answer:Given |vec{a}| = 5, |vec{b}| = 3, and vec{a} cdot vec{b} = -9, to find the magnitude of the projection of vec{a} onto vec{b}, we use the definition of the scalar projection. The scalar projection of vec{a} onto vec{b}, denoted as text{proj}_{vec{b}}vec{a}, is given by: text{proj}_{vec{b}}vec{a} = frac{vec{a} cdot vec{b}}{|vec{b}|}. Substituting the given values into the formula: text{proj}_{vec{b}}vec{a} = frac{-9}{3} = -3. Thus, the magnitude of the projection is -3. In this context, the negative sign indicates that the direction of the projection is opposite to the direction of vec{b}. So, the magnitude of the projection of vec{a} onto vec{b} is boxed{-3}. This problem is an application of vector operations in the plane. To solve it, one must apply the definition of the scalar projection to find the answer, which is a fundamental concept in vector calculus.

answer:To solve the problem, we start with the given equation: [ frac{x}{3} = 50 + frac{x}{4}. ] We want to isolate x, so we'll first get all the x terms on one side. To do this, we subtract frac{x}{4} from both sides: [ frac{x}{3} - frac{x}{4} = 50. ] To combine the fractions, we find a common denominator, which is 12. This gives us: [ frac{4x}{12} - frac{3x}{12} = 50. ] Simplifying the left side by combining like terms: [ frac{x}{12} = 50. ] To solve for x, we multiply both sides by 12: [ x = 50 times 12. ] This simplifies to: [ x = 600. ] Therefore, the number we were looking for is boxed{600}.

answer:Since a=int_{0}^{pi}(sin x+cos x)dx=2, the general term T_{r+1}=(-1)^{r} C_{6}^{6-r}(2 sqrt {x})^{6-r} frac {1}{ sqrt {x}^{r}}=(-1)^{r}C_{6}^{r}2^{6-r}x^{3-r} Let 3-r=2, we get r=1, therefore, the coefficient of the term containing x^{2} in the expansion is -192. Hence, the answer is boxed{-192}. By using the properties of definite integrals, we can find the value of a, and then by applying the binomial theorem to expand the binomial (a sqrt {x}- frac {1}{ sqrt {x}})^{6}, setting the power series of x to 2, we solve for r, thereby finding the solution. This problem tests the basic methods of calculating simple definite integrals and finding the specified term in the expansion of a binomial expression.

answer:1. We start with the square (ABCD) and note that point (E) lies on the segment ([BD]) such that (EB = AB). Given that (ABCD) is a square, this implies (EB = AB = BC), making (E) the midpoint of (BD). 2. Since (ABCD) is a square, each of its internal angles is (90^circ). 3. We observe that (EB = AB). Therefore, triangles (ABE) and (CEB) are both isosceles triangles with (AB = BE = EB) and (CB = BE), respectively. This occurs because (BE) is half of the diagonal of the square due to point (E) being the midpoint. 4. In triangle (ABE): - We know that (angle ABE = 45^circ), since the angles in the square meet at (90^circ) and the triangle (ABE) being isosceles divides it evenly. - Given that the sum of angles in any triangle adds up to (180^circ), we calculate the remaining angles: [ angle ABE + 2 cdot angle AEB = 180^circ ] [ 45^circ + 2 cdot angle AEB = 180^circ ] [ 2 cdot angle AEB = 135^circ ] [ angle AEB = frac{135^circ}{2} = 67.5^circ ] 5. Similarly, in triangle (CEB): - We know that (angle CBE = 45^circ), for the same reason as above. - Using the triangle sum property, we have: [ angle CBE + 2 cdot angle CEB = 180^circ ] [ 45^circ + 2 cdot angle CEB = 180^circ ] [ 2 cdot angle CEB = 135^circ ] [ angle CEB = frac{135^circ}{2} = 67.5^circ ] 6. From points 4 and 5, we observe that (angle AEB = angle CEB = 67.5^circ). 7. Finally, considering the alignment of points (F), (E), and (C) on the line (CE) intersecting with line (AD) at point (F), we calculate the angle (angle FEA). This involves knowing: - (angle AEC = angle AEB + angle CEB = 67.5^circ + 67.5^circ = 135^circ). - Therefore, [ angle FEA = 180^circ - angle AEC = 180^circ - 135^circ = 45^circ ] # Conclusion: [ boxed{45^circ} ]