3 years ago

I need help please . . . . i even upped the points . . .

Mathematics

2 answers:

Darina [25.2K]3 years ago

8 0

The answer is b hope this helps

larisa86 [58]3 years ago

4 0

Answer:

Step-by-step explanation:

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2 years ago

Given: x+2y=-6. Solve for y. y=x-6/2

Snowcat [4.5K]

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3 years ago

Help me, I don't understand this problem

Crank

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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>

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

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

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

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3 years ago

Which of the following shows 2x3y-3x2+7yx-12y written in standard form

AysviL [449]

2x^(3)y+7xy-3x^(2)-12y

is in standard form

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3 years ago