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3 years ago

I need help with this problem. Please answer and explain! (Ignore selected answer)

Mathematics

1 answer:

MrMuchimi3 years ago

8 0

Answer:

g'(0) = 1

Step-by-step explanation:

The derivative of a function g at a number a, denoted by g'(a), is given by the definition of the derivative:

$\displaystyle g'(a) = \lim_{h\to0} \dfrac{g(a+h) - g(a)}{h}$

In the definition of the derivative, "h" can be understood to be the horizontal change in the function with respect to a number <em>a</em>.

So g'(0) is

$\displaystyle g'(0) = \lim_{h\to0} \dfrac{g(0+h) - g(0)}{h}=\lim_{h\to0} \dfrac{g(h) - g(0)}{h}$

The variable in the limit is bound; without any other variables around, we can change the variable name to another reasonable variable name and it will still be the same. Hence we can change h to x and it will be equivalent:

$\displaystyle g'(0) = \lim_{x\to0} \dfrac{g(x) - g(0)}{x} = 1$

thus the given limit implies g'(0) = 1.

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If a train travels at 80 mph for 12 min, what is the distance traveled?

AnnyKZ [126]

Answer:

<h2>ANSWER→ </h2>

16 miles
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2 years ago

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Solve for XX. Assume XX is a 2×22×2 matrix and II denotes the 2×22×2 identity matrix. Do not use decimal numbers in your answer.

sveticcg [70]

The question is incomplete. The complete question is as follows:

Solve for X. Assume X is a 2x2 matrix and I denotes the 2x2 identity matrix. Do not use decimal numbers in your answer. If there are fractions, leave them unevaluated.

$\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ =<em>I</em>.

First, we have to identify the matrix <em>I. </em>As it was said, the matrix is the identiy matrix, which means

<em>I</em> = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

So, $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ · X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Isolating the X, we have

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{cc}2&8\\-6&-9\end{array}\right]$ - $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

Resolving:

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}2-1&8-0\\-6-0&-9-1\end{array}\right]$

X· $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ = $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, we have a problem similar to A.X=B. To solve it and because we don't divide matrices, we do X=A⁻¹·B. In this case,

X= $\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ ⁻¹· $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Now, a matrix with index -1 is called Inverse Matrix and is calculated as: A . A⁻¹ = I.

So,

$\left[\begin{array}{ccc}9&-3\\7&-6\end{array}\right]$ · $\left[\begin{array}{ccc}a&b\\c&d\end{array}\right]$ = $\left[\begin{array}{ccc}1&0\\0&1\end{array}\right]$

9a - 3b = 1

7a - 6b = 0

9c - 3d = 0

7c - 6d = 1

Resolving these equations, we have a= $\frac{2}{11}$ ; b= $\frac{7}{33}$ ; c= $\frac{-1}{11}$ and d= $\frac{-3}{11}$ . Substituting:

X= $\left[\begin{array}{ccc}\frac{2}{11} &\frac{-1}{11} \\\frac{7}{33}&\frac{-3}{11} \end{array}\right]$ · $\left[\begin{array}{ccc}1&8\\-6&-10\end{array}\right]$

Multiplying the matrices, we have

X= $\left[\begin{array}{ccc}\frac{8}{11} &\frac{26}{11} \\\frac{39}{11}&\frac{198}{11} \end{array}\right]$

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3 years ago

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Lostsunrise [7]

Answer:

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Step-by-step explanation:

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3 years ago

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MArishka [77]

Answer:

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Step-by-step explanation:

this is the sample response

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Amy: (3 pages/min)(x min) + 9 pages
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3 years ago