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Kay [80]
3 years ago
12

How many distinct and real roots can an nth-degree polynomial have?

Mathematics
1 answer:
Vlad [161]3 years ago
8 0
The fundamental theorem of algegra is often cited to say that an n-th degree polynomial has n roots. They may not always be distinct, and they may not always be real.

An n-th degree polynomial may have up to n distinct real roots.
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What negative angle is equivalent to a 75 angle
eimsori [14]

Answer:

-75

Step-by-step explanation:

4 0
3 years ago
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Find all values of c in the open interval (a, b) such that f'(c)=(f(b)-f(a))/(b-a)
timama [110]
<h3>Answer:   c = 7/4</h3>

================================================

Work Shown:

Compute the function value at the endpoints

f(x) = \sqrt{4-x}\\\\f(-5) = \sqrt{4-(-5)} = 3\\\\f(4) = \sqrt{4-4} = 0\\\\

With a = -5 and b = 4, we have

f'(c) = \frac{f(b)-f(a)}{b-a}\\\\f'(c) = \frac{f(4)-f(-5)}{4-(-5)}\\\\f'(c) = \frac{0-3}{9}\\\\f'(c) = -\frac{1}{3}\\\\

So,

f(x) = \sqrt{4-x}\\\\f'(x) = -\frac{1}{2\sqrt{4-x}}\\\\f'(c) = -\frac{1}{3}\\\\-\frac{1}{2\sqrt{4-c}} = -\frac{1}{3}\\\\

Use algebra to solve for c

-\frac{1}{2\sqrt{4-c}} = -\frac{1}{3}\\\\\frac{1}{2\sqrt{4-c}} = \frac{1}{3}\\\\3 = 2\sqrt{4-c}\\\\2\sqrt{4-c} = 3\\\\\sqrt{4-c} = \frac{3}{2}\\\\4-c = \frac{9}{4}\\\\c = 4-\frac{9}{4}\\\\c = \frac{16-9}{4}\\\\c = \frac{7}{4}\\\\

6 0
3 years ago
What's the angle of y and x ??
victus00 [196]
Y is going to be the number 40
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6 0
3 years ago
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Step-by-step explanation:

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4 0
3 years ago
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yaroslaw [1]

Answer:

r = 15

Step-by-step explanation:

Volume of a sphere is given by

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Divide each side by pi

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Multiply each side by 3/4

3/4 *4500  = 3/4 *4/3  r^3

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Take the cube root of each side

(3375)^(1/3) = (r^3)^(1/3)

15 = r

7 0
2 years ago
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