Five minus three fourths

1 answer:

8 0

5-3 5 3 20 3
__= _-_ = __-__=
4 1 4 4 4

17
___
4
I divide it the number to get the same denominator and do it on the top too.

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Which function has the same domain as y=2 square root x

matrenka [14]

We are given the function
y = 2 √x

We are asked for the function that has the same domain as the given function

First, the domain for the given function is this
{ x | x ≥ 0 }

So, any function that has a domain of x that is equal to or greater than 0. Some examples:
f(x) = √x - 2
f(x) = 5 √x

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What are the domain and range of the function below?

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Answer:

Domain: all real numbers

Range: all real numbers greater than or equal to -2

Step-by-step explanation:

The domain is the set of x-coordinates.

It is the set of real numbers since x can be any real number.

The range is the set of y-coordinates.

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Domain: all real numbers

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Marla owes her sister $15. She pays her $6 today. Which integer represents the amount of debt she has remaining?

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Answer:

The integer that represents the amount of debt she has remaining is $9

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

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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.