answer:Since for the function f(x) = sin(πx + φ), if there exists a non-zero real number t, such that for any x ∈ R, f(x + t) = f(x), Therefore, t is a period of the function. Hence, it could be t = frac{2π}{π} = 2. So, the answer is boxed{2}. First, determine that t is a period of the function, then use the periodicity of trigonometric functions to reach the conclusion. This problem primarily tests the understanding of the periodicity of trigonometric functions and is considered a basic question.

answer:1. **Identify and Name Points of Intersection:** Let ( K ) and ( M ) be the points of intersection of the line ( CB ) with the line ( AD ) and the circumcircle of triangle ( ACD ), respectively. 2. **Determine Key Angles:** Since (angle BCA = 10^circ), and given that ( AC perp BD ), we have the following angle calculations: [ angle MDA = angle MCA = 10^circ ] This implies that ( DM ) is the angle bisector of (angle KDB). 3. **Calculations Involving Other Angles:** The angle (angle BAD = 20^circ). Given (angle BAC = 40^circ), that implies: [ angle DAB = 180^circ - angle BAC - angle BAD = 180^circ - 40^circ - 20^circ = 120^circ ] Therefore: [ angle ABD = angle ABCD - angle DAB = 180^circ - 120^circ = 60^circ ] Furthermore: [ angle CBD = 80^circ, quad text{therefore} quad angle KBA = 50^circ ] 4. **Analyze Bisectors and Point of Incenter:** Since ( BA ) is the bisector of (angle KBD) and ( DM ) is the bisector of (angle KDB), let ( I ) be the point of intersection of these bisectors ( BA ) and ( DM ). Knowing that (angle BID = 120^circ), we can calculate (angle BKI): [ angle BKI = 60^circ ] 5. **Determine ( angle BDC ):** Since quadrilateral ( KAIM ) is circumscribed and given that ( KI ) is the angle bisector of (angle AKM). By the angle bisector theorem and properties of the inscribed angles: [ angle ACD = angle AMD = angle AMI = angle AKI = 30^circ ] Therefore: [ angle BDC = 60^circ ] # Conclusion: [ boxed{60^circ} ]

answer:Let mathbf{A} = begin{pmatrix} 1 & 3 & a 0 & 1 & 5 0 & 0 & 1 end{pmatrix}. Then we can write mathbf{A} = mathbf{I} + mathbf{B}, where [ mathbf{B} = begin{pmatrix} 0 & 3 & a 0 & 0 & 5 0 & 0 & 0 end{pmatrix}. ] Calculating mathbf{B}^2 and mathbf{B}^3: [ mathbf{B}^2 = begin{pmatrix} 0 & 3 & a 0 & 0 & 5 0 & 0 & 0 end{pmatrix} begin{pmatrix} 0 & 3 & a 0 & 0 & 5 0 & 0 & 0 end{pmatrix} = begin{pmatrix} 0 & 0 & 15 0 & 0 & 0 0 & 0 & 0 end{pmatrix}, ] [ mathbf{B}^3 = mathbf{B} mathbf{B}^2 = begin{pmatrix} 0 & 3 & a 0 & 0 & 5 0 & 0 & 0 end{pmatrix} begin{pmatrix} 0 & 0 & 15 0 & 0 & 0 0 & 0 & 0 end{pmatrix} = mathbf{0}. ] By the Binomial Theorem, [ mathbf{A}^n = mathbf{I} + n mathbf{B} + frac{n(n - 1)}{2} mathbf{B}^2 = begin{pmatrix} 1 & 3n & an + 5n + 7.5n(n-1) 0 & 1 & 5n 0 & 0 & 1 end{pmatrix}. ] Matching this to begin{pmatrix} 1 & 15 & 1010 0 & 1 & 25 0 & 0 & 1 end{pmatrix}, we have: 1. 3n = 15 Rightarrow n = 5. 2. 5n = 25 Rightarrow n = 5. 3. an + 5n + 7.5n(n-1) = 1010 Rightarrow a5 + 25 + 7.5 times 5 times 4 = 1010 Rightarrow 5a + 175 = 1010 Rightarrow a = 167. Thus, a + n = 167 + 5 = boxed{172}.

answer:Let's calculate the total cost of the coloring books first. Linda bought two coloring books at 4 each, so the total cost for the coloring books is: 2 coloring books * 4/coloring book = 8 Now, let's subtract the cost of the coloring books and the stuffed animal from the total amount Linda gave to the cashier to find out how much she spent on packs of peanuts. Total amount given to cashier = 25 Cost of coloring books = 8 Cost of stuffed animal = 11 Amount spent on packs of peanuts = Total amount given to cashier - (Cost of coloring books + Cost of stuffed animal) Amount spent on packs of peanuts = 25 - (8 + 11) Amount spent on packs of peanuts = 25 - 19 Amount spent on packs of peanuts = 6 Now, let's find out how many packs of peanuts Linda bought. Each pack of peanuts costs 1.50, so we divide the amount spent on peanuts by the cost per pack: Number of packs of peanuts = Amount spent on packs of peanuts / Cost per pack Number of packs of peanuts = 6 / 1.50 Number of packs of peanuts = 4 Linda bought boxed{4} packs of peanuts.