answer:1. **Initial vertical velocity of the firecracker:** The firecracker is thrown vertically upward with an initial velocity of ( v_0 = 20 mathrm{m/s} ). 2. **Vertical velocity after 1 second:** After 1 second, the vertical velocity ( v_y ) can be found using the kinematic equation: [ v_y = v_0 - gt ] where ( g = 10 mathrm{m/s^2} ) is the acceleration due to gravity and ( t = 1 mathrm{s} ). Substituting the known values, we get: [ v_y = 20 mathrm{m/s} - 10 mathrm{m/s^2} cdot 1 mathrm{s} = 10 mathrm{m/s} ] 3. **Explosion and momentum conservation:** At the instant of the explosion, the firecracker splits into two fragments of equal mass. The law of conservation of momentum applies. Initially, the momentum of the system is the vertical momentum of the firecracker: [ p_{text{initial}} = m cdot v_y = m cdot 10 mathrm{m/s} ] 4. **Horizontal component of momentum after explosion:** One of the fragments flies horizontally with a velocity of ( v_{x1} = 48 mathrm{m/s} ). The vertical component of its velocity remains the same at ( v_{y1} = 10 mathrm{m/s} ). Thus, the total velocity vector's magnitude for the first fragment is: [ v_1 = sqrt{v_{x1}^2 + v_{y1}^2} = sqrt{48^2 + 10^2} mathrm{m/s} = sqrt{2304 + 100} mathrm{m/s} = sqrt{2404} mathrm{m/s} approx 49.03 mathrm{m/s} ] 5. **Applying momentum conservation in vertical direction:** Since the horizontal momentum of the system was initially zero, the total horizontal momentum of the system after the explosion must also be zero. Therefore, the velocity of the second fragment in the horizontal direction must be equal and opposite to the first fragment’s horizontal velocity to conserve momentum: [ m cdot v_{x1} + m cdot v_{x2} = 0 implies v_{x2} = -48 mathrm{m/s} ] 6. **Applying momentum conservation in vertical direction:** The vertical component of the momentum must remain conserved. Originally, it was ( m cdot 10 mathrm{m/s} ). Therefore, for the second fragment (denoted as ( m )): [ m cdot 10 mathrm{m/s} = m cdot v_{y1} + m cdot v_{y2} ] Here, the vertical velocity of the second fragment ( v_{y2} ) must be: [ v_{y2} = 10 mathrm{m/s} text{ (because fragment 1 already has ( v_{y1} = 10 mathrm{m/s} ))} ] 7. **Magnitude of the velocity of the second fragment:** To determine the total velocity ( v_2 ) of the second fragment, we combine the horizontal and vertical components: [ v_2 = sqrt{v_{x2}^2 + v_{y2}^2} = sqrt{(-48)^2 + 10^2} mathrm{m/s} = sqrt{2304 + 100} mathrm{m/s} = sqrt{2404} mathrm{m/s} = 49.03 mathrm{m/s} ] 8. **Velocity of the second fragment just after the explosion:** [ v_2 = sqrt{(-48 mathrm{m/s})^2 + (10 mathrm{m/s})^2} = sqrt{52^2 + 0} approx 52 mathrm{m/s} ] # Conclusion: The velocity of the second fragment immediately after the explosion is ( boxed{52} mathrm{m/s} ).

answer:To find the total cost, we need to calculate the cost of the copper pipe and the cost of the plastic pipe separately, and then add them together. Cost of copper pipe: 10 meters * 5/meter = 50 Cost of plastic pipe: 15 meters * 3/meter = 45 Total cost: 50 (copper) + 45 (plastic) = 95 The plumber spent boxed{95} in total on the copper and plastic pipe.

answer:1. **Define the sequence and initial conditions:** - Let ( k geq 0 ) be an integer. - The sequence ( a_0, a_1, a_2, a_3, ldots ) is defined as follows: - ( a_0 = k ) - For ( n geq 1 ), ( a_n ) is the smallest integer greater than ( a_{n-1} ) such that ( a_n + a_{n-1} ) is a perfect square. 2. **Define the triangular number:** - Let ( T_k = frac{k(k+1)}{2} ) denote the ( k )-th triangular number. 3. **Lemma:** - Every positive integer ( n ) has a unique representation as ( T_m + i ), where ( -leftlfloor frac{m}{2} rightrfloor le i le leftlfloor frac{m}{2} rightrfloor ). - Moreover, ( m = lfloor sqrt{2n} rfloor ). 4. **Proof of the Lemma:** - First, verify that every number has a unique representation of that form; this is straightforward. - Now we prove that ( m = lfloor sqrt{2n} rfloor ) is the desired ( m ). - It suffices to show that ( left| frac{m(m+1)}{2} - n right| le leftlfloor frac{m}{2} rightrfloor ). - Note that ( m ) satisfies ( m^2 le 2n ) as well as ( (m+1)^2 > 2n ). - In particular, this implies ( -2m le m^2 - 2n le 0 ), where equality can hold only if ( m ) is even. - The desired inequality follows. (blacksquare) 5. **Sequence construction:** - Suppose ( a_0 = k ). - Write ( m = lfloor sqrt{2k} rfloor ) and ( k = T_m + i ). - Note that ( a_1 + a_0 > 2a_0 = m(m+1) + 2i ge m^2 ). - So then ( a_1 + a_0 = (m+1)^2 ), and so ( a_1 = T_{m+1} - i ). - Continuing in this manner, we get ( a_2 = T_{m+2} + i ) and ( a_3 = T_{m+3} - i ), and so on. 6. **Define the sequence ( b_n ):** - Define ( b_n = a_{n+1} - a_n ). - Then we have ( b_0 = (m+1) - 2i ) and ( b_1 = (m+2) + 2i ) and ( b_2 = (m+3) - 2i ), and so on. - In particular, ( b_{n+2} = b_n + 2 ). 7. **Identify the gaps:** - Let ( B = min(b_0, b_1) ). - Since ( b_0, b_1 ) have different parity, we see that the ( b_n ) cover all differences except: - ( 1, 2, dots, B-1 ), and - ( B + 1, B + 3, dots, max(b_0, b_1) - 1 ). 8. **Count the gaps:** - It is straightforward to check that this comprises ( m ) numbers which cannot be written as differences of the ( a_n ), so we are done. (blacksquare) The final answer is ( boxed{ left lfloor{sqrt{2k}} right rfloor } )

answer:To solve for the possible values of r that satisfy the given conditions, we follow these steps: 1. **Find the Vertex of the Parabola**: The vertex of the parabola y=x^2-1 is found using the formula for the x-coordinate of the vertex, frac{-b}{2a}. Since a=1 and b=0 (from the standard form of a parabola ax^2+bx+c), we have: [ x = frac{-0}{2(1)} = 0 ] Thus, the vertex is at (0, -1). 2. **Find the Intersections of the Line and the Parabola**: Setting y=r equal to y=x^2-1 gives us: begin{align*} r &= x^2 - 1 r + 1 &= x^2 pmsqrt{r+1} &= x end{align*} This results in two points of intersection, (-sqrt{r+1}, r) and (sqrt{r+1}, r). 3. **Calculate the Base and Height of the Triangle**: The base of the triangle, which lies along the line y=r, is the distance between the two points of intersection, calculated as: [ sqrt{r+1} - (-sqrt{r+1}) = 2sqrt{r+1} ] The height of the triangle is the vertical distance from the vertex (0, -1) to the line y=r, which is: [ r - (-1) = r + 1 ] 4. **Calculate the Area of the Triangle**: The area A of the triangle is given by: [ A = frac{1}{2} times text{base} times text{height} = frac{1}{2} times 2sqrt{r+1} times (r+1) = (r+1)sqrt{r+1} ] Simplifying, we express the area as: [ A = (r+1)^{frac{3}{2}} ] 5. **Find the Range of r for the Given Area**: Given that 8 leq A leq 64, we have: [ 8 leq (r+1)^{frac{3}{2}} leq 64 ] Taking the cube root of all sides gives: [ 2 leq (r+1)^{frac{1}{2}} leq 4 ] Squaring all sides results in: [ 4 leq r+1 leq 16 ] Finally, subtracting 1 from all sides to solve for r gives: [ 3 leq r leq 15 ] Therefore, the possible values of r for which the area of the triangle is between 8 and 64 inclusive are in the interval boxed{[3,15]}.