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Mariana [72]
4 years ago
13

Which is the domain of the function f(x) = -5/6(3/5)^x?

Mathematics
1 answer:
lutik1710 [3]4 years ago
7 0

the domain of f is {\bf R} = (-\infty, \infty)

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If Carbon Monoxide had a concentration in the air of 0.000518% how many particles out of 331,002,651 would be Carbon Monoxide?
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Answer:

171,459 particles of CO

Step-by-step explanation:

331,002,651×0.000518 = 171459

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I believe it is B. If you were to solve, you would get x=4i, x=-4i. Those are not real numbers, so if I were you I'd go with B, but you should get a second opinion.
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Korolek [52]

Answer:3/8

Step-by-step explanation:

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1. Determine the intervals on which each function is increasing or
motikmotik

Answer:

\text{$f(x)$ is increasing for $(-\infty, -\frac{1}{3})$ and $(4, \infty)$} \\ \text{$f(x)$ is decreasing for $(-\frac{1}{3}, 4)$}

Step-by-step explanation:

We have the function:

f(x)=x^3-\frac{11}{2}x^2-4x

And we want to determine the intervals for which the function is increasing or decreasing.

We will need to first find the critical points of the function. Note that our original function is continuous across the entire x-axis.

To find the critical points, we need to find the first derivative and then solve for the <em>x. </em>So:

f^\prime(x)=\frac{d}{dx}[x^3-\frac{11}{2}x^2-4x]

Differentiate:

f^\prime(x)=(3x^2)-\frac{11}{2}(2x)-(4) \\ f^\prime(x)=3x^2-11x-4

Set the derivative equal to 0 and solve for <em>x: </em>

<em />0=3x^2-11x-4<em />

Factor:

0=3x^2-12x+x-4 \\ 0=3x(x-4)+(x-4) \\ 0=(3x+1)(x-4)

Zero Product Property:

3x+1=0\text{ or } x-4=0 \\ x=-\frac{1}{3}\text{ or } x=4

Therefore, our critical points are -1/3 and 4.

We can thus sketch the following number line:

<————-(-1/3)——————————————(4)—————>

Now, let’s test values for the three intervals: less than -1/3, between -1/3 and 4, and greater than 4.

For less then -1/3, we can use -1. So, substitute -1 for <em>x</em> for our first derivative and see which we get:

f^\prime(-1)=3(-1)^2-11(-1)-4=3+11-4=10>0

The result is <em>positive</em>.

Therefore, for all numbers less than -1/3,<em>f(x)</em> is increasing (since its derivative is positive).

For between -1/3 and 4, we can use 0. Substitute 0 for our derivative:

f^\prime(0)=3(0)^2-11(0)-4=-4

The result is <em>negative</em>.

Therefore, for all numbers between -1/3 and 4, <em>f(x) </em>is decreasing (since its derivative is negative).

And finally, for greater than 4, we can use 5:

f^\prime(5)=3(5)^2-11(5)-4=16>0

The result is positive.

Therefore, for all numbers greater than 4, <em>f(x)</em> is increasing (since its derivative is positive.

Therefore:

\text{$f(x)$ is increasing for $x4$}\\ \text{ $f(x)$ is decreasing for $-\frac{1}{3}

In interval notation:

\text{$f(x)$ is increasing for $(-\infty, -\frac{1}{3})$ and $(4, \infty)$} \\ \text{$f(x)$ is decreasing for $(-\frac{1}{3}, 4)$}

4 0
3 years ago
G^5/G^5 <br><br> Please help!!
aev [14]

Answer:

1

Step-by-step explanation:

If a thing is divided by itself then it equals 1.

7 0
4 years ago
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