5em}\cfrac{(x-0)^2}{48^2}-\cfrac{(y-0)^2}{14^2}=1\implies \cfrac{x^2}{48^2}-\cfrac{y^2}{14^2}=1

8 0
3 years ago
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