This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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5. The triangle ABC with vertices A(1,0), B(3,0) and C(1,2) is transformed by the matrix T = 3k & 1.6 \\ 3k & -0.9 onto triangle A'B'C'. Given that the area of triangle A'B'C' is 6 square units, determine the value of k. Step 1: Calculate the area of the original triangle ABC. The vertices are A(1,0), B(3,0), C(1,2). The base of the triangle can be taken as the distance between A(1,0) and B(3,0) along the x-axis. Base length = 3 - 1 = 2 units. The height of the triangle is the perpendicular distance from C(1,2) to the x-axis. Height = 2 units. Area of triangle ABC = (1)/(2) × base × height Area_ABC = (1)/(2) × 2 × 2 = 2 square units Step 2: Calculate the determinant of the transformation matrix T. T = 3k & 1.6 \\ 3k & -0.9 The determinant of T is det(T) = (3k)(-0.9) - (1.6)(3k). det(T) = -2.7k - 4.8k det(T) = -7.5k Step 3: Use the relationship between the areas of the original and transformed triangles. The area of the transformed triangle is given by the absolute value of the determinant of the transformation matrix multiplied by the area of the original triangle. Area_A'B'C' = |det(T)| × Area_ABC Given Area_A'B'C' = 6 square units. 6 = |-7.5k| × 2 6 = 7.5|k| × 2 6 = 15|k| Step 4: Solve for k. |k| = (6)/(15) |k| = (2)/(5) |k| = 0.4 Therefore, k can be 0.4 or -0.4. k = 0.4 or k = -0.4 9. Solve the equation 6 ^2 x + x = 4 for 0^ x 180^, giving the answer correct to 2 decimal places. Step 1: Rewrite the equation in terms of a single trigonometric function using the identity ^2 x = 1 - ^2 x. 6(1 - ^2 x) + x = 4 6 - 6 ^2 x + x = 4 Step 2: Rearrange the equation into a quadratic form. 0 = 6 ^2 x - x + 4 - 6 6 ^2 x - x - 2 = 0 Step 3: Let y = x and solve the quadratic equation 6y^2 - y - 2 = 0. Using the quadratic formula y = -b ± sqrt(b^2 - 4ac)2a: y = -(-1) ± sqrt((-1)^2 - 4(6)(-2))2(6) y = 1 ± sqrt(1 + 48)12 y = 1 ± sqrt(49)12 y = (1 ± 7)/(12) This gives two possible values for y: y_1 = (1 + 7)/(12) = (8)/(12) = (2)/(3) y_2 = (1 - 7)/(12) = (-6)/(12) = -(1)/(2) Step 4: Find the values of x for each solution of x within the range 0^ x 180^. Case 1: x = (2)/(3) Since x is positive, x is in the first or second quadrant. The reference angle is ((2)/(3)). x_1 = ((2)/(3)) ≈ 41.8103^ To 2 decimal places, x_1 = 41.81^. The second quadrant solution is 180^ - x_1: x_2 = 180^ - 41.8103^ ≈ 138.1897^ To 2 decimal places, x_2 = 138.19^. Both 41.81^ and 138.19^ are within the range 0^ x 180^. Case 2: x = -(1)/(2) Since x is negative, x is in the third or fourth quadrant. The reference angle is ((1)/(2)) = 30^. Third quadrant solution: 180^ + 30^ = 210^. Fourth quadrant solution: 360^ - 30^ = 330^. Both 210^ and 330^ are outside the given range 0^ x 180^. Therefore, there are no solutions from this case. The solutions for x are: x = 41.81^, 138.19^ What's next?