All categories

2 years ago

Six times a number ia greater than 20 more than that number. What are possible values of that number?

Mathematics

1 answer:

VMariaS [17]2 years ago

3 0

Answer:

6x<x+20 , any number less than 20 more than the unknown value.

You might be interested in

Classify the number as rational or irrational.

lara [203]

- Rational
-Irrational
-Rational
-Rational

Just took the test

6 0

3 years ago

Read 2 more answers

Converges or Diverges: Please help me

bagirrra123 [75]

Use a/ 1-r

Where a = 18 and r = 1.2

18/ 1 - 1.2 = 18/-.2 = -90

Because the answer is below 1. The sum does not exist therefore it diverges.

8 0

3 years ago

Write the equation of the lines resented below in slope-intercept form

scoray [572]

Slope intercept form: y = mx + b

mx = slope

b = y-intercept

We know the y intercept is 0, so nothing will be written there.

To find the slope of this line, we can use the slope formula.

$\frac{y.2 - y.1}{x.2 - x.1} = m$

We'll use the points (1, 0) and (3, 1) to find the slope.

Now we can just plug these values into the equation to find the slope.

1 - 0 / 3 - 1

1 / 2

The slope of the line is 1/2, or 0.5.

The slope-intercept form of this line can be written as:

y = 0.5x

6 0

3 years ago

What are the points of discontinutity y=(x-3)/x^2-12x+27

madreJ [45]

Answer:

$(3, -\frac{1}{6})$

Step-by-step explanation:

We can rewrite the equation as

$y = \frac{x - 3}{(x - 3)(x - 9)}$

Notice that we have $x - 3$ in both the numerator and the denominator, so it looks like we can divide it out. However, what if $x - 3$ is $0$ ? Then we would have $y = \frac{0}{0 \times (x - 9)} = \frac{0}{0}$ , which is undefined. So although it looks like the numerator and denominator can be simplified, the resulting function we would get from simplification would not have the same behavior as this one (since such a function would be defined for $x = 3$ , but this one is not).

A point of discontinuity refers to a particular point which is included in the simplified function, but which is not included in the original one. In this case, the point which is not included in the unsimplified function is at $x = 3$ . In the simplified version of the function, if we plug in $x = 3$ , we get

$y = \frac{1}{((3) - 9)} = -\frac{1}{6}$

So the point $(3, -\frac{1}{6})$ is our only point of discontinuity.

It's also important to distinguish between specific points of discontinuity and vertical asymptotes. This function also has a vertical asymptote at $x = 9$ (since it causes the denominator to be 0), but the difference in behavior is that in the case of the asymptote, only the denominator becomes 0 for a specific value of $x$