Example Question - equation of line
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Multiple Mathematics Problems Involving Algebra, Geometry, and Complex Numbers
<p>First question: Find the 4th term from the end in the expansion of \( \left( \frac{3}{x^2} - x^3 \right)^7 \).</p> <p>The 4th term from the end is the same as the 4th term from the beginning, which corresponds to \( T_4 \) in the expansion:</p> <p>\( T_k = \binom{n}{k-1} \cdot (a)^{n-(k-1)} \cdot (b)^{k-1} \) where \( n = 7 \), \( a = \frac{3}{x^2} \), and \( b = -x^3 \).</p> <p>\( T_4 = \binom{7}{3} \cdot \left(\frac{3}{x^2}\right)^{7-3} \cdot (-x^3)^3 \)</p> <p>\( T_4 = 35 \cdot \left(\frac{3}{x^2}\right)^4 \cdot (-x^3)^3 \)</p> <p>\( T_4 = 35 \cdot \frac{81}{x^8} \cdot (-x^9) \)</p> <p>\( T_4 = 35 \cdot 81 \cdot \frac{-x^9}{x^8} \)</p> <p>\( T_4 = -2835 \cdot \frac{1}{x} \)</p> <p>Second question: Find the equation of lines passing through (1,2) and making angle 30° with y-axis.</p> <p>The slope of the line making a 30° angle with the y-axis is the tangent of (90°-30°), which is \( \tan(60°) \).</p> <p>Slope (m) of the desired line: \( m = \tan(60°) = \sqrt{3} \)</p> <p>Use the point-slope form \( y - y_1 = m(x - x_1) \) to find the equation of the line passing through (1,2).</p> <p>\( y - 2 = \sqrt{3}(x - 1) \)</p> <p>Equation of the line: \( y = \sqrt{3}x + (2 - \sqrt{3}) \)</p> <p>Third question: Find the domain and range of the function \( f(x) = \frac{1}{\sqrt{9 - x^2}} \).</p> <p>The domain is where the function is defined and the denominator is not zero.</p> <p>\( 9 - x^2 > 0 \)</p> <p>\( -3 < x < 3 \), so the domain is \( (-3, 3) \).</p> <p>Since the function is the reciprocal of a square root, its range is all positive real numbers, \( (0, \infty) \).</p> <p>Fourth question: Solve the system of equations \( Re(z) = 0, |z| = 2 \).</p> <p>A complex number \( z = x + yi \) satisfies \( Re(z) = 0 \) when \( x = 0 \).</p> <p>Using \( |z| = 2 \), we get \( |0 + yi| = 2 \), which means \( \sqrt{0^2 + y^2} = 2 \).</p> <p>\( y = \pm 2 \), so \( z = 0 \pm 2i \).</p> <p>Fifth question: Write the complex number \( 1 + 7i \) in polar form.</p> <p>Let \( z = 1 + 7i \).</p> <p>Magnitude \( r = |z| = \sqrt{1^2 + 7^2} = \sqrt{50} = 5\sqrt{2} \).</p> <p>Angle \( \theta = \tan^{-1}\left(\frac{7}{1}\right) \), which is in the first quadrant.</p> <p>Polar form: \( z = r(\cos \theta + i \sin \theta) \)</p> <p>\( z = 5\sqrt{2}\left(\cos \left(\tan^{-1} 7\right) + i \sin \left(\tan^{-1} 7\right)\right) \)</p>
Equation of Perpendicular Line Passing Through a Point
To find the equation of the line h, which is perpendicular to line g, and passes through the point (3, −9), we first need to determine the slope of line g. The equation for line g is given by: \[ y = \frac{3}{10}x - \frac{8}{10} \] The slope of line g is the coefficient of x, which is \(\frac{3}{10}\). Two lines that are perpendicular to each other have slopes that are negative reciprocals of one another. So, the slope (m) of line h will be: \[ m = -\frac{1}{\text{slope of g}} = -\frac{1}{\frac{3}{10}} = -\frac{10}{3} \] Now we know the slope of line h, we can use the point-slope form of the equation of a line to find its equation. The point-slope form is: \[ y - y_1 = m(x - x_1) \] Plugging in our values for the slope and the coordinates of the given point (3, −9), we get: \[ y - (-9) = -\frac{10}{3}(x - 3) \] \[ y + 9 = -\frac{10}{3}x + 10 \] To write this in slope-intercept form, which is \( y = mx + b \), we solve for y: \[ y = -\frac{10}{3}x + 10 - 9 \] \[ y = -\frac{10}{3}x + 1 \] Here \( b \), the y-intercept, is 1. The numbers in the equation are already simplified proper fractions, improper fractions, or integers. So, the equation for line h in slope-intercept form is: \[ y = -\frac{10}{3}x + 1 \]
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