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schepotkina [342]
2 years ago
8

The formula for the distance, d, an accelerating car can travel is d = 5.6t square. how many minutes, t, doe the car take to tra

vel 1,814.4 m?
Mathematics
1 answer:
Likurg_2 [28]2 years ago
6 0
Your distance(d) is 1,814.4
1814.4 m = 5.6t square
then you square root 5.6t
\sqrt{1814.4} = \sqrt{ {5.6t}^{2} }
the square root eliminates the square
\sqrt{1814.4} = 5.6t
you square root 1,814.4 m
42.6 = 5.6t
(I rounded 42.6) then divide 42.6 by 5.6
7.6 = t
(I rounded 7.6)
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Answer:

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Step-by-step explanation:

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Multiply the row of the first matrix to the column of the second matrix

\begin{pmatrix}-3&4\end{pmatrix}\begin{pmatrix}3\\ 1\end{pmatrix}=\left(-3\right)\cdot \:3+4\cdot \:1

\begin{pmatrix}-3&4\end{pmatrix}\begin{pmatrix}-2\\ 0\end{pmatrix}=\left(-3\right)\left(-2\right)+4\cdot \:0

\begin{pmatrix}2&-5\end{pmatrix}\begin{pmatrix}3\\ 1\end{pmatrix}=2\cdot \:3+\left(-5\right)\cdot \:1\

\begin{pmatrix}2&-5\end{pmatrix}\begin{pmatrix}-2\\ 0\end{pmatrix}=2\left(-2\right)+\left(-5\right)\cdot \:0\

=\begin{pmatrix}\left(-3\right)\cdot \:3+4\cdot \:1&\left(-3\right)\left(-2\right)+4\cdot \:0\\ 2\cdot \:3+\left(-5\right)\cdot \:1&2\left(-2\right)+\left(-5\right)\cdot \:0\end{pmatrix}

simplifying them we get

=\begin{pmatrix}-5&6\\ 1&-4\end{pmatrix}

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3 years ago
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Answer:

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we have.

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4 0
3 years ago
A. Plot the data for the functions f(x) and g(x) on a grid and connect the points.
labwork [276]

Answer:

  a) see the plots below

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Step-by-step explanation:

a) a graph of the two function values is attached

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b) Adjacent values of f(x) have a common ratio of 3, so f(x) is exponential (with a base of 3). Adjacent values of g(x) have a common difference of 2, so g(x) is linear (with a slope of 2).

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4 0
3 years ago
A wheel initially has an angular velocity of 18 rad/s, but it is slowing at a constant rate of 2 rad/s 2 . By the time it stops,
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3 years ago
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jok3333 [9.3K]

Answer:

Step-by-step explanation:

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