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Rom4ik [11]
2 years ago
14

If a is an arbitrary nonzero constant, what happens to a/b as b approaches 0

Mathematics
1 answer:
Stolb23 [73]2 years ago
7 0

It depends on how b approaches 0

If b is positive and gets closer to zero, then we say b is approaching 0 from the right, or from the positive side. Let's say a = 1. The equation a/b turns into 1/b. Looking at a table of values, 1/b will steadily increase without bound as positive b values get closer to 0.

On the other side, if b is negative and gets closer to zero, then 1/b will be negative and those negative values will decrease without bound. So 1/b approaches negative infinity if we approach 0 on the left (or negative) side.

The graph of y = 1/x shows this. See the diagram below. Note the vertical asymptote at x = 0. The portion to the right of it has the curve go upward to positive infinity as x approaches 0. The curve to the left goes down to negative infinity as x approaches 0.

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If a histogram of a sample of​ men's ages is​ skewed, what do you expect to see in the normal quantile​ plot?
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Answer: Points are not following a straight-line pattern

Step-by-step explanation:

5 0
2 years ago
Suppose again that we are counting the ways to distribute exams to TAs and it matters which students' exams go to which TAs. The
abruzzese [7]

The question is incorrect.

The correct question is:

Three TAs are grading a final exam.

There are a total of 60 exams to grade.

(c) Suppose again that we are counting the ways to distribute exams to TAs and it matters which students' exams go to which TAs. The TAs grade at different rates, so the first TA will grade 25 exams, the second TA will grade 20 exams and the third TA will grade 15 exams. How many ways are there to distribute the exams?

Answer: 60!/(25!20!15!)

Step-by-step explanation:

The number of ways of arranging n unlike objects in a line is n! that is ‘n factorial’

n! = n × (n – 1) × (n – 2) ×…× 3 × 2 × 1

The number of ways of arranging n objects where p of one type are alike, q of a second type are alike, r of a third type are alike is given as:

n!/p! q! r!

Therefore,

The answer is 60!/25!20!15!

6 0
2 years ago
Which oh the following are among the five basic postulates of Euclidean geometry
igomit [66]

1. A straight line segment can be drawn joining any two points.

2. Any straight line segment can be extended indefinitely in a straight line.

3. Given any straight line segment, a circle can be drawn having the segment as radius and one endpoint as center.

4. All right angles are congruent.

5. If two lines are drawn which intersect a third in such a way that the sum of the inner angles on one side is less than two right angles, then the two lines inevitably must intersect each other on that side if extended far enough. This postulate is equivalent to what is known as the parallel postulate.

8 0
2 years ago
A- 5.57<br> B- 41<br> C-12<br> D-21.71
Kryger [21]

Answer:

Step-by-step explanation:

If KM bisects angle NKL, the angle 3 is congruent to angle 4.  

We are given that angle 1 is congruent to angle 2, so that means that angle JKP is congruent to angle PKN.  By the definition of an angle bisector, we know then that angle NKM is congruent to angle MKL.  By the definition of a straight angle formed by opposite rays, all those angles named above add up to equal 180 degrees.  So if angle JKN = 8x + 2 and angle MKL = 3x + 5 and angles NKM and MKL are congruent, then angle NKL = 2(3x + 5) which is 6x + 10.  Again, if all those angles above add up to equal 180, then

8x + 2 + 6x + 10 = 180 and

14x + 12 = 180 and

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x = 12.

Angle MKN = 3x + 5 so if x = 12, then

Angle MKN = 3(12) + 5 and

Angle MKN = 41 degrees

4 0
2 years ago
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They are 5.8 units apart
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