Either Table E or Table F shows a proportional relationship.

Mathematics

1 answer:

Alborosie2 years ago

7 0

your answer is correct

Step-by-step explanation:

because when you plot all the points they line up

You might be interested in

Which relation is a function

olya-2409 [2.1K]

Answer:

The relation that is a function is{ (-1,5), (-2,6), (-3,7) }. A function will not have any repeating x-values, they all have to be different.

3 0

2 years ago

Read 2 more answers

Explain how you can find the missing number and 3 4/5 divided by something equals 2 5/7 then find the missing number

enyata [817]

let's firstly, convert the mixed fractions to improper, and then do equation.

$\bf \stackrel{mixed}{3\frac{4}{5}}\implies \cfrac{3\cdot 5+4}{5}\implies \stackrel{improper}{\cfrac{19}{5}}
~\hfill
\stackrel{mixed}{2\frac{5}{7}}\implies \cfrac{2\cdot 7+5}{7}\implies \stackrel{improper}{\cfrac{19}{7}}
\\\\[-0.35em]
\rule{34em}{0.25pt}$

$\bf \cfrac{~~\frac{19}{5}~~}{x}=\cfrac{19}{7}\implies \cfrac{~~\frac{19}{5}~~}{\frac{x}{1}}=\cfrac{19}{7}\implies \cfrac{19}{5}\cdot \cfrac{1}{x}=\cfrac{19}{7}\implies \cfrac{19}{5x}=\cfrac{19}{7}
\\\\\\
\stackrel{}{\cfrac{7}{5x}=\cfrac{19}{19}}\implies \cfrac{7}{5x}=1\implies 7=5x\implies \cfrac{7}{5}=x\implies 1\frac{2}{5}=x$

5 0

3 years ago

Suppose x is a real number and epsilon > 0. Prove that (x - epsilon, x epsilon) is a neighborhood of each of its members; in

kaheart [24]

Answer:

See proof below

Step-by-step explanation:

We will use properties of inequalities during the proof.

Let $y\in (x-\epsilon,x+\epsilon)$ . then we have that $x-\epsilon$ . Hence, it makes sense to define the positive number delta as $\delta=\min\{x+\epsilon-y,y-(x-\epsilon)\}$ (the inequality guarantees that these numbers are positive).

Intuitively, delta is the shortest distance from y to the endpoints of the interval. Now, we claim that $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ , and if we prove this, we are done. To prove it, let $z\in (y-\delta,y+\delta)$ , then $y-\delta$ . First, $\delta \leq y-(x-\epsilon)$ then $-\delta \geq -y+x-\epsilon$ hence $z>y-\delta \geq x-\epsilon$

On the other hand, $\delta \leq x+\epsilon-y$ then $z$ hence $z$ . Combining the inequalities, we have that $x-\epsilon$ , therefore $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ as required.