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2 years ago

11

using the figure below complete the proof. Which of the following sequence of reasons may be used to fill the blanks in the two-

columns proof above?

1 answer:

dlinn [17]2 years ago

3 0

Answer: Definition of parallelogram, ASA Theorem, CPCTC

Step-by-step explanation:

I just did the math nation and I go that question correct

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frosja888 [35]

Answer:

word

Step-by-step explanation:

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3 years ago

Read 2 more answers

Please help ASAP show work^

Art [367]

Answer:x=6

Step-by-step explanation: Remove the radical by raising each side to the index of the radical

4 0

3 years ago

2x^2+5x-3<br> Show steps

lbvjy [14]

2x^2+(6-1)x-3
2x^2+6x-x-3
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3 years ago

Let f (x) = 3x − 1 and ε > 0. Find a δ > 0 such that 0 < ∣x − 5∣ < δ implies ∣f (x) − 14∣ < ε. (Find the largest

s344n2d4d5 [400]

Answer:

This proves that f is continous at x=5.

Step-by-step explanation:

Taking f(x) = 3x-1 and $\varepsilon>0$ , we want to find a $\delta$ such that $|f(x)-14|$

At first, we will assume that this delta exists and we will try to figure out its value.

Suppose that $|x-5|$ . Then

$|f(x)-14| = |3x-1-14| = |3x-15|=|3(x-5)| = 3|x-5|< 3\delta$ .

Then, if $|x-5|$ , then $|f(x)-14|$ . So, in this case, if $3\delta \leq \varepsilon$ we get that $|f(x)-14|$ . The maximum value of delta is $\frac{\varepsilon}{3}$ .

By definition, this procedure proves that $\lim_{x\to 5}f(x) = 14$ . Note that f(5)=14, so this proves that f is continous at x=5.

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2 years ago

Is y=2x + 1 a function

shepuryov [24]

Yes. It's a linear function.

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2 years ago