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2 years ago

Which number is irrational? OA. 0.45 OB. 0.636363... O C. √25 OD. √6

Mathematics

1 answer:

pashok25 [27]2 years ago

7 0

Answer:

D. Sqroot6

Step-by-step explanation:

6 is not a perfect square. So it is a non-repeating decimal that never ends

sqroot6 ~=

2.4494897428...

this is an irrational number.

The rest on the numbers can be written as a ratio:

0.45=45/100=9/20

0.63636363

= 63/99 = 7/11

sqrt25 = 5 = 5/1

that means they are rational.

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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:

$f''(x)=\frac{10}{x^3}-\frac{24}{x^4}$

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

$0=10x-24$

$-10x=-24$

$x=\frac{24}{10}$

$x=\frac{12}{5}$

We then test where $f''(x)$ is negative or positive by plugging in test values. I will use -1 and 3 to test this:

$f''(-1)=\frac{10}{(-1)^3}-\frac{24}{(-1)^4}=-34$ , 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 $(\frac{12}{5},\infty)$

The inflection point is where concavity changes, which can be determined by plugging in $x=\frac{12}{5}$ into the original function, which would be $f(\frac{12}{5})=1+\frac{5}{\frac{12}{5}}+\frac{4}{\frac{12}{5}^2 }=\frac{43}{18}$ , or $(\frac{12}{5},\frac{43}{18})$ .

E) See attached graph

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3 years ago

Steven combines 7.49 liters of red paint with 6.26 liters of blue paint to make purple paint. He pours the paint equally into 4

kondaur [170]

Answer:

3.3075 liters of paint

Step-by-step explanation:

7.40 (red paint) + 6.26 (blue paint ) = 13.66 (purple paint)

He pours 13.66 liters of purple paint equally into 4 containers and has 0.43 liters left.

x + x + x + x = 13.66 - 0.43

4x = 13.23

x = 13.23/4

x = 3.3075

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3 years ago

Find the smallest natural number that is a multiple of 45 and where all of its digits are zeros and ones.

Sergio039 [100]

Answer:

1111111110

Step-by-step explanation:

45 is 9, and 5.

9 divisibility rule: Digits have to add up to 9.

We know this has to be 1's and 0's.

So there has to be 9 1's.

111111111.

5 divisibility rule: Last digit has to be 0 or 5.

1111111110 will be our final answer.

Wait- let's check if it's divisible. Using a calculator, this determines it does work.

Enjoy.

6 0

3 years ago