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3 years ago

Please help I can’t figure this out it’s been a very long time since I was in 5th grade. Thanks

Mathematics

2 answers:

Liula [17]3 years ago

7 0

books off challenge

Step-by-step explanation:

2×2×6=n

antoniya [11.8K]3 years ago

4 0

I guess it is 3.03 because there are three wholes and 3 little boxes

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How do you solve this step by step?

kolezko [41]

Answer:

Step-by-step explanation:

Evaluate x^2 + 2 x^2 + 3 x^2 + 4 x^2 where x = 3:

x^2 + 2 x^2 + 3 x^2 + 4 x^2 = 3^2 + 2×3^2 + 3×3^2 + 4×3^2

3^2 = 9:

3^2 + 2×9 + 3×3^2 + 4×3^2

3^2 = 9:

3^2 + 2×9 + 3×9 + 4×3^2

3^2 = 9:

3^2 + 2×9 + 3×9 + 4×9

3^2 = 9:

9 + 2×9 + 3×9 + 4×9

2×9 = 18:

9 + 18 + 3×9 + 4×9

3×9 = 27:

9 + 18 + 27 + 4×9

4×9 = 36:

9 + 18 + 27 + 36

| 3 |

| 3 | 6

| 2 | 7

| 1 | 8

+ | | 9

| 9 | 0:

Answer: 90

5 0

3 years ago

Read 2 more answers

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.