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atroni [7]
3 years ago
7

Hello I need help with this algebra problem. 23X negative 4=65​

Mathematics
1 answer:
Dmitrij [34]3 years ago
4 0

Answer:

If it 23x - 4=65​ then it x=3, but if it 23x(-4)=65​ then it x=−0.707

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Can someone plz answer this question
nata0808 [166]
So you basically add all the numbers together multiply them by 100& divide by 54
6 0
3 years ago
Read 2 more answers
Show all worked identify the asymptotes and state the end behavior of the function F(x)=6x over x-36
astraxan [27]

Solution

Asymptote:

Vertical Asymptote

- The vertical asymptotes of a rational function are determined by the denominator expression.

- The expression given is:

f(x)=\frac{6x}{x-36}

- The denominator of (x- 36) determines the asymptote line.

- The vertical asymptote defines where the rational function isundefined. Iin order for a rational function to be undefined, its denominator must be zero.

- Thus, we can say:

\begin{gathered} x-36=0 \\ Add\text{ 36 to both sides} \\  \\ \therefore x=36 \end{gathered}

- Thus, the vertical asymptote is

x=36

Horizontal Asymptote:

- The horizontal asymptote exists in two cases:

1. When the highest degree of the numerator is less han the degree of the demnominator. In this case, the horizontal asymptote is y = 0

2. When the highest degee sof the numerator and tdenominator are the same. In this case, the horizontal asymptote is

\begin{gathered} y=\frac{N}{D} \\ where, \\ N=\text{ Coefficient of the highest degree of the numerator} \\ D=\text{ Coefficient of the highest degree of the denominator} \end{gathered}

- For our question, we can see that the highest degrees of the numerator and denominator are the same. Thus, we have the Horizontal Asymptote to be:

y=\frac{6}{1}=6

End behavior:

- The end behavior is examining the y-values of the function as x tendsto negative and positive infinity.

- Thus, we have that:

\begin{gathered} f(x)=\frac{6x}{x-36} \\  \\ \text{ Divide top and bottom by }x \\ f(x)=\frac{6x}{x-36}\times\frac{x}{x} \\  \\ f(x)=\frac{\frac{6x}{x}}{\frac{x-36}{x}}=\frac{6}{1-\frac{36}{x}} \\  \\ As\text{ }x\to-\infty \\ f(-\infty)=\frac{6}{1-\frac{36}{-\infty}}=\frac{6}{1+\frac{36}{\infty}}=\frac{6}{1+0}=6 \\  \\ \text{ Thus, we can say: }x\to-\infty,f(x)\to6 \\  \\ Also, \\ As\text{  }x\to\infty \\ f(\infty)=\frac{6}{1-\frac{36}{\infty}}=\frac{6}{1-0}=6 \\  \\ \text{ Thus, we can also say: }x\to\infty,f(x)\to6 \end{gathered}

Final Answers

Asymptotes:

\begin{gathered} \text{ Vertical:} \\ x=36 \\  \\ \text{ Horizontal:} \\ y=6 \end{gathered}

End behavior:

\begin{gathered} As\text{  }x\to-\infty,f(x)\to6 \\  \\ As\text{  }x\to\infty,f(x)\to6 \end{gathered}

7 0
1 year ago
Jake opens a credit card with 18% annual interest compounded monthly. If he charges $800 on the card and never pays it off, how
Makovka662 [10]
<h3>Answer:  $1,143.60</h3>

Work Shown:

A = P*(1+r/n)^(n*t)

A = 800*(1+0.18/12)^(12*2)

A = 1,143.60224954321

A = 1,143.60

Notes:

  • P = 800 is the principal. It is the amount loaned to Jake.
  • r = 0.18 is the decimal form of the 18% interest rate
  • n = 12 because we're compounding monthly, i.e. 12 times a year
  • t = 2 = number of years
  • For more information, check out the compound interest formula.

4 0
2 years ago
Is this correct and if not what am I missing?
olga2289 [7]
It would be B. because the others aren’t correct
6 0
4 years ago
Read 2 more answers
2 |d+5| - 1 &lt; 3 please show the work
natka813 [3]
<span>2 |d+5| - 1 < 3  

Start by adding 1 to  both sides...

2 | d + 5 | <  4 

Divide both sides by 2 

|  d +  5 |  <   2 

Now make to sections...

d + 5 <  2  or   d +  5 >   - 2
    
Subtract both sides 5 


d  <  - 3  or  d >  - 7 is the answer

Or you can write is as - 7 < d < - 3</span>
4 0
4 years ago
Read 2 more answers
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