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4 years ago

10 pts. Write yes or no for which is rational or irrational. Thank you!

Mathematics

1 answer:

Sophie [7]4 years ago

5 0

1) rational

2)rational

3) irrational

4)irrational

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Please help 10 points

jok3333 [9.3K]

Answer:

Step-by-step explanation:

Let's first find the slope of the linear function represented by the table:

As we move from x = 4 to x = 6 *an increase of 2), y increases by 6 from -2 to +4. Thus, the slope of this line is m = rise / run = m = 6/2, or 3.

Now we compare the two slopes: -2 and 3.

The postive difference between them is 5. This is Answer B.

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4 years ago

The 4 in 400,000 has a value of 400 yes or no

Oduvanchick [21]

I'm pretty sure its not.

8 0

4 years ago

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A line has a rise of 6 and a slope of 1/20 . What is the run?

Lerok [7]

Answer:

120

Step-by-step explanation:

$\text{ Slope }=\frac{\text{ Rise }}{\text{ Run }}$ .

Plug in the information given:

$\frac{1}{20}=\frac{6}{\text{ Run }}$ .

Now we want to solve this equation for Run.

If you don't like Run as a variable, let's just use x in it's place:

$\frac{1}{20}=\frac{6}{x}$

Cross multiply:

$1(x)=20(6)$

Simplify both sides:

$x=120$

The run is 120.

Let's check our work.

$\text{ Slope }=\frac{\text{ Rise }}{\text{ Run }}$ .

So if our rise is 6 and our run is 120.

Does 6/120 reduce to 1/20?

6 and 120 do have a common factor of 6.

So if you divide top and bottom of 6/120 by 6 you do get 1/20.

8 0

3 years ago

Read 2 more answers

"assume that s is a non-empty set of real numbers which is bounded above and lambda is the least upper bound. Prove that for all

s2008m [1.1K]

By definition, if $\lambda$ is the least upper bound of the set $S$ , it means two thing:

$\forall x \in S,\ x \leq \lambda$
$\forall \varepsilon>0,\ \exists x \in S:\ x>\lambda-\varepsilon$

In other words, the least upper bound of a set is greater than or equal to every single element of the set, but it is "close enough" to the elements of the set, because you guaranteed to find elements in the set between $\lambda-\varepsilon$ and $\lambda$

For example, pick $S = [1,10)$ . Obvisouly, the least upper bound is $\lambda = 10$ . In fact, every number in $[1,10)$ is smaller than 10, but as soon as you take away something from 10, say 0.01, you get 9.99, and there are elements in $S$ greater than 9.99, say 9.9999.

So, the claim is basically proven by definition: if $\beta < \lambda$ , let $0 < \delta = \lambda - \beta$ . By definition, there exists $\alpha \in S:\ \alpha > \lambda - \delta$ .