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3 years ago

Brenda school is selling tickets to a spring musical

Mathematics

1 answer:

Phantasy [73]3 years ago

8 0

I don't see a picture on here.

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What are three consecutive odd integers such that the third is nine more than twice the sum of the first and second

Natasha_Volkova [10]

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540.1 is 55% of what amount

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3 years ago

Write the linearization of the function at the points indicated. (Enter your answer as an equation. Let x be the independent var

Natalija [7]

Answer:

$\displaystyle y=5+\frac{x}{10}$

$\displaystyle y=10+\frac{(x-75)}{20}$

Step-by-step explanation:

<u>Linearization</u>

It consists of finding an approximately linear function that behaves as close as possible to the original function near a specific point.

Let y=f(x) a real function and (a,f(a)) the point near which we want to find a linear approximation of f. If f'(x) exists in x=a, then the equation for the linearization of f is

$y=f(x)=f(a)+f'(a)(x-a)$

Let's find the linearization for the function

$y=\sqrt{25+x}$

at (0,5) and (75,10)

Computing f'(x)

$\displaystyle f'(x)=\frac{1}{2\sqrt{25+x}}$

At x=0:

$\displaystyle f'(0)=\frac{1}{2\sqrt{25+0}}=\frac{1}{10}$

We find f(0)

$f(0)=\sqrt{25+0}=5$

Thus the linearization is

$\displaystyle y=f(0)+f'(0)(x-0)=5+\frac{1}{10}x$

$\displaystyle y=5+\frac{x}{10}$

Now at x=75:

$\displaystyle f'(75)=\frac{1}{2\sqrt{25+75}}=\frac{1}{20}$

We find f(75)

$f(75)=\sqrt{25+75}=10$

Thus the linearization is

$\displaystyle y=f(75)+f'(75)(x-75)=10+\frac{1}{20}(x-75)$

$\displaystyle y=10+\frac{(x-75)}{20}$

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3 years ago