answer:The domain of f(x) = log (x^2) includes all real numbers except 0, as x^2 is always non-negative. However, the domain of g(x) = 2log x is restricted to positive real numbers, as the logarithm of a negative number is undefined. This difference in domain causes the graph of f(x) to extend to both sides of the y-axis, while the graph of g(x) is only defined for positive x-values.

answer:The cross product of two vectors vec{a} = (a_1, a_2, a_3) and vec{b} = (b_1, b_2, b_3) is defined as: vec{a} times vec{b} = left( begin{array}{c} a_2b_3 - a_3b_2 a_3b_1 - a_1b_3 a_1b_2 - a_2b_1 end{array} right) Using this formula, we can compute the cross product of vec{a} and vec{b} as follows: vec{a} times vec{b} = left( begin{array}{c} frac{7}{8} cdot frac{23}{4} - frac{15}{8} cdot (-4) frac{15}{8} cdot frac{7}{4} - left(-frac{21}{8}right) cdot frac{23}{4} left(-frac{21}{8}right) cdot (-4) - frac{7}{8} cdot frac{7}{4} end{array} right) Simplifying this expression, we get: vec{a} times vec{b} = left( begin{array}{c} frac{401}{32} frac{147}{8} frac{287}{32} end{array} right) Therefore, the cross product of vec{a} and vec{b} is left( begin{array}{c} frac{401}{32} frac{147}{8} frac{287}{32} end{array} right). The answer is vec{a} times vec{b} = left( begin{array}{c} frac{401}{32} frac{147}{8} frac{287}{32} end{array} right)

answer:To solve for x and y using elimination, we can eliminate y by adding the two equations together. First, we multiply the second equation by 3 to make the coefficients of y opposite: 2x + 3y = 23 9x - 3y = -12 Now we can add the two equations: 11x = 11 Dividing both sides by 11, we get: x = 1 Now that we know x = 1, we can substitute this value back into one of the original equations to solve for y. Let's use the first equation: 2(1) + 3y = 23 2 + 3y = 23 3y = 21 y = 7 Therefore, the solution to the system of equations is x = 1 and y = 7.

answer:The completed square form of the given quadratic is: [7 left(x + frac{3}{14}right)^2 + frac{75}{28}]