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kozerog [31]
4 years ago
14

The function h(t)=-4.87t^2+18.75t is used to model the height of an object projected in the air, where h(t) is the height in met

ers and t is the time in seconds. Rounded to the nearest hundredth, what are the domain and range of the function h(t)?

Mathematics
1 answer:
rodikova [14]4 years ago
3 0
Looking at the graph you can see that the domain of the function is:
 [0, 3.85]
 To find the range of the function, we must follow the following steps:
 Step 1)
 
Evaluate for t = 0
 h (0) = - 4.87 (0) ^ 2 + 18.75 (0)
 h (0) = 0
 Step 2) 
 find the maximum of the function:
 h (t) = - 4.87t ^ 2 + 18.75t
 h '(t) = - 9.74 * t + 18.75
 -9.74 * t + 18.75 = 0
 t = 18.75 / 9.74
 t = 1.925051335
 We evaluate the function at its maximum point:
 h (1.925051335) = - 4.87 * (1.925051335) ^ 2 + 18.75 * (1.925051335)
 h (1.93) = 18.05
 The range of the function is:
 [0, 18.05]
 Answer:
 
Domain: [0, 3.85]
 Range: [0, 18.05]
 option 1
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Frank was skiing down big bear mountain. He finished 22% of his skiing in the morning. Then he still had 23.4 miles left. How ma
gtnhenbr [62]

Answer: 6.6 miles

Step-by-step explanation:

Since he finished 22% of his skiing in the morning, that means he still has 78% (100% - 22%) of his skiing left.

The remaining 78% is equivalent to 23.4 miles. We need to know the total miles of the skiing. This will be:

78% of x = 23.4

0.78 × x = 23.4

0.78x = 23.4

x = 23.4 / 0.78

x = 30 miles

Since the total skiing is 30 miles and he has 23.4 miles left, the miles covered in the morning will be:

= 30 - 23.4

= 6.6 miles

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3 years ago
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Find d for the arithmetic series with S17=-170 and a1=2
Irina18 [472]
So, we know the sum of the first 17 terms is -170, thus S₁₇ = -170, and we also know the first term is 2, well

\bf \textit{ sum of a finite arithmetic sequence}\\\\
S_n=\cfrac{n(a_1+a_n)}{2}\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
----------\\
n=17\\
S_{17}=-170\\
a_1=2
\end{cases}
\\\\\\
-170=\cfrac{17(2+a_{17})}{2}\implies \cfrac{-170}{17}=\cfrac{(2+a_{17})}{2}
\\\\\\
-10=\cfrac{(2+a_{17})}{2}\implies -20=2+a_{17}\implies -22=a_{17}

well, since the 17th term is that much, let's check what "d" is then anyway,

\bf n^{th}\textit{ term of an arithmetic sequence}\\\\
a_n=a_1+(n-1)d\qquad 
\begin{cases}
n=n^{th}\ term\\
a_1=\textit{first term's value}\\
d=\textit{common difference}\\
----------\\
n=17\\
a_{17}=-22\\
a_1=2
\end{cases}
\\\\\\
-22=2+(17-1)d\implies -22=2+16d\implies -24=16d
\\\\\\
\cfrac{-24}{16}=d\implies -\cfrac{3}{2}=d
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