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Kaylis [27]
4 years ago
5

Find the linear approximation of the function g(x)=(1+x)^(1/3) at a=0 and use it to approximate the numbers (0.95)^(1/3) and (1.

1)^(1/3). Illustrate by graphing g and the tangent line.
Mathematics
1 answer:
Leya [2.2K]4 years ago
8 0
So the question ask to calculate the linear approximation of the function g(x) = (1+x)^(1/3) at a = 0 and base on the given in the problem and further computation of the said problem, the linear approximation of the function is y = 1/3x + 1  I hope you are satisfied with my answer and feel free to ask for more 
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Answer:

3r3

Step-by-step explanation:

7 times 3 is as high as you can go w/o over multiplying, and you have a remainder of 3 until you can actually get 24

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Compute the determinant using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion do
V125BC [204]

Answer:

Step-by-step explanation:

It is given that

\Delta=\begin{vmatrix}3&0&3\\2 &3&3\\0 &4&-2\end{vmatrix}

By cofactor expansion across the first row, we get

\Delta=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}

\Delta=3\left[(-1)^{1+1}\begin{vmatrix}3&3\\4&-2\end{vmatrix}\right]+0\left[(-1)^{1+2}\begin{vmatrix}2&3\\0&-2\end{vmatrix}\right]+3\left[(-1)^{1+3}\begin{vmatrix}2&3\\0&4\end{vmatrix}\right]

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\Delta=-54+0+24

\Delta=-30

Therefore, the value of determinant is -30.

By cofactor expansion across the second column, we get

\Delta=a_{12}C_{12}+a_{22}C_{22}+a_{32}C_{32}

\Delta=0\left[(-1)^{2+1}\begin{vmatrix}2&3\\0&-2\end{vmatrix}\right]+3\left[(-1)^{2+2}\begin{vmatrix}3&3\\0&-2\end{vmatrix}\right]+4\left[(-1)^{3+2}\begin{vmatrix}3&3\\2&3\end{vmatrix}\right]

\Delta=0\left[(-1)(-4)\right]+3\left[(-6)\right]+4\left[(-1)3\right]

\Delta=-18-12

\Delta=-30

Therefore, the value of determinant is -30.

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4 years ago
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Answer:

0.4(−0.45)=9.2

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