If we are talking about whole numbers, the only number in between 1 and 3 is 2 but if we are talking about numbers in general, there are an infinite amount of numbers (for example, 1.1, 1.12, 1.004, and 1.80543285 are all in between 1 and 3).

5 0

3 years ago

What is the quotient (3x^3 10x^2 10x 4) divided by (x 2)?

lozanna [386]

Answer:

3x^2 + 16x + 42

3x^2 + 4x + 2

3x^2 – 16x + 42

3x^2 – 4x + 2

Step-by-step explanation:

3 0

3 years ago

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>

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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