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Andru [333]
2 years ago
5

If f'(0) = 5 and F(x) = f(3x), what is F'(0)?

Mathematics
2 answers:
Norma-Jean [14]2 years ago
8 0

Answer:

F'(0)=15

Step-by-step explanation:

Differentiate F(x) with respect to x.

F'(x)=3f'(x)

Subsitute 0 for x in the above equation.

F'(0)=3f'(0)\\=3\cdot5\\=15

Llana [10]2 years ago
7 0

Answer:

\displaystyle F'(0) = 15

Step-by-step explanation:

We are given that:

f'(0) = 5 \text{ and } F(x) = f(3x)

And we want to find F'(0).

First, find F(x):

\displaystyle F'(x) = \frac{d}{dx}\left[ f(3x)]

From the chain rule:

\displaystyle \begin{aligned} F'(x) &= f'(3x) \cdot \frac{d}{dx} \left[ 3x\right] \\ \\ &= 3f'(3x)\end{aligned}

Then:

\displaystyle \begin{aligned} F'(0) & = 3f'(3(0)) \\ \\ & = 3f'(0) \\ \\ & = 3(5) \\ \\ & = 15\end{aligned}

In conclusion, F'(0) = 15.

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Answer:

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Step-by-step explanation:

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Solve the Law of Cosine: c^2 = a^2+ b^2 - 2abcosC for cos C.
Andrew [12]

Answer:

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Step-by-step explanation:

See the figure to understand the proof :

Let A Triangle ABC with sides a,b,c,

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Divide base length b as AD = x -b   and    CD = x

By Pythagoras Theorem

In Triangle BDC             And     In Triangle BDA

a² = h² + x²     (  1  )                        c² = h² + (x-b)²

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From above eq 1 and 2

c² = (a² - x²) + x² + b² - 2xb

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cos C = \frac{BD}{BC}

Or, cos C = \frac{x}{a}

∴ x= a cos C

Now put ht value of x in eq 3

I.e, c² = a² + b² - 2ab cos C

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Answer:

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x^2 = 13

x = √13.

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