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pogonyaev
3 years ago
14

Ratio question I will mark brainliest

Mathematics
1 answer:
goldfiish [28.3K]3 years ago
6 0

Answer:

27

Step-by-step explanation:

3 : 8 cannot be simplified into integers

neither can 6 : 11

because decimals are not integers

so 3 + 8 + 6 + 11 = 27

Im not sure why there are two b's so I added both of them

if you dont need both b's then the answer is 25

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Find the intercepts and the vertical and horizontal asymptotes, and then use them to sketch a graph of the function. f(x)=x+2/x^
hoa [83]

We have the function:

f(x)=x+\frac{2}{x^2}-16.

We must find:

0. the intercepts,

,

1. the vertical and horizontal asymptotes.

1) x-intercepts

The x-intercepts are given by the x values such that f(x) = 0. So we must find the values of x that satisfies the equation:

f(x)=x+\frac{2}{x^2}-16=0.

Solving for x, we get:

\begin{gathered} x+\frac{2}{x^2}-16=0 \\ x\cdot x^2+2-16\cdot x^2,\text{ }x\ne0, \\ x^3-16x^2+2=0. \end{gathered}

The real roots of this equation are:

\begin{gathered} x_1\approx15.9922, \\ x_2\approx0.35757, \\ x_3\approx-0.34975. \end{gathered}

So the x-intercepts are the points:

\begin{gathered} P_1=(15.9922,0), \\ P_2=(0.35757,0), \\ P_3=(-0.34975,0)\text{.} \end{gathered}

2) y-intercepts

The y-intercepts are given by the y values such that x = 0. Replacing x = 0 in the definition f(x), we see that the denominator of the second term diverges. So we conclude that there are no y-intercepts.

3) Vertical asymptotes

Vertical asymptotes are vertical lines near which the function grows without bound. From point 2, we know that the function grows without limit when x goes to zero. So one vertical asymptote is:

x=0.

4) Horizontal asymptotes

Horizontal asymptotes are horizontal lines that the graph of the function approaches when x → ±∞. We consider the limit of the function f(x) when x → ±∞:

\lim _{x\rightarrow\pm\infty}f(x)=\lim _{x\rightarrow\pm\infty}(x+\frac{2}{x^2}-16)\rightarrow\pm\infty.

We see that the function does not tend to any constant value when x → ±∞. So we conclude that there are no horizontal asymptotes.

5) Oblique asymptotes

When a linear asymptote is not parallel to the x- or y-axis, it is called an oblique asymptote or slant asymptote.

A function ƒ(x) is asymptotic to the straight line y = mx + n (m ≠ 0) if

{\displaystyle\lim _{x\to+\infty}\mleft[f(x)-(mx+n)\mright]=0\, {\mbox{ or }}\lim _{x\to-\infty}\mleft[f(x)-(mx+n)\mright]=0.}

We consider the line given by:

y=mx+n=x-16.

We compute the limit:

\begin{gathered} \lim _{x\rightarrow\pm\infty}(f(x)-(x-16)) \\ =\lim _{x\rightarrow\pm\infty}((x+\frac{2}{x^2}-16)-(x-16)) \\ =\lim _{x\rightarrow\pm\infty}(\frac{2}{x^2}) \\ =0. \end{gathered}

So we have proven that f(x) has the oblique asymptote:

y=x-16.

6) Graph

Plotting the intercepts and the asymptotes, we get the following graph:

Answer

1) x-intercepts: (-0.34975, 0), (0.35757, 0), (15.9922, 0)

2) y-intercepts: none

3) Vertical asymptotes: x = 0

4) Horizontal asymptotes: none

5) Oblique asympsotes: y = x -16

6) Graph

6 0
1 year ago
Parabola with vertex at (4, -2) and passes through<br> the point (2, 10)
amid [387]

Answer:

x2-x1/y2-y1=2-4/10--2=2-4/10+2=-2/12

=-6

6 0
4 years ago
Solve the system of linear equations by substitution.<br> 3x + y = -9<br> y=5x+7
11Alexandr11 [23.1K]

Answer:

The solution is x = -2, y = -3

Step-by-step explanation:

Let's start with the equation 3x + y = -9

Substitute the y for 5x+7 seen in the second equation

Now you have 3x + (5x+7) = -9

Lets solve:

Let's add the x's

Now we have 8x + 7 = -9

Substracy seven from each side

Now we have 8x = -16

Now let's divide 8 on each side

Now we have x = -2

Now lets plug in -2 to the second equation

The second equation is y = 5x+7

When we plug in -2 we get y=5(-2)+7

We multiply

We now have y = -10+7

Add

y = -3

The solution is x = -2, y = -3

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