All categories

4 years ago

Can someone answer this for me

Mathematics

1 answer:

Alexxandr [17]4 years ago

7 0

2. For this, you find -1 on the number line and jump backward three whole numbers.

3. For 2+5, you go forward five spaces. However, with -2+-5, you have to go backward five spaces because of the keep change change rule. -2+-5 is practically subtraction.

4. -2+-5=-7 so the sum of -1+-6 is the same.

You might be interested in

how much interest is earned on a principal of $575 invested at an interest rate of 7% compounded annually for nine years?

kogti [31]

Answer:

$482.11 in interest

Step-by-step explanation:

Use the compound amount formula. Calculate the compound amount and then subtract the principle to find the amount of interest earned.

A = P(1 + r)^t, where r is the interest rate as a decimal fraction

This becomes:

A = $575(1 + 0.07)^9, or A = $1057.11.

Subtracting the principal: A - P: $1057.11 - $575 = $402.11

3 0

3 years ago

Read 2 more answers

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: