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AURORKA [14]
3 years ago
5

The graph of a line and an exponential can intersect twice, once or not at all. Describe the possible number of intersections fo

r each of following pairs of graphs. Your solution to each part should include all possibilities and a quickly sketched example of each one. Part 2; (SHOW WORK)
c. A parabola and a circle

d. A parabola and the hyperbola y = 1/x
Mathematics
1 answer:
Katyanochek1 [597]3 years ago
7 0

Answer:

  c) parabola and circle: 0, 1, 2, 3, 4 times

  d) parabola and hyperbola: 1, 2, 3 times

Step-by-step explanation:

c. A parabola can miss a circle, be tangent to it in 1 or 2 places, intersect it 2 places and be tangent at a 3rd, or intersect in 4 places.

__

d. A parabola must intersect a hyperbola in at least one place, but cannot intersect in more than 3 places. If the parabola is tangent to the hyperbola, the number of intersections will be 2.

If the parabola or the hyperbola are "off-axis", then the number of intersections may be 0 or 4 as well. Those cases seem to be excluded in this problem statement.

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Answer:

The three chords are AB , BD and EF

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2 years ago
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Select the exclusion that fits best with this problem.<br><br><br><br> 13x^6/51x^4
Damm [24]

ANSWER

x \ne0

EXPLANATION

The given problem is

\frac{13 {x}^{6} }{ 51 {x}^{4}  }

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Even the given expression can be simplified to obtain:

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5 0
3 years ago
2.
Fantom [35]

Answer:

the mode remains the same

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the mode remains the same.

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3 years ago
A small combination lock on a suitcase has 55 ​wheels, each labeled with the 10 digits 0 to 9. How many 55 digit combinations ar
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Suppose a_n is the number of possible combinations for a suitcase with a lock consisting of n wheels. If you added one more wheel onto the lock, there would only be 9 allowed possible digits you can use for the new wheel. This means the number of possible combinations for n+1 wheels, or a_{n+1} is given recursively by the formula

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For example, if the combination for a 3-wheel lock is 282, then a 4-wheel lock can be any one of 2820, 2821, 2823, ..., 2829 (nine possibilities depending on the second-to-last digit).

By substitution, you have

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