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Katen [24]
2 years ago
6

Lots of points Please helpi add a photo ​

Mathematics
2 answers:
satela [25.4K]2 years ago
5 0
The answer is 6*p.
Thanks

valkas [14]2 years ago
3 0
C = 6*P

Use that formula and plug in the x-axis values for P and plot the results (C) on the graph
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Sebastian earned $250 dollars from helping his dad at the store. He has plans to spend one fifth of his earnings. How much money
Akimi4 [234]

Answer:

50

Step-by-step explanation:

you multiply 1/5 and 250/1 to get 250/5 then simplify to get 50

3 0
3 years ago
Read 2 more answers
I would like to give some people points and also pray for everyone to have a great life
Karo-lina-s [1.5K]
Thank u!!!!!! :) have a good day
3 0
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

5 0
3 years ago
Given the two points determine the equation of the line.
deff fn [24]
The answer is y= 2x + 1
6 0
3 years ago
The diameter of the base of the cone measures 8 units. The height measures 6 units
daser333 [38]

For this case we can find the volume and area of the cone:

V = \frac {\pi * r ^ 2 * h} {3}

Where:

V: It's the volume

A: It's the radius of the base

h: It's the height

We have to:

r = \frac {8} {2} = 4 \ units\\h = 6 \ units

Substituting:

V = \frac {\pi * 4 ^ 2 * 6} {3}\\V = \frac {96 * \pi} {3}\\V = 32 \pi \ units ^ 3

On the other hand, the area of the cone is given by:

A = \pi * r * g +  \pi * r ^ 2

Where:

A: It's the radio

g: It is the generator of the cone.

g = \sqrt {h ^ 2 + r ^ 2} = \sqrt {6 ^ 2 + 4 ^ 2} = \sqrt {36 + 16} = \sqrt {52} = 7.2

SW:

A = \pi * 4 * 7.2 + \pi * 4 ^ 2\\A = 28.8 \pi + 16 \pi\\A = 44.8 \pi \ units ^ 2

Answer:

V = 32 \pi \ units ^ 3\\A = 44.8 \pi \ units ^ 2

7 0
3 years ago
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