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alina1380 [7]
1 year ago
10

The graph of the cube root parent function y = RootIndex 3 StartRoot x EndRoot is translated to form f(x) shown on the graph.

Mathematics
1 answer:
Mars2501 [29]1 year ago
3 0

Step-by-step explanation:

The graph of the cube root parent function y = RootIndex 3 StartRoot x EndRoot is translated to form f(x) shown on the graph.

On a coordinate plane, a cube root function goes through (negative 7, 0), has an inflection point at (negative 6, 1), and goes through (2, 3).

Which equation represents f(x)?

f(x) = RootIndex 3 StartRoot x + 6 EndRoot + 1

f(x) = RootIndex 3 StartRoot x minus 6 EndRoot + 1

f(x) = RootIndex 3 StartRoot x + 6 EndRoot minus 1

f(x) = RootIndex 3 StartRoot x minus 6 EndRoot minus 1

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Answer:

\theta=36.43^{\circ}

Step-by-step explanation:

Given that,

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We need to find the value of \theta. It can be calculated using trigonometry as follows :

\tan\theta=\dfrac{P}{B}\\\\\theta=\tan^{-1}(\dfrac{P}{B})

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\theta=\tan^{-1}(\dfrac{31}{42})\\\\\theta=36.43^{\circ}

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Felix

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The graph contains 3 segments,

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second one is for the next 2 minutes (standing still)

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Am amount of $36,000 is borrowed for 12 year of 5% interest,compounded annually.If the loan is paid in full at the end of that p
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2 years ago
Given the matrices: 1 2 A= 1 -1 2 1 1 B= 3 4 Calculate AB: C11 C12 [2.1] х 1 2 3 4 C21 C22 C11 = C12 = -2 C22 - C215 DONE​
eduard

Answer:

\:c_{11}=-2,\:\:\:c_{12}=-2

\:c_{21}=5,\:\:\:c_{22}=8

Step-by-step explanation:

Given the matrices

A=\begin{pmatrix}1&-1\\ 2&1\end{pmatrix}

B=\begin{pmatrix}1&2\\ \:3&4\end{pmatrix}

Calculating AB:

\begin{pmatrix}1&-1\\ \:\:2&1\end{pmatrix}\times \:\begin{pmatrix}1&2\\ \:\:3&4\end{pmatrix}=\begin{pmatrix}c_{11}&c_{12}\\ \:\:\:c_{21}&c_{22}\end{pmatrix}

Multiply the rows of the first matrix by the columns of the second matrix

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

                                   =\begin{pmatrix}-2&-2\\ 5&8\end{pmatrix}

Hence,

\begin{pmatrix}c_{11}&c_{12}\\ \:\:\:c_{21}&c_{22}\end{pmatrix}=\begin{pmatrix}-2&-2\\ \:5&8\end{pmatrix}

Therefore,

\:c_{11}=-2,\:\:\:c_{12}=-2

\:c_{21}=5,\:\:\:c_{22}=8

5 0
2 years ago
Read 2 more answers
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