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stich3 [128]
4 years ago
9

How resolve this equation 2x^2-3x-y=-5 -x+y=5

Mathematics
1 answer:
Lelechka [254]4 years ago
7 0
-x+y=5\Rightarrow y=x+5

Substitute y in the other equation:
2x^{2}-3x-(x+5)=-5
\Rightarrow 2x^{2}-3x-x-5=-5
\Rightarrow 2x^{2}-4x-5=-5
\Rightarrow 2x^{2}-4x=0
\Rightarrow 2x(x-2)=0

So either
2x=0\Rightarrow x=0\Rightarrow y=0+5=5
or
x-2=0\Rightarrow x=2\Rightarrow y=2+5=7
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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

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Answer:

I hope this helps 40 divided by 4 is 10 10 per person 10-2=8 8 coins in each players bag and the total amount of coins in bags is 8 times 4 which is 32

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What is 5.7916 rounded to the nearest thousandth?​
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Answer:

5,792

Step-by-step explanation:

I'm pretty sure tell me if I'm wrong

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Miguel hiked 3miles in 54minutes. At this rate, how long will it take him to hike 5miles
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Answer:

90 minutes

Step-by-step explanation:

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