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andrezito [222]
4 years ago
15

In a different plan for area​ codes, the first digit could be any number from 4 through 8​, the second digit was either 3, 4, 5,

or 6​, and the third digit could be any number exceptnbsp 2 or 5. with this​ plan, how many different area codes are​ possible?
Mathematics
1 answer:
Snezhnost [94]4 years ago
7 0

For the first digit, we have 5 options that are 4,5,6,7,8 . For the second digit, we have 4 options which are 3,4,5 or 6 and for the third digit, we have the options of all numbers except 2 or 5 that is 1,3,4,6,7,8,9,0 . SO we have 8 options for third digit . So to find the total number of options, we need to multiply all the possible options for each digit that is 5 times 4 times 8 = 160 . So the number of possible options are 160 .

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Answer:

See Below.

Step-by-step explanation:

We are given the graph of <em>y</em> = f'(x) and we want to determine the characteristics of f(x).

Part A)

<em>f</em> is increasing whenever <em>f'</em> is positive and decreasing whenever <em>f'</em> is negative.

Hence, <em>f</em> is increasing for the interval:

(-\infty, -2) \cup (-1, 1)\cup (3, \infty)

And <em>f</em> is decreasing for the interval:

\displaystyle (-2, -1) \cup (1, 3)

Part B)

<em>f</em> has a relative maximum at <em>x</em> = <em>c</em> if <em>f'</em> turns from positive to negative at <em>c</em> and a relative minimum if <em>f'</em> turns from negative to positive to negative at <em>c</em>.

<em>f'</em> turns from positive to negative at <em>x</em> = -2 and <em>x</em> = 1.

And <em>f'</em> turns from negative to positive at <em>x</em> = -1 and <em>x</em> = 3.

Hence, <em>f</em> has relative maximums at <em>x</em> = -2 and <em>x</em> = 1, and relative minimums at <em>x</em> = -1 and <em>x</em> = 3.

Part C)

<em>f</em> is concave up whenever <em>f''</em> is positive and concave down whenever <em>f''</em> is negative.

In other words, <em>f</em> is concave up whenever <em>f'</em> is increasing and concave down whenever <em>f'</em> is decreasing.

Hence, <em>f</em> is concave up for the interval (rounded to the nearest tenths):

\displaystyle (-1.5 , 0) \cup (2.2, \infty)

And concave down for the interval:

\displaystyle (-\infty, -1.5) \cup (0, 2.2)

Part D)

Points of inflections are where the concavity changes: that is, <em>f''</em> changes from either positive to negative or negative to positive.

In other words, <em>f </em>has an inflection point wherever <em>f'</em> has an extremum point.

<em>f'</em> has extrema at (approximately) <em>x</em> = -1.5, 0, and 2.2.

Hence, <em>f</em> has inflection points at <em>x</em> = -1.5, 0, and 2.2.

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