Q03 Test 43 English

Test No. 43 – Complete Solutions

172

Convert to radians and write answer in terms of \(\pi\): 142.5°

🔑 Key Concept – Degrees to Radians: \[ \theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180} \] Simplify the resulting fraction fully to express in terms of \(\pi\).

Step 1 – Apply the formula: \[ 142.5 \times \frac{\pi}{180} = \frac{142.5\pi}{180} \]

Step 2 – Eliminate the decimal (multiply numerator and denominator by 2): \[ \frac{142.5\pi}{180} = \frac{285\pi}{360} \]

Step 3 – Simplify by dividing by 15: \[ \frac{285\pi}{360} = \frac{19\pi}{24} \]

\( \boxed{142.5° = \dfrac{19\pi}{24}\ \text{rad} \approx 2.4871\ \text{rad}} \)

173

Convert \(\dfrac{11\pi}{16}\) radians to degrees and write answer in DMS.

🔑 Key Concept – Radians to DMS: Step 1: Multiply by \(\dfrac{180°}{\pi}\) to get decimal degrees.
Step 2: Multiply the decimal part by 60 → whole minutes.
Step 3: Multiply the decimal part of minutes by 60 → seconds.

Step 1 – Convert to decimal degrees: \[ \frac{11\pi}{16} \times \frac{180°}{\pi} = \frac{11 \times 180°}{16} = \frac{1980°}{16} = 123.75° \]

Step 2 – Separate whole degrees: Whole degrees \(= 123°\),  Decimal part \(= 0.75°\)

Step 3 – Convert decimal degrees to minutes: \[ 0.75 \times 60 = 45' \] Whole minutes \(= 45'\),  Decimal part \(= 0'\) (exact)

Step 4 – No seconds needed (decimal of minutes = 0): \[ 0 \times 60 = 0'' \]

\( \boxed{\dfrac{11\pi}{16} = 123°\ 45'\ 0''} \)

174

Find equation of line with x-intercept \(-9\) and slope \(4\).

🔑 Key Concept – Point-Slope Form: The x-intercept gives a point on the line: \((a,\ 0)\).
With a known point and slope, use: \[ y - y_1 = m(x - x_1) \]

Given: x-intercept \(= -9\) → point \((-9,\ 0)\),  slope \(m = 4\)

Step 1 – Apply point-slope form: \[ y - 0 = 4\bigl(x - (-9)\bigr) \] \[ y = 4(x + 9) \]

Step 2 – Expand: \[ y = 4x + 36 \]

Step 3 – Write in standard form: \[ 4x - y + 36 = 0 \]

\( \boxed{4x - y + 36 = 0} \)

175

Find the unknown volume \(V_1\) for similar cubes with edges 5 cm and 7 cm, given \(V_2 = 686\ \text{cm}^3\).

🔑 Key Concept – Volumes of Similar Solids: \[ \frac{V_1}{V_2} = \left(\frac{l_1}{l_2}\right)^3 \] Volumes scale as the cube of the ratio of corresponding edge lengths.

Given: Cube 1: edge \(l_1 = 5\ \text{cm}\),  \(V_1 = ?\)
Cube 2: edge \(l_2 = 7\ \text{cm}\),  \(V_2 = 686\ \text{cm}^3\)

Step 1 – Write the formula: \[ \frac{V_1}{V_2} = \left(\frac{l_1}{l_2}\right)^3 \]

Step 2 – Substitute known values: \[ \frac{V_1}{686} = \left(\frac{5}{7}\right)^3 \]

Step 3 – Cube the ratio: \[ \left(\frac{5}{7}\right)^3 = \frac{125}{343} \]

Step 4 – Solve for \(V_1\): \[ V_1 = 686 \times \frac{125}{343} = \frac{686 \times 125}{343} = 2 \times 125 = 250\ \text{cm}^3 \] Note: \(686 \div 343 = 2\) exactly.

\( \boxed{V_1 = 250\ \text{cm}^3} \)