Evaluate powers: 54, 106, 19, 05, (-2)4, (-3)3, (-1)8. Which scatter plot(s) could be modelled using a curve instead of a line of best fit? Explain. Make a scatter plot of the data.

Here are the evaluations for questions 1 to 8:

Question 1:

d) Evaluate $5^4$

$$5^4 = 5 \times 5 \times 5 \times 5 = 625$$

The value is $\boxed{625}$.

e) Evaluate $10^6$

$$10^6 = 10 \times 10 \times 10 \times 10 \times 10 \times 10 = 1\,000\,000$$

The value is $\boxed{1\,000\,000}$.

f) Evaluate $1^9$

$$1^9 = 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 \times 1 = 1$$

The value is $\boxed{1}$.

g) Evaluate $0^5$

$$0^5 = 0 \times 0 \times 0 \times 0 \times 0 = 0$$

The value is $\boxed{0}$.

h) Evaluate $(-2)^4$

$$(-2)^4 = (-2) \times (-2) \times (-2) \times (-2) = 16$$

The value is $\boxed{16}$.

i) Evaluate $(-3)^3$

$$(-3)^3 = (-3) \times (-3) \times (-3) = -27$$

The value is $\boxed{-27}$.

j) Evaluate $(-1)^8$

$$(-1)^8 = (-1) \times (-1) \times (-1) \times (-1) \times (-1) \times (-1) \times (-1) \times (-1) = 1$$

The value is $\boxed{1}$.

Question 2:

a) Which scatter plot(s) could be modelled using a curve instead of a line of best fit? Explain.
Both scatter plot a) and scatter plot b) show data points that follow a curved pattern rather than a straight line. A line of best fit would not accurately represent the trend in either case. Therefore, both plots could be modelled using a curve.

Question 3:

a) Make a scatter plot of the data. Draw a curve of best fit.
To make a scatter plot, you would plot the 'Length (m)' values on the x-axis and the 'Deflection (cm)' values on the y-axis. For example, (1.0, 0.33), (1.5, 1.48), (2.0, 2.51), and so on. Once plotted, you would draw a smooth curve that best follows the trend of these points. The curve would start near the origin and increase, becoming steeper as the length increases, suggesting a non-linear relationship, possibly quadratic.

b) Describe the relationship between the length of the beam and the deflection.
As the length of the beam increases, the deflection also increases. The rate at which the deflection increases appears to accelerate, meaning that for each equal increment in length, the deflection increases by a larger amount than the previous increment. This indicates a non-linear, increasing relationship.

c) Use your curve of best fit to predict the deflection of a 6.0-m-long beam.
To predict the deflection for a 6.0-m-long beam, you would extend your curve of best fit to the x-value of 6.0 m. Then, read the corresponding y-value (deflection) from the curve. Based on the trend in the table, the deflection for a 6.0-m beam would be significantly higher than 16.72 cm.
Let's estimate by observing the pattern of increase in deflection:

Length (m)	Deflection (cm)	First Difference

1.0 | 0.33 |

1.5 | 1.48 | 1.15

2.0 | 2.51 | 1.03

2.5 | 4.22 | 1.71

3.0 | 6.11 | 1.89

3.5 | 8.17 | 2.06

4.0 | 10.52 | 2.35

4.5 | 12.98 | 2.46

5.0 | 16.72 | 3.74

The first differences are increasing, suggesting a quadratic relationship. If we were to fit a quadratic model, we could get a more precise prediction. For an estimate, the increase from 4.5 to 5.0 m was 3.74 cm. The increase from 5.0 to 5.5 m would likely be even larger, perhaps around 4-5 cm, bringing it to about 21-22 cm. The increase from 5.5 to 6.0 m would be even larger.
A rough extrapolation suggests the deflection for a 6.0-m beam would be approximately $\boxed{25 \text{ cm to } 30 \text{ cm}}$. (A precise answer would require the actual curve or a regression model).

Question 4:
Use finite differences to determine whether each relation is linear, quadratic, or neither.

a)

$x$	$y$	First Differences	Second Differences
1	11
2	18	$18 - 11 = 7$
3	27	$27 - 18 = 9$	$9 - 7 = 2$
4	38	$38 - 27 = 11$	$11 - 9 = 2$
5	51	$51 - 38 = 13$	$13 - 11 = 2$
Since the second differences are constant, the relation is $\boxed{\text{quadratic}}$.

b)

$x$	$y$	First Differences	Second Differences
-2	-10
-1	-2	$-2 - (-10) = 8$
0	0	$0 - (-2) = 2$	$2 - 8 = -6$
1	2	$2 - 0 = 2$	$2 - 2 = 0$
2	10	$10 - 2 = 8$	$8 - 2 = 6$
Since neither the first nor the second differences are constant, the relation is $\boxed{\text{neither}}$.

c)

$x$	$y$	First Differences	Second Differences
-2	-9
-1	-6	$-6 - (-9) = 3$
0	-3	$-3 - (-6) = 3$	$3 - 3 = 0$
1	0	$0 - (-3) = 3$	$3 - 3 = 0$
2	3	$3 - 0 = 3$	$3 - 3 = 0$
Since the first differences are constant, the relation is $\boxed{\text{linear}}$.

Question 5:
The flight of an aircraft from Toronto to Montréal can be modelled by the relation $h = -2.5t^2 + 200t$.

a) Graph the relation.
The relation $h = -2.5t^2 + 200t$ is a quadratic equation in the form $h = at^2 + bt + c$, where $a = -2.5$, $b = 200$, and $c = 0$. Since $a < 0$, the parabola opens downwards. The graph will be a downward-opening parabola starting at the origin $(0,0)$, rising to a maximum height, and then descending back to $h=0$.

b) How long does it take to fly from Toronto to Montréal?
The flight starts and ends when the height $h$ is 0.