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lora16 [44]
3 years ago
6

The distance versus time plot for a particular object shows a quadratic relationship. Which column of distance data is possible

for this situation?
Time (s) / A. Distance (m) /B. Distance (m) /C. Distance (m) / D. Distance (m) / E. Distance(m)
0 / 0 / 2.00 / 9.00 / infinity / infinity
1 / 1.00 / 4.00 / 18.00 / 1.00 / 1.00
2 / 4.00 / 6.00 / 27.00 / 0.50 / 0.25
3 / 9.00 / 8.00 / 36.00 / 0.33 / 0.11
4 / 16.00 / 10.00 / 45.00 / 0.25 / 0.06
5 / 25.00 / 12.00 / 54.00 / 0.20 / 0.04
6 / 36.00 / 14.00 / 63.00 / 0.16 / 0.02


(A) column A
(B) column B
(C) column C
(D) column D
(E) column E
Mathematics
1 answer:
Viktor [21]3 years ago
3 0

Answer:

(A) column A

Step-by-step explanation:

The distance values in column A are:

0, 1, 4, 9, 16, 25 and 36

If you just look at these values, these are the squares of 0, 1, 2, 3, 4, 5 and 6 which shows that distance in column A versus time would show a quadratic plot.

For the values showing a quadratic relationship, the second differences in the values are constant. This means, if we find the differences of two consecutive terms for all the terms and then find the difference of the answers resulted in previous step, these should be a constant for the quadratic function.

Differences in the values of Columns A are:

1, 3, 5, 7, 9, 11

The difference in these differences are:

2, 2, 2, 2, 2

which is a constant. Since, the difference is being calculated two times, it is known as second difference. The second difference must be constant for a quadratic relationship. For the other columns, the second differences are not constant.

Therefore, the correct answer is option A.

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What does 5 over 100 =
Nuetrik [128]

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\frac{5}{100}

To reduce this fraction, you need to divide the numerator and the denominator by the Greatest Common Factor (GCF) between them.

In this case, to find the GCF, you need to:

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8 0
11 months ago
Consider the infinite geometric series (∞/Σ/n=1) * -4(1/3)^{n-1}.
raketka [301]
The first four terms of the series probably refer to the first four partial sums.

\displaystyle\sum_{n=1}^1-4\left(\frac13\right)^{n-1}=-4\left(\frac13\right)^{1-1}=-4
\displaystyle\sum_{n=1}^2-4\left(\frac13\right)^{n-1}=-\frac{16}3
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\displaystyle\sum_{n=1}^4-4\left(\frac13\right)^{n-1}=-\frac{160}{27}

which you can compute either by adding one term at a time, or using the well-known formula,

\displaystyle\sum_{n=1}^kar^{n-1}=a\frac{1-r^{k+1}}{1-r}

(I can provide a link to a derivation I gave in a nearly identical question in the comments)

The series converges as k\to\infty if and only if |r|, which is certainly the case here, since -1.

Extrapolating from the formula above, the sum of the convergent series is

\displaystyle\sum_{n=1}^\infty ar^{n-1}=\frac a{1-r}

so the sum for this series is \dfrac{-4}{1-\frac13}=-6.
5 0
3 years ago
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