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3 years ago

Tell me exactly what to do I have no idea wut this is and I have no tips please help

Mathematics

1 answer:

timurjin [86]3 years ago

8 0

The independent variable is a constant that does not rely on anything. This will be $5 because she has to pay it every month regardless.

The dependent variable change and affect the outcome. This will be $1.50. She has to pay per movie. If she doesn't watch a movie she doesn't pay anything on top of the $5. If she watches 10 movies, she payed $15 on top of the $5.

The equation:

x = Number of movies.

y = price

y = 5 + 1.5x

Table:

y x

11 4

12.5 5

14 6

15.5 7

17 8

18.5 9

20 10

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The following numerically ordered sequence of numbers has a mean

Vladimir79 [104]

Median = middle number

Total ten numbers so
median =   (17 + b)/2
given median = 18
so
(17 + b)/ 2 = 18
17+b = 36
b = 36 - 17
b = 19

Now you have the data set
7, a , 12, 15 , 17, 19, 20, 22, 24 , 25

Mean =  (7 + a + 12+ 15 +17 + 19 + 20 + 22 +24  +25)/ 10
Given mean = 17

so
17 = (7 + a + 12+ 15 +17 + 19 + 20 + 22 +24  +25)/ 10
17 = (a + 161) / 10
a + 161 = 170
a = 170 - 161
a = 9

answer
a = 9
b = 19

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3 years ago

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True or false 798.7 divided by 1,000= 7.987

aleksandrvk [35]

False. 798.7 divided by 1000 is 0.7987 because 1000 has three zeros, so you move the decimal point three units/numbers to the left

False. 73.09 divided by 100 is 0.7309, you are not multiplying by 100

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3 years ago

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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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yarga [219]

3+6, I believe. (I have to explain it a little better, so, 9-6=3, which means that 3+6=9)

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3 years ago

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A direct variation function contains the points (–9, –3) and (–12, –4). Which equation represents the function?

Alinara [238K]

A direct variation equation is one that requires y varies directly as x and looks like this in equation form:

$\frac{y}{x} =k$

where k is the constant of variation. If we solve this for y, we have y = kx, which happens to be a linear function... a line. k here, then, serves as the slope. So what we are given as points on a direct variation function are actually points on a line. The equation for this requires that we find the slope and then rewrite the formula accordingly. First the slope:

$m(k)=\frac{-4-(-3)}{-12-(-9)}=\frac{-4+3}{-12+9}=\frac{1}{3}$

Now we need to write the equation by using one of the points' coordinates. I picked the first point that has an x coordinate of -9 and a y coordinate of -3. Fitting those into the slope-intercept form of a line,

$-3=\frac{1}{3}(-9)+b$

which simplifies to

-3 = -3 + b and b = 0. That means that the equation of direct variation is

$y=\frac{1}{3}x+0$ or just

$y=\frac{1}{3}x$

8 0

3 years ago

Tell me exactly what to do I have no idea wut this is and I have no tips please help​

Tell me exactly what to do I have no idea wut this is and I have no tips please help