Describe the pattern. Draw what might be the next three figures in the pattern

Mathematics

1 answer:

Dennis_Churaev [7]3 years ago

6 0

You add another row and the row has to be one taller than the row before it

You might be interested in

Write each percent as a fraction in simplest form 2 2/5%

vovikov84 [41]

Answer:

12/5%

Step-by-step explanation:

12/5 %

that's it cuz u can't divide them

7 0

3 years ago

Eliza falls asleep at 9:35 pm and wakes up 8 hours and 45 miunets later, what time does she wake uo

alexdok [17]

Answer: Eliza wakes up at 6:25 A.M.

Step-by-step explanation:

What I did is first used the 45 minutes to make the time 10:00 P.M. which leaves you with 8 hours and 25 minutes, then i added the 8 hours onto 10:00 P.M. which leaves you at 6:00 A.M. with 25 minutes left, then you add the 25 minutes onto 6:00 A.M. which puts it at 6:25 A.M.

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 $x$