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Katyanochek1 [597]
3 years ago
5

Need an answer show work please thank you

Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
7 0

Answer:

\text{C. }1

Step-by-step explanation:

In the question, we're given that the notation \#\#(a,b,c) produces a number a less than the product of b and c raised to the a power. Let the number produced be n. As a mathematical equation, we can write this production as n=(bc)^a-a

For \#\#(2, 5, x), we can assign:

  • a\implies 2
  • b\implies 5
  • c\implies x

Substituting these values into n=(bc)^a-a, we get:

23=(5x)^2-2

Add 2 to both sides:

25=(5x)^2

Take the square root of both sides:

5=|5x|

For y=|z|, there are two cases:

\begin{cases}y=z,\\y=-z\end{cases}

Therefore, we have:

\begin{cases}5=5x, x=\boxed{1}\\5=-(5x), 5=-5x, x=\boxed{-1}}\end{cases}

The only answer choice applicable is \boxed{\text{C. }1}.

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Step-by-step explanation:

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Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = x3 − 6x2 − 15x + 2 (a) Find the interval on which
Burka [1]

Answer:

a) (-\infty, -1) \cup (5, \infty)

b) (-1,5)

Step-by-step explanation:

The first step to solve this question is finding the roots of the derivative of x.

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

ax^{2} + bx + c, a\neq0.

This polynomial has roots x_{1}, x_{2} such that ax^{2} + bx + c = a(x - x_{1})*(x - x_{2}), given by the following formulas:

x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}

x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}

\bigtriangleup = b^{2} - 4ac

In this question:

f(x) = x^{3} - 6x^{2} - 15x + 2

So

f'(x) = 3x^{2} - 12x - 15

Finding the roots:

3x^{2} - 12x - 15 = 0

Simplifying by -3

x^{2} - 4x - 5 = 0

So a = 1, b = -4, c = -5

Then

\bigtriangleup = (-4)^{2} - 4*1*(-5) = 36

x_{1} = \frac{-(-4) + \sqrt{36}}{2} = 5

x_{2} = \frac{-(-4) - \sqrt{36}}{2} = -1

So the function can be divided in three intervals.

They are:

Less than -1

Between -1 and 5

Higher than 5

In which it increases and which it decreases?

Less than -1

Lets find the derivative in a point in this interval, for example, -2

f'(x) = 3x^{2} - 12x - 15

f'(-2) = 3*(-2)^{2} - 12*(-2) - 15 = 21

Positive.

So in the interval of (-\infty, -1), the function increases.

Between -1 and 5

Will choose 0.

f'(x) = 3x^{2} - 12x - 15

f'(0) = 3*(0)^{2} - 12*(0) - 15 = -15

Negative.

So in the interval of (-1,5), the function decreases.

Higher than 5

Will choose 6.

f'(x) = 3x^{2} - 12x - 15

f'(6) = 3*(6)^{2} - 12*(6) - 15 = 21

Positive

So in the interval of (5, \infty), the function increases.

(a) Find the interval on which f is increasing.

Using interval notation

(-\infty, -1) \cup (5, \infty)

b) Find the interval on which f is decreasing.

(-1,5)

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4 years ago
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Answer:

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Step-by-step explanation:

|x-4|<9 ➡ -9 < x-4 < 9

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Answer:

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