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dexar [7]
4 years ago
6

Drag the dots to plot (7 1/2, -2) (-3,7) and (2,0)

Mathematics
1 answer:
kirill115 [55]4 years ago
4 0

Step-by-step explanation:

7 1/2 = 7 + 1/2 = 14/2 + 1/2 = 15/2 = 7.5

check the attached picture.

:)

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What is the solution to this system of linear equations? X-3y=-2 and x-3y=16
Katen [24]

Answer:

No solution

Step-by-step explanation:

Equations:

x - 3y = -2

x - 3y = 16

Subtract:

0 - 0 = -18

Answer:

No solution

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3 years ago
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lord [1]
The answer is A, c, e. It is since the one first listed is always on the x- axis and the second is always the y-axis. So u just apply that and u get a, c, and e. The answer is A.C.E.
6 0
4 years ago
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What is the slope of the line that passes through the points (1, 1) and (9,7)?
Zarrin [17]
The answer is 3/4.

Step-By-Step Explanation

Slope Formula:m=y2-y1/x2-x1
Substitute: 7-1/9-1
Answer: 3/4
5 0
3 years ago
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
6(40 - 2b) – 2195-70)
Naddika [18.5K]

Answer:

B= -6,255

Step-by-step explanation:

7 0
3 years ago
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