Ask question

Ask question

All categories

2 years ago

14

Which function is increasing and has a vertical asymptote at x = 5?

1 answer:

marusya05 [52]2 years ago

3 0

C. f(x)= ln(x-5) hope that helps

You might be interested in

What is the answer to this problem (x+3)(x+5)

klasskru [66]

Easy, showing work,

using the method FOIL (First,Outer,Inner, Last)

Take, (x+3)(x+5) and multiply,

First,

x*x= $x^{2}$

Outer,

x*5=5x

Inner,

3*x=3x

Last,

3*5=15

Add the terms together,

$x^{2}$ +8x+15 ( $x^{2}$ +5x+3x+15)

Thus, your answer.

4 0

3 years ago

Read 2 more answers

If the answer to the square root is 5, then what is the number that gets you 5

tensa zangetsu [6.8K]

$\sqrt x=5\\
x=25$

4 0

3 years ago

Plz plz answer this question <br> NO LINKS :)

masya89 [10]

Answer:

did you forget a picture?

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Burn is to ointment as grief is to

Nitella [24]

Answer:The answer is Sympathy

Step-by-step explanation:

4 0

2 years ago

Use quadratic formula to find both solutions to the quadratic equation given below . 3x^2-5x-1=0

Rudiy27

Answer:

$x1=\frac{5+\sqrt{37}} {6}$

$x2=\frac{5-\sqrt{37}} {6}$

Step-by-step explanation:

we know that

The formula to solve a quadratic equation of the form $ax^{2} +bx+c=0$ is equal to

$x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$3x^{2} -5x-1=0$

so

$a=3\\b=-5\\c=-1$

substitute in the formula

$x=\frac{5(+/-)\sqrt{(-5)^{2}-4(3)(-1)}} {2(3)}$

$x=\frac{5(+/-)\sqrt{37}} {6}$

$x1=\frac{5+\sqrt{37}} {6}$

$x2=\frac{5-\sqrt{37}} {6}$

8 0

2 years ago

Read 2 more answers