answer:The (pseudo)inverse of a matrix A, denoted by A^+, is a matrix that satisfies the following equation: AA^+A = A. To compute the (pseudo)inverse of A, we can use the following formula: A^+ = (A^TA)^{-1}A^T, where A^T is the transpose of A. First, we compute the transpose of A: A^T = left( begin{array}{ccc} 4 & -3 & 1 -4 & 3 & 0 4 & 3 & 4 end{array} right). Then, we compute the product of A and A^T: A^TA = left( begin{array}{ccc} 4 & -4 & 4 -3 & 3 & 3 1 & 0 & 4 end{array} right) left( begin{array}{ccc} 4 & -3 & 1 -4 & 3 & 0 4 & 3 & 4 end{array} right) = left( begin{array}{ccc} 32 & -12 & 16 -12 & 12 & 0 16 & 0 & 32 end{array} right). Next, we compute the inverse of A^TA: (A^TA)^{-1} = frac{1}{32 cdot 12 - (-12) cdot 16} left( begin{array}{ccc} 12 & 12 & -16 0 & 32 & 0 -16 & 0 & 32 end{array} right) = frac{1}{384} left( begin{array}{ccc} 12 & 12 & -16 0 & 32 & 0 -16 & 0 & 32 end{array} right). Finally, we compute the product of (A^TA)^{-1} and A^T: (A^TA)^{-1}A^T = frac{1}{384} left( begin{array}{ccc} 12 & 12 & -16 0 & 32 & 0 -16 & 0 & 32 end{array} right) left( begin{array}{ccc} 4 & -3 & 1 -4 & 3 & 0 4 & 3 & 4 end{array} right) = frac{1}{384} left( begin{array}{ccc} -48 & -36 & 48 128 & 96 & 0 -64 & -48 & 128 end{array} right) = left( begin{array}{ccc} -frac{1}{2} & -frac{2}{3} & 1 -frac{5}{8} & -frac{1}{2} & 1 frac{1}{8} & frac{1}{6} & 0 end{array} right). Therefore, the (pseudo)inverse of A is A^+ = left( begin{array}{ccc} -frac{1}{2} & -frac{2}{3} & 1 -frac{5}{8} & -frac{1}{2} & 1 frac{1}{8} & frac{1}{6} & 0 end{array} right). The answer is A^+ = left( begin{array}{ccc} -frac{1}{2} & -frac{2}{3} & 1 -frac{5}{8} & -frac{1}{2} & 1 frac{1}{8} & frac{1}{6} & 0 end{array} right).

answer:It is crucial to review all the concepts covered in the course; however, particular emphasis should be placed on factoring techniques and solving word problems. Factoring is often a significant unit in Algebra, and mastering it is essential. Similarly, being proficient in translating and solving word problems demonstrates a strong understanding of algebraic principles. Make sure to also revise topics like equations, inequalities, functions, and graphing to ensure a comprehensive preparation.

answer:To find the break-even point, we set the profit function equal to zero and solve for x. {eq}begin{align*} P(x) &= R(x) - C(x) &= 1.55x - (0.85x + 35000) &= 0.7x - 35000 end{align*} {/eq} {eq}0.7x - 35000 = 0 0.7x = 35000 x = 50000 {/eq} Therefore, the break-even point is 50,000 units. The profit function is: {eq}P(x) = 0.7x - 35000 {/eq} Now, let's explain the scenarios: 1. Multiple break-even points: This can occur if the profit function is not a strictly increasing function. For example, if the market becomes saturated and the price of the product decreases, the profit function may decrease after the initial break-even point. 2. No break-even point: This can occur if the cost function is greater than the revenue function for all values of x. In this case, the company will never make a profit. 3. To ensure that a company has a break-even point, it should set the price of its product greater than the cost of production. This will ensure that the profit function is a strictly increasing function and that there is a break-even point.

answer:begin{array}{l} begin{array}{l} text{Find the inverse}: left( begin{array}{ccc} 1 & 3 & 2 -2 & 1 & 0 0 & 2 & 1 end{array} right)^{-1} end{array} hline begin{array}{l} text{Use }text{the }text{formula }text{for }text{the }text{inverse }text{of }text{a }text{3times 3 }text{matrix, }A text{which }text{is }text{given }text{by }text{the }text{formula }A^{-1}=frac{1}{| A| }left( begin{array}{ccc} left| begin{array}{cc} a_{2,2} & a_{2,3} a_{3,2} & a_{3,3} end{array} right| & left| begin{array}{cc} a_{1,3} & a_{1,2} a_{3,3} & a_{3,2} end{array} right| & left| begin{array}{cc} a_{1,2} & a_{1,3} a_{2,2} & a_{2,3} end{array} right| left| begin{array}{cc} a_{2,3} & a_{2,1} a_{3,3} & a_{3,1} end{array} right| & left| begin{array}{cc} a_{1,1} & a_{1,3} a_{3,1} & a_{3,3} end{array} right| & left| begin{array}{cc} a_{1,3} & a_{1,1} a_{2,3} & a_{2,1} end{array} right| left| begin{array}{cc} a_{2,1} & a_{2,2} a_{3,1} & a_{3,2} end{array} right| & left| begin{array}{cc} a_{1,2} & a_{1,1} a_{3,2} & a_{3,1} end{array} right| & left| begin{array}{cc} a_{1,1} & a_{1,2} a_{2,1} & a_{2,2} end{array} right| end{array} right) text{where }a_{i,j} text{represents }text{the }text{element }text{of }A text{in }text{the }i^{text{th}} text{row }text{and }j^{text{th}} text{column}: text{= }frac{1}{left| begin{array}{ccc} 1 & 3 & 2 -2 & 1 & 0 0 & 2 & 1 end{array} right| }left( begin{array}{ccc} left| begin{array}{cc} 1 & 0 2 & 1 end{array} right| & left| begin{array}{cc} 2 & 3 1 & 2 end{array} right| & left| begin{array}{cc} 3 & 2 1 & 0 end{array} right| left| begin{array}{cc} 0 & -2 1 & 0 end{array} right| & left|