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3 years ago

Find the product. simplify if possible. -1/3 times (8 1/7)

2 answers:

8 0

You could change the second fraction into an improper one making it 57/7. When you multiply fractions you multiply straight across so the product would be -57/21. This can be simplified to -19/7 or -2 5/7 if you want to write it as a mixed number. That makes the answer B.

Inessa05 [86]3 years ago

6 0

The answer is b -25/7

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Which of the binomials below is a factor of this trinomial? x2 + 5x + 6

lianna [129]

Factor out the trinomial. Find factors of x² and 6 that would, when added, give you 5x

x² = x * x

6 = 3 * 2

(x + 3)(x + 2) are your binomials.

To solve for x, set each parenthesis = 0

x + 3 = 0

x + 3 (-3) = 0 (-3)

x = -3

x + 2 = 0

x + 2 (-2) = 0 (-2)

x = -2

x = -3, -2

hope this helps

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3 years ago

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Artemon [7]

Answer:

I think its B.

Step-by-step explanation:

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3 years ago

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Answer:

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3 years ago

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zvonat [6]

Answer:

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2 years ago

10. Determine the open interval(s) on which the function f(x) = 12x - x^3 is increasing.

Gre4nikov [31]

Answer:

$\large\boxed{x\in(-2,\ 2)}$

Step-by-step explanation:

$f(x)=12x-x^3$

Calculate the derivative of the function:

$f'(x)=\left(12x-x^3\right)'=(12x)'-\left(x^3\right)'=12-3x^2$

Calculate the zeros of the derivative:

$f'(x)=0\Rightarrow12-3x^2=0\qquad\text{add}\ 3x^2\ \text{to both sides}\\\\3x^2=12\qquad\text{divide both sides by 3}\\\\x^2=4\to x=\pm\sqrt4\\\\x=\pm2$

Let's check the derivative sign:

$f'(x)>0\iff12-3x^2>0\qquad\text{use the distributive property}\\\\-3(x^2-4)>0\\\\-3(x^2-2^2)>0\qquad\text{use}\ a^2-b^2=(a-b)(a+b)\\\\-3(x-2)(x+2)>0\\^{\qquad x=2\qquad x=-2$

Sketch a parabola open down

(before the brackets there is the number -3 < 0)

Look at the picture.

Where the derivative is negative, the function decreases.

Where the derivative is positive, the function increases.

Therefore your answer is:

$x\in(-2,\ 2)$

3 0

3 years ago