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3 years ago

Round 9594.20475485 to the nearest thousand.

Mathematics

2 answers:

EleoNora [17]3 years ago

8 0

Answer:

9594.205

Step-by-step explanation:

The thousandth place is three digits after the decimal.

Since a 7 follows the 4, it gets rounded up to 5.

I hope this helps! :)

Katena32 [7]3 years ago

7 0

Answer:

9594.205

Step-by-step explanation:

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Please help! I will mark a brainliest! There is a picture! :)

ella [17]

Answer: No mistakes was made

Step-by-step explanation:

Because It all equals the same as the answer

4 0

3 years ago

Read 2 more answers

Write a sentence to represent the solution set in the context of the problem.

Nikolay [14]

Answer:

6 sales

Step-by-step explanation:

(I MADE THIS QUESTION I DIDINT KNOW IF YOU NEEDED IT)

Jake wants to but the new computer(it costs $920). Jake has $560 saved in his bank. Jake also gets $60 whenever he makes a sale at his job. How many sale(s) at his job does Jake need to make? <em>Answer </em>-- Jake needs to make 6 sales of $60 to buy the new computer.

MATH -----

$6 x $60 = $360

$360 + $560 = $920

6 0

2 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.