answer:To rewrite the equation in the form y = mx + b, we need to isolate y on one side of the equation. Starting with: 3x - y = -3 Add y to both sides: 3x = -3 + y Now, to solve for y, subtract 3x from both sides: - y = -3 + 3x Finally, multiply both sides by -1 to get y alone: y = 3x - 3 So, the equation in the form y = mx + b is: y = 3x - 3

answer:Freeness: If F is a free R-module with a linearly ordered basis {v_1,ldots,v_n}, then bigwedge^k F is free over {v_{i_1}wedgecdotswedge v_{i_k} | i_1<ldots<i_k} and {mathfrak S}^n F is free over {v_{i_1}cdots v_{i_k} | i_1leq ldotsleq i_k}. This also holds for infinitely generated free modules Fcong R^{(I)} with a linearly ordered index set I. Projectivity: Since bigwedge^n(-) and {mathfrak S}^n(-) preserve retracts and freeness, they preserve projectivity because projective modules are retracts of free modules. Flatness: Since bigwedge^n(-) and {mathfrak S}^n(-) preserve direct limits and finite rank free modules, they preserve flatness because flat modules are direct limits of finite rank free modules.

answer:Substituting {eq}x = 2 - 3sin t{/eq} into the equation of the circle, we get: {eq}begin{align*} (x + 2)^2 + (y + 1)^2 &= 16 (2 - 3sin t + 2)^2 + (y + 1)^2 &= 16 (4 - 3sin t)^2 + (y + 1)^2 &= 16 16 - 24sin t + 9sin^2 t + (y + 1)^2 &= 16 (y + 1)^2 &= 24sin t - 9sin^2 t (y + 1)^2 &= 9sin t(3 - sin t) y + 1 &= pm 3sqrt{sin t(3 - sin t)} end{align*} Since the circle is traced counterclockwise as the parameter increases, {eq}y{/eq} increases initially. Therefore, we have: {eq}y = 3sqrt{sin t(3 - sin t)} - 1{/eq}

answer:The cross product of two vectors, say vec{u}=langle a_1, b_1, c_1rangle and vec{v}=langle a_2, b_2, c_2rangle, produces a vector that is perpendicular to both vec{u} and vec{v}. This can be shown using the definition of the cross product and the property that two vectors are orthogonal when their dot product is zero. The cross product is calculated as: vec{u} times vec{v} = begin{vmatrix} mathbf{i} & mathbf{j} & mathbf{k} a_1 & b_1 & c_1 a_2 & b_2 & c_2 end{vmatrix} = langle b_1c_2 - b_2c_1, -a_1c_2 + a_2c_1, a_1b_2 - a_2b_1 rangle. To prove that the cross product is perpendicular to vec{u}, we evaluate their dot product: vec{u} cdot (vec{u} times vec{v}) = (a_1, b_1, c_1) cdot (b_1c_2 - b_2c_1, -a_1c_2 + a_2c_1, a_1b_2 - a_2b_1). Expanding the dot product, we get: a_1(b_1c_2 - b_2c_1) + b_1(-a_1c_2 + a_2c_1) + c_1(a_1b_2 - a_2b_1). After simplification, all terms cancel out: 0 = 0. Therefore, vec{u} cdot (vec{u} times vec{v}) = 0, which implies that vec{u} is perpendicular to the cross product. Similarly, it can be shown that vec{v} is also perpendicular to the cross product. Thus, the cross product is a vector that is orthogonal to both vec{u} and vec{v}.