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3 years ago

Quick Please :) Exponents and powers

Mathematics

2 answers:

MrRissso [65]3 years ago

6 0

Step-by-step explanation:

$( {4}^{ - 3} ) ^{ - 3} = {4}^{9}$

$( { - 4}^{9} ) ^{2} = ( - 4) {}^{18}$

$( {4}^{6} ) ^{ - 3} = \frac{1}{{4}^{18}}$

$( { - 4}^{9} ) {}^{2} = ( { - 4})^{18}$

$({4}^{0} ) = {4}^{0}$

Hope this is correct and helpful

HAVE A GOOD DAY!

n200080 [17]3 years ago

3 0

It would be 4 and a little 0 on the top

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4 years ago

A plane traveled 2565 miles with the wind in 4.5 hours and 2205 miles against the wind in the same amount of time. Find the spee

kherson [118]

Answer:

plane: 530 mi/h
wind: 40 mi/h

Step-by-step explanation:

Let p and w represent the speeds of the plane and the wind. The relation between time, speed, and distance is ...

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p +w = (2565 mi)/(4.5 h) = 570 mi/h

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3 years ago

What is the solution of log3x − 5 16 = 2?

Stolb23 [73]

Answer:

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Step-by-step explanation:

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3 years ago

Consider the following function.

Kryger [21]

Answer:

See below

Step-by-step explanation:

I assume the function is $f(x)=1+\frac{5}{x}-\frac{4}{x^2}$

A) The vertical asymptotes are located where the denominator is equal to 0. Therefore, $x=0$ is the only vertical asymptote.

B) Set the first derivative equal to 0 and solve:

$f(x)=1+\frac{5}{x}-\frac{4}{x^2}$

$f'(x)=-\frac{5}{x^2}+\frac{8}{x^3}$

$0=-\frac{5}{x^2}+\frac{8}{x^3}$

$0=-5x+8$

$5x=8$

$x=\frac{8}{5}$

Now we test where the function is increasing and decreasing on each side. I will use 2 and 1 to test this:

$f'(2)=-\frac{5}{2^2}+\frac{8}{2^3}=-\frac{5}{4}+\frac{8}{8}=-\frac{5}{4}+1=-\frac{1}{4}$

$f'(1)=-\frac{5}{1^2}+\frac{8}{1^3}=-\frac{5}{1}+\frac{8}{1}=-5+8=3$

Therefore, the function increases on the interval $(0,\frac{8}{5})$ and decreases on the interval $(-\infty,0),(\frac{8}{5},\infty)$ .

C) Since we determined that the slope is 0 when $x=\frac{8}{5}$ from the first derivative, plugging it into the original function tells us where the extrema are. Therefore, $f(\frac{8}{5})=1+\frac{5}{\frac{8}{5}}-\frac{4}{\frac{8}{5}^2 }=\frac{41}{16}$ , meaning there's an extreme at the point $(\frac{8}{5},\frac{41}{16})$ , but is it a maximum or minimum? To answer that, we will plug in $x=\frac{8}{5}$ into the second derivative which is $f''(x)=\frac{10}{x^3}-\frac{24}{x^4}$ . If $f''(x)>0$ , then it's a minimum. If $f''(x)$ , then it's a maximum. If $f''(x)=0$ , the test fails. So, $f''(\frac{8}{5})=\frac{10}{\frac{8}{5}^3}-\frac{24}{\frac{8}{5}^4}=-\frac{625}{512}$ , which means $(\frac{8}{5},\frac{41}{16})$ is a local maximum.

D) Now set the second derivative equal to 0 and solve:

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$0=\frac{10}{x^3}-\frac{24}{x^4}$

$0=10x-24$

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$f''(-1)=\frac{10}{(-1)^3}-\frac{24}{(-1)^4}=-34$ , so the function is concave down on the interval $(-\infty,0)\cup(0,\frac{12}{5})$

$f''(3)=\frac{10}{3^3}-\frac{24}{3^4}=\frac{2}{27}>0$ , so the function is concave up on the interval $(\frac{12}{5},\infty)$

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E) See attached graph

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Answer:

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Step-by-step explanation:

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3 years ago