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Verdich [7]
3 years ago
7

How many vertical asymptotes exist for this function?

Mathematics
1 answer:
marin [14]3 years ago
5 0
Remember that to find the vertical asymptotes of rational functions, we just need to set the denominator equal to zero and solve for the variable, in this case x, so lets do it:
(x-4)(x+1)=0
x=4,x=-1

As you can see, our rational functions has 2 vertical asymptotes: x=4 and x=-1
You might be interested in
Solve the following: - 3 raised to 1 by 5 the whole raised to 4 (3^1/5)^4
Diano4ka-milaya [45]

Answer:

8.30256

Step-by-step explanation:

Step 1: Write out expression

((-3)^{\frac{1}{5} })^{4(3^{\frac{1}{5} })^4

Step 2: Use BPEMDAS to evaluate

(-1.24573)^{4(3^{\frac{1}{5} })^4

(-1.24573)^{4(1.24573)^4

(-1.24573)^{4(2.40822)

(-1.24573)^{9.6329}

= 8.30256

And we have our answer!

8 0
4 years ago
One hundred sixty-two guests attended a banquet. Three servers provided their beverages. The first server helped three times as
Rudiy27

Answer:

  • Below given.

Step-by-step explanation:

  • 162 total guests.
  • 3 servers

Let the number of guests the second server helped be x.

then the first server helped 3x

then the third server helped [(3x)*2] = 6x

Equation in total:

                               6x + 3x + x = 162

                                10x = 162

                                   x = 16.2

  • First server helped: 3 * 16.2 = 49 guests
  • Second server helped: 16 guests
  • Third server helped: 6x = 6(16.2) = 97 guests
6 0
3 years ago
Which equation has the solutions x = startfraction negative 3 plus-or-minus startroot 3 endroot i over 2 endfraction? 2x2 6x 9 =
Gennadij [26K]

The quadratic equation which matches given solution is x² + 3x + 3 = 0. Then the correct option is C.

<h3>What is a quadratic equation?</h3>

It is a polynomial that is equal to zero. Polynomial of variable power 2, 1, and 0 terms are there. Any equation having one term in which the power of the variable is a maximum of 2 then it is called a quadratic equation.

We know that the formula

\rm x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}

The solutions are given below.

\rm x = \dfrac{-3 \pm \sqrt3 \iota }{2}

Then

1)  2x² + 6x + 9 = 0, the zeroes of the equation will be

\rm x = \dfrac{-6\pm 6\iota}{4}\\\\x = \dfrac{-3\pm 3 \iota}{2}

2)  x² + 3x + 12 = 0, the zeroes of the equation will be

\rm x = \dfrac{-3\pm \sqrt{39}\iota}{2}

3)  x² + 3x + 3 = 0, the zeroes of the equation will be

\rm x = \dfrac{-3\pm \sqrt3\iota}{2}

4)  2x² + 6x + 3 = 0, the zeroes of the equation will be

\rm x = \dfrac{-6\pm 2\sqrt{3}}{4}\\\\x = \dfrac{-3\pm \sqrt3 }{2}

More about the quadratic equation link is given below.

brainly.com/question/2263981

4 0
2 years ago
Draw any circle and mark (a) its centre<br> (b) a radius<br> (c) a diameter<br> (d) a sector
Nonamiya [84]

Answer:

I hope it's helpful for you

4 0
3 years ago
Dakota is asked to find E. What is her error
Sever21 [200]

Answer:

There is not enough information

to find EF because you need to

know either the length of DF or

one of the other angle measures.

Step-by-step explanation:

7 0
2 years ago
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