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romanna [79]
3 years ago
14

Solve the equation by completing the square. Round to the nearest hundredth if necessary. x2 – 6x = 20

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
8 0

x^2 – 6x = 20

b=-6

to complete the square, add (b/2)^2 to each side

(b/2)^2 = (-6/2)^2=(-3)^2 =9

x^2 – 6x +9 = 20+9

(x+b/2) ^2 =29

(x-3)^2=29

take the square root of each side

x-3 = +- sqrt (29)

add 3 to each side

x = 3+- sqrt (29)

x = 3 + sqrt(29), x- sqrt(29)

Answer: -2.39, 8.39

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Yes this is the correct answer

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Let f(x) = (x − 3)−2. Find all values of c in (1, 7) such that f(7) − f(1) = f '(c)(7 − 1). (Enter your answers as a comma-separ
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Answer:

This contradicts the Mean Value Theorem since there exists a c on (1, 7) such that f '(c) = f(7) − f(1) (7 − 1) , but f is not continuous at x = 3

Step-by-step explanation:

The given function is

f(x)=(x-3)^{-2}

When we differentiate this function with respect to x, we get;

f'(x)=-\frac{2}{(x-3)^3}

We want to find all values of c in (1,7) such that f(7) − f(1) = f '(c)(7 − 1)

This implies that;

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(c-3)^3=63.15789

c-3=\sqrt[3]{63.15789}

c=3+\sqrt[3]{63.15789}

c=6.98

If this function satisfies the Mean Value Theorem, then f must be continuous on  [1,7] and differentiable on (1,7).

But f is not continuous at x=3, hence this hypothesis of the Mean Value Theorem is contradicted.

 

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