Detailed Answer

a) \( w = 0.39 \)

b) We know that a has to be between 1 and 10, so the only possibility is:

\[ 0.3862901 = 3.862901 * 10^{-1}\]

That means that \( k = -1 \).

c) Using the formula for percentage error, we get that:

\[ ε= |\frac{v_a - v_e}{v_e}| * 100\% = |\frac{0.3862901- 0.427}{0.427}| * 100\% = 9.53%\]

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Detailed Answer

a)

\[ p = \frac{8cos(45°)}{3x + 2y} + z = \frac{8cos⁡(45°)}{3*6+2*9} + 4 = \frac{8 * \frac{\sqrt{2}}{2}}{18 + 18} = \frac{\sqrt{2}}{9} + 4\]

b) Using GDC we can see that:

\[ \frac{\sqrt{2}}{9} + 4 \approx 4.157\]

c) Since \( k = 2 \), we know that:

\[ 4.157 = 0.04157 * 10^2\]

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Detailed Answer

a) The volume of a cuboid can be calculated by multiplying the area of its base by its length, so:

\[ 4.8 * 5.8 * 9.1 = 253.344 \ cm^3\]

b) As we know that \( 1dm = 10cm \), the conversion factor is \( 10^3 = 1000 \), so we get:

\[253.344 \ cm^3 = 0.253344 \ dm^3\]

c) 250 cm³

d) Let's start by calculating the area of the base:

\[ A_{base} = 4.8 * 5.8 = 27.84 \ cm^2\]

Now, the area of the larger side:

\[ A_{large \ side} = 5.8 * 9.1 = 52.78 \ cm^2\]

Finally, the area of the smaller side

\[ A_{small \ side} = 4.8 * 9.1 = 43.68 \ cm^2\]

We know that we have two of each side, so all of them have to be multiplied by two.

Let's calculate how much will be spent on the top and bottom of the cuboid. We know that it costs 0.15$ per cm², so:

\[ C_{top + bottom} = 27.84 * 2 * 0.15 = $8.352\]

Now, the cost for the sides:

\[ C_{sides} = (52.78 * 2 + 43.68*2)*0.12 = $23.1504\]

So, the total cost is:

\[ C_{total} = 23.1504 + 8.352 = 31.5024 \approx $31.50\]

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Detailed Answer

a)

\[ q = \frac{12x^2sin(α)}{5x^2 + 2y} = \frac{12 * 9 * \frac{\sqrt{3}}{2}}{45 + 18} = \frac{54\sqrt{3}}{63} = \frac{6\sqrt{3}}{7}\]

b)

\[\frac{6\sqrt{3}}{7} \approx 1.48\]

c) We know that k = 4, so:

\[1.48 = 0.000148*10^4\]

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Detailed Answer

a) To find the volume we need to apply the volume formula:

\[ V = \pi r^2 * h\]

As it can be seen from the figure, the radius will be a half of 8.4cm, so 4.2cm.

\[ V = \pi 4.2^2 * 12.5 = 692.721... \approx 693cm^3\]

b) The area of a cylinder is given by the formula:

\[ A = 2 \pi rh + 2 \pi r^2\]

We know that the top paert should not be paint, meaning that the formula we will actually use is:

\[ A = 2 \pi rh + \pi r^2\]

\[ A = 2 \pi 4.2 * 12.5 + \pi 4.2^2 = 385.285... \approx 385cm^2\]

c) To calculate the percentage error we first need to find the exact value, so the area will need to be multiplied by the dollar amount per each cm². It is important to here take the exact value, not the rounding to 3 significant figures:

\[ 0.05 * 385.285... \approx 19.26\]

Now, using the formula for percentage error, we get that:

\[ ε= |\frac{v_a - v_e}{v_e}| * 100\% = |\frac{20 - 19.26}{19.26}| * 100\% = 3.84\%\]

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Detailed Answer

a) By plugging the values into the formula, we get:

\[ V = \frac{1}{2}(2.4 + 3.6) * 3.1 * 8 = 74.4cm^3\]

b)

\[ V = 74.4cm^3 \approx 74cm^3\]

c) Knowing that a has to be between 1000 and 10000, we know that the only possible value it can take is 7440, so:

\[ 74.4 = 7440 * 10^{-2}\]

That means that k = -2

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Detailed Answer

a)

Given: Point A: (3, 5), Point B: (6, 1)

\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Substitute the coordinates of points A and B into the formula:

\[ \text{Distance}_{AB} = \sqrt{(6 - 3)^2 + (1 - 5)^2} \]

\[ = \sqrt{3^2 + (-4)^2} \]

\[ = \sqrt{9 + 16} \]

\[ = \sqrt{25} \]

\[ = 5 \]

So, the distance between points A and B is 5.

b)

We first need to find the distance between points C and D. So, given: Point C: (2, 4), Point D: (8, 9)

Using the distance formula:

\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

Substitute the coordinates of points C and D into the formula:

\[ \text{Distance}_{CD} = \sqrt{(8 - 2)^2 + (9 - 4)^2} \]

\[ = \sqrt{6^2 + 5^2} \]

\[ = \sqrt{36 + 25} \]

\[ = \sqrt{61}\]

\[ \approx 7.81\]

Now, let's check Alex's claim that the distance between points C and D is twice the distance between points A and B:

Alex's claim: \(\text{Distance}_{CD} = 2 \times \text{Distance}_{AB}\)

\[ 7.81 = 2 \times 5 \]

Since \(7.81 \neq 10\), Alex's claim is incorrect.

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Detailed Answer

a) (i) Using the GPS measurements and the formula for area:

\[\text{Area} = 98.4 \, \text{m} \times 26.2 \, \text{m} = 2578.08 \, \text{m}^2\]

a) (ii) Using the formula for percentage error:

\[\epsilon = \left| \frac{2578.08-2646}{2646} \right| \times 100\% = 2.56689 \ldots \approx 2.567\%\]

b) (i) We use the length and width given together with the Pythagorean theorem:

\[l = \sqrt{98.4^2 + 26.2^2} \, \text{m} = 101.828287 \ldots \, \text{m} \approx 101.83 \, \text{m}\]

b) (ii)

\[l \approx 1.0183 \times 10^2 \, \text{m}\]

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Detailed Answer

a) \(\text{t} = 0.00219864 \, \text{ms} \approx 0.00220 \, \text{ms}\)

b) Since \( t \) is 0.00219864 ms, we can express it as:

\[\text{t} = 2.19864 \times 10^{-3} \, \text{ms} \approx 2.199 \times 10^{-3}\]

Then, \(\text{a} = 2.199\)

c) First, we notice that \( t = 2.19864μs \). Using the formula for percentage error:

\[ \epsilon = \left|\frac{2.19864 - 2.1969811}{2.1969811}\right| \times 100 \% \approx 0.092 \% \]

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Detailed Answer

a) Simplifying the fraction:

\[\frac{I}{\kappa} = \frac{8281}{2028} = \frac{49}{12}\]

Replacing the given values in the formula, we get:

\[\tau = 2 \pi \sqrt{\frac{49}{12}} \text{ ms} = \frac{7 \pi \sqrt{3}}{3} \text{ ms}\]

b) Since \( 1 ms= 10^{-3}s\), it follows:

\[\tau = \frac{7 \pi \sqrt{3}}{3} \text{ ms} = 12.6965955 \ldots \text{ ms} \approx 12.7 \text{ ms} \approx 12.7 \times 10^{-3} \text{ s} \approx 1.27 \times 10^{-2} \text{ s}\]

c) Using the formula for percentage error:

\[\epsilon = \left|\frac{12.5 - 12.6966}{12.6966}\right| \times 100 \% \approx 1.5728 \% \approx 1.573 \%\]

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Detailed Answer

a) Find first the absolute error in each dimension:

\[ \epsilon_{L} = 83 \times 5 \% = 4.15 \, \text{cm} \]

\[ \epsilon_{W} = 57 \times 2 \% = 1.14 \, \text{cm} \]

Calculate the lower bounds for both length and width in meters:

\[ L_{l} = 83 - 4.15 = 78.85 \, \text{cm} = 0.7885 \, \text{m} \]

\[ W_{l} = 57 - 1.14 = 55.86 \, \text{cm} = 0.5586 \, \text{m} \]

\[ A_{l} = L_{l} \times W_{l} = 0.4404561 \, \text{m}^{2} \approx 0.440 \, \text{m}^{2} \]

b) Calculate the upper bounds for both length and width in meters:

\[ L_{u} = 83 + 4.15 = 87.15 \, \text{cm} = 0.8715 \, \text{m} \]

\[ W_{u} = 57 + 1.14 = 58.14 \, \text{cm} = 0.5614 \, \text{m} \]

\[ A_{u} = L_{u} \times W_{u} = 0.50669 \, \text{m}^{2} \approx 0.51 \, \text{m}^{2} \]

c) Using the formula for absolute maximum and percentage error:

\[ e = \frac{0.51 - 0.44}{2} = 0.2244 \, \text{m}^{2} \]

\[ \epsilon = \frac{e}{83 \times 57} \times 100 \% = 23.71592 \ldots \% \approx 23.716 \% \]

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Detailed Answer

a)

\[ \alpha = 789.654 \approx 790 \]

b) (i)

\[ \alpha = 789.654 \approx 789.65 \]

b) (ii)

\[ \alpha \approx 789.65 \approx 7.8965 \times 10^{2} \]

c) Using the formula for percentage error:

\[ \epsilon \approx \left|\frac{790 - 789.654}{789.654}\right| \times 100 \% \approx 0.043817 \ldots \% \approx 0.044 \% \]

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Detailed Answer

a) Begin by converting \(\phi\) to radians and plug in the values in the formula:

\[ \phi = \frac{60 \pi}{180} = \frac{\pi}{3}\]

\[ s = \ln \left(\tan \left(\frac{\pi + \frac{2 \pi}{3}}{4}\right)\right) \approx 1.316957897 \ldots \approx 1.317 \]

b) \( s \approx 1.317 = 131.7 \times 10^{-2} \)

c) (i) Find the new value of \(\phi\):

\[ \phi^{*} = 1.05 \times \phi = \frac{1.05 \pi}{3} \]

\[ s = \ln \left(\tan \left(\frac{\pi + \frac{2.1 \pi}{3}}{4}\right)\right) \approx 1.426788247 \ldots \approx 1.427 \]

c) (ii) Using the formula for percentage error:

\[ \epsilon \approx \left|\frac{1.427 - 1.317}{1.317}\right| \times 100 \% \approx 8.3397 \ldots \% \approx 8.34 \% \]

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