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eduard
4 years ago
13

(x-3)^2+ (y-5)^2 = 144 is it (3,5) or (-3,-5)

Mathematics
2 answers:
Alenkasestr [34]4 years ago
8 0
(3, 5) I believe.

Is that all?
True [87]4 years ago
3 0
The answer is (3,5), pretty sure.
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Since he wanted to run farther this week so the last week 26.1 which is the distance he ran is less than the distance of this week

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Step-by-step explanation:

Electricians usually charge between $50 to $100 per hour. For the entire project, you'll spend an average of $323, or within a range between $160 and $508 or more. Both hourly and project rates vary depending on the type of project, license and experience of the service provider.

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The first two terms of a geometric sequence are a1 = 1/3 and a2 = 1/6. What is a8, the eighth term?
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B. 1/384

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3 years ago
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Please help me with the below question.
tresset_1 [31]

We have the following three conclusions about the <em>piecewise</em> function evaluated at x = 14.75:

  1. \lim_{t \to 14.75^{-}} f(t) = 66.
  2. \lim_{t \to 14.75^{+}} f(t) = 10.
  3. \lim_{t \to 14.75} f(t) does not exist as \lim_{t \to 14.75^{-}} f(t) \ne  \lim_{t \to 14.75^{+}} f(t).

<h3>How to determinate the limit in a piecewise function</h3>

In a <em>piecewise</em> function, the limit for a given value exists when the two <em>lateral</em> limits are the same and, thus, continuity is guaranteed. Otherwise, the limit does not exist.  

According to the definition of <em>lateral</em> limit and by observing carefully the figure, we have the following conclusions:

  1. \lim_{t \to 14.75^{-}} f(t) = 66.
  2. \lim_{t \to 14.75^{+}} f(t) = 10.
  3. \lim_{t \to 14.75} f(t) does not exist as \lim_{t \to 14.75^{-}} f(t) \ne  \lim_{t \to 14.75^{+}} f(t).

To learn more on piecewise function: brainly.com/question/12561612

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2 years ago
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776 - 747 = 29 grams

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