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maksim [4K]
4 years ago
9

Below is the graph of a trigonometric function. It has a minimum point at \left(-5.4,1.25\right)(−5.4,1.25)left parenthesis, min

us, 5, point, 4, comma, 1, point, 25, right parenthesis and a maximum point at \left(7.4,6.75\right)(7.4,6.75)left parenthesis, 7, point, 4, comma, 6, point, 75, right parenthesis.
Mathematics
1 answer:
HACTEHA [7]4 years ago
6 0

Answer:

Multiple and divide

Step-by-step explanation:

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Salina’s age is one-quarter the age of her aunt. If a represents her aunt’s age, which expression represents Salina’s age?
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Salina is 1/4 of age of unt
s=1/4 times a
s=a/4
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3 years ago
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Kyle divided two numbers in scientific notation using the interactive calculator and found a solution of 3.5E4. If the numerator
Maru [420]

Answer:

a=2, b=5

Step-by-step explanation:

Let

n -----> the denominator

we have

3.5*10^4=\frac{7*10^9}{n}

Solve for n

That means -----> isolate the variable n

Multiply by n both sides

3.5*10^4(n)=7*10^9

Divide by 3.5*10^4 both sides

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Atmospheric pressure decreases by about 11.8% for every 1000 meters you climb. The pressure at sea level is 1013 millibars (a un
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At sea level, the pressure is 1013, that's when the altitude is at 0, sea level, let's see

\bf \textit{Periodic Exponential Decay}\\\\
A=I(1 - r)^{\frac{t}{p}}\qquad 
\begin{cases}
A=\textit{accumulated amount}\to &1013\\
I=\textit{initial amount}\\
r=rate\to 11.8\%\to \frac{11.8}{100}\to &0.118\\
t=\textit{meters climbed}\to &0\\
p=period\to &1000
\end{cases}
\\\\\\
1013=I(1-0.118)^{\frac{0}{1000}}\implies 1013=I\cdot 1\implies 1013=I

so, the inital amount is 1013, when t = 0,

\bf \textit{Periodic Exponential Decay}\\\\
A=I(1- r)^{\frac{t}{p}}\qquad 
\begin{cases}
A=\textit{accumulated amount}\\
I=\textit{initial amount}\to &1013\\
r=rate\to 11.8\%\to \frac{11.8}{100}\to &0.118\\
t=\textit{meters climbed}\to &t\\
p=period\to &1000
\end{cases}
\\\\\\
A=1013(1-0.118)^{\frac{t}{1000}}\implies A=1013(0.882)^{\frac{t}{1000}}

now, to check the atmospheric pressure at 4000, simply set t = 4000, to get A.


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