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cestrela7 [59]
3 years ago
5

The hourly temperature at Portland, Oregon, on a particular day is recorded below.

Mathematics
1 answer:
adell [148]3 years ago
3 0

Answer:

a. Amplitude: 19.65446

b. Vertical Shift: 58.00713

c. Period: 2\pi/0.22797 = 27.56146

d. 19.65446 * sin(0.22797x + 1.79552) + 58.00713

e. Model's Temperature at 10 A.M.: f(-7) = 61.91

    This value is 9.91 degrees higher than the actual value.

Step-by-step explanation:

Used graphing calculator's stat calculations

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3 years ago
between 1995 and 2006, the population of samburg, usa(in thousands) can be modeled by f(x) = 0.35x(squared) - 2.1x+15.8 where x=
Juli2301 [7.4K]
\bf f(x)=0.35x^2-2.1x+15.8\qquad 
\begin{cases}
x=\textit{year since 1995}\\
f(x)=\textit{population amount}
\end{cases}

the equation is a quadratic one, and it has a positive coefficient on the leading term, meaning, is opening upwards, so it has a "burrow" for the vertex.

the minimum or lowest point for a quadratic opening upwards is, well, the vertex point :),   the "x" value is the year, the "y" or f(x) value is the population, we're asked for the year, or the x-coordinate of the vertex

well   \bf \begin{array}{llll}
f(x)=&0.35x^2&-2.1x&+15.8\\
&\quad \uparrow &\quad \uparrow&\uparrow \\
&\quad  a&\quad  b &c
\end{array}
\\\\

\\\\
\qquad  \textit{vertex of a parabola}\\ \quad \\
\qquad 

\left(\boxed{-\cfrac{{{ b}}}{2{{ a}}}}\quad ,\quad  {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)
5 0
3 years ago
Which system of linear inequalities is graphed?
iragen [17]
I am sure that This one would be c
6 0
3 years ago
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If you divide 3/4 by 4/7 what is the answer in fraction form?
victus00 [196]
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8 0
3 years ago
What is the equation for the hyperbola shown?
amid [387]

Answer:

D. y²/5² - x²/8² = 1

Step-by-step explanation:

A and B are both incorrectly oriented, and D is the only hyperbola that contains the points (0,5) and (0,-5).

Verification (0,5) and (0,-5) are in the hyperbola:

First replace x and y with corresponding x and y values (We will start with x=0 and y=5)

\frac{5^{2}}{5^{2}}-\frac{0^{2}}{8^{2}}=1

Then simplify.

\frac{25}{25}-\frac{0}{16}=1

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1=1

If the result is an equation (where both sides are equal to each other) then the original x and y values inputted are valid. The same is true with x and y inputs x=0 and y=-5, or any other point along   the hyperbola.

6 0
3 years ago
Read 2 more answers
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