answer:To simplify the expression sqrt{8} - sqrt{2}, we follow these steps:1. Recognize that sqrt{8} can be simplified further because 8 is a perfect square times a smaller number. Specifically, 8 = 4 times 2, so sqrt{8} = sqrt{4 times 2}. Since the square root of 4 is 2, this simplifies to 2sqrt{2}.2. Substitute the simplified form of sqrt{8} into the original expression: [ sqrt{8} - sqrt{2} = 2sqrt{2} - sqrt{2}. ]3. Notice that both terms involve sqrt{2}. This allows us to combine like terms by subtracting their coefficients, just as we would with algebraic terms. In this case, we have 2 times sqrt{2} minus 1 times sqrt{2}, which simplifies to (2-1)sqrt{2}.4. Perform the subtraction in the coefficient: [ (2-1)sqrt{2} = 1sqrt{2}. ]5. Recognize that 1sqrt{2} is simply sqrt{2}, as multiplying by 1 does not change the value of an expression.Therefore, the simplified form of sqrt{8} - sqrt{2} is boxed{sqrt{2}}.

answer:To solve the given problem, we start by analyzing the general term of the series, which is frac{k}{(k+1)!}. We can rewrite this term by adding and subtracting 1 in the numerator as follows:[frac{k}{(k+1)!} = frac{k+1-1}{(k+1)!} = frac{(k+1)-1}{(k+1)!}.]This allows us to split the fraction into two parts:[frac{(k+1)-1}{(k+1)!} = frac{k+1}{(k+1)!} - frac{1}{(k+1)!} = frac{1}{k!} - frac{1}{(k+1)!}.]Applying this transformation to each term of the series, we get:[begin{align*}frac{1}{2!} + frac{2}{3!} + frac{3}{4!} + cdots + frac{n}{(n+1)!} &= left(frac{1}{1!} - frac{1}{2!}right) + left(frac{1}{2!} - frac{1}{3!}right) + left(frac{1}{3!} - frac{1}{4!}right) + cdots + left(frac{1}{n!} - frac{1}{(n+1)!}right) &= 1 - frac{1}{2!} + frac{1}{2!} - frac{1}{3!} + frac{1}{3!} - frac{1}{4!} + cdots + frac{1}{n!} - frac{1}{(n+1)!}.end{align*}]Notice that all terms except the first and the last cancel out in a telescoping manner:[1 - frac{1}{2!} + frac{1}{2!} - frac{1}{3!} + frac{1}{3!} - frac{1}{4!} + cdots + frac{1}{n!} - frac{1}{(n+1)!} = 1 - frac{1}{(n+1)!}.]Therefore, the sum of the series is:[boxed{1 - frac{1}{(n+1)!}}.]

answer:1. The answer is mathbf{( B )}.We observe that 45=3^{2} cdot 5 certainly divides N=3^{5} cdot 4^{4} cdot 5^{3}. We also note that 42 is not a divisor of N because it is a multiple of 7, while N is not divisible by 7; the same applies to 105=7 cdot 15. Similarly: 52 is a multiple of 13, 85 is a multiple of 17, and neither 13 nor 17 divide N, because they are prime numbers that do not appear in the prime factorization of N.[Problem proposed by A. Colesanti.]

answer:We use simple case work to solve this problem.Case 1: even, even, even = 4 cdot 5 cdot 5 = 100Case 2: even, odd, odd = 4 cdot 5 cdot 5 = 100Case 3: odd, even, odd = 5 cdot 5 cdot 5 = 125Case 4: odd, odd, even = 5 cdot 5 cdot 5 = 125Simply sum up the cases to get your answer. 100 + 100 + 125 + 125 = boxed{textbf{(D)~}450}.- Wesseywes7254

answer:1. Understanding the Problem: We are given three sets of lines: [ y = k, quad y = sqrt{3}x + 2k, quad y = -sqrt{3}x + 2k ] where ( k ) ranges from (-10) to (10). These lines form equilateral triangles with side length (frac{2}{sqrt{3}}).2. Finding Intersections: To form equilateral triangles, we need to find the intersections of these lines. The horizontal lines ( y = k ) intersect with the slanted lines ( y = sqrt{3}x + 2a ) and ( y = -sqrt{3}x + 2b ).3. Intersection Points: Consider the intersection of ( y = sqrt{3}x + 2a ) and ( y = -sqrt{3}x + 2b ): [ sqrt{3}x + 2a = -sqrt{3}x + 2b ] Solving for ( x ): [ 2sqrt{3}x = 2b - 2a implies x = frac{b - a}{sqrt{3}} ] Substituting ( x ) back into one of the equations to find ( y ): [ y = sqrt{3} left( frac{b - a}{sqrt{3}} right) + 2a = b + a ] Thus, the intersection point is ( left( frac{b - a}{sqrt{3}}, a + b right) ).4. Range of ( y ): The value of ( y ) must be within the range of the horizontal lines ( y = k ), which are from ( k = -10 ) to ( k = 10 ). Since the height of an equilateral triangle with side length (frac{2}{sqrt{3}}) is (1), the possible values of ( y ) for the vertices of the triangles must be between ( -9 ) and ( 11 ).5. Counting the Intersections: We need to count the number of pairs ((a, b)) such that ( a + b ) falls within the range ([-9, 11]). - For ( y = 11 ), ( a ) can range from ( 10 ) to ( 1 ), giving ( 10 ) possible pairs. - For ( y = 10 ), ( a ) can range from ( 10 ) to ( 0 ), giving ( 11 ) possible pairs. - This pattern continues until ( y = 0 ), where there are ( 21 ) possible pairs. - Then, the number of pairs decreases symmetrically until ( y = -9 ), where there are ( 12 ) possible pairs. Summing these, we get: [ 10 + 11 + 12 + ldots + 21 + ldots + 12 = 330 ]6. Triangles Facing Downward: The same logic applies for triangles facing downward, giving another ( 330 ) triangles.7. Total Number of Triangles: The total number of equilateral triangles is: [ 2 times 330 = 660 ]The final answer is (boxed{660})

answer:Solution.begin{gathered}text { 1) } P(x, y)=yleft(e^{x y}+6right) ; frac{partial P}{partial y}=left(1+x y e^{x y}right)+6 Q(x, y)=xleft(e^{x y}+6right)end{gathered} begin{aligned}frac{partial Q}{partial x}=6+left(1+x y e^{x y}right) ; frac{partial P}{partial y}=frac{partial Q}{partial x} ; text { for all }(x, y) in E^{2} text { 2) } f(x, y)=int P d x=e^{x y}+6 x y+U(y) ; f(x, y)= =int Q d x=e^{x y}+6 x y+V(x)end{aligned}Since all known terms in the two found expressions for f(x, y) coincide, then, consequently,f(x, y)=e^{x y}+6 x y+C, quad(C=text { const })## 3.4. DIFFERENTIATION OF COMPOSITE AND IMPLICIT FUNCTIONSLet the function fleft(x_{1}, x_{2}, ldots, x_{n}right) be differentiable at the point Pleft(x_{1}^{0}, x_{2}^{0}, ldots, x_{n}^{0}right), and the functions x_{i}=varphi_{i}left(t_{1}, t_{2}, ldots, t_{m}right),(i=1, 2, ldots, k) be differentiable at the point M_{0}left(t_{1}^{0}, t_{2}^{0}, ldots, t_{m}^{0}right) and such that x_{i}^{0}=varphi_{i}left(t_{1}^{0}, t_{2}^{0}, ldots, t_{m}^{0}right). Then the composite function y= =fleft(x_{1}, x_{2}, ldots, x_{n}right), x_{i}=varphi_{i}left(t_{1}, t_{2}, ldots, t_{m}right), i=1,2, ldots, k is differentiable at the point M_{0} and its partial derivatives are determined by the formulafrac{partial f}{partial t_{j}}=sum_{i=1}^{n} frac{partial f}{partial x_{i}} frac{partial x_{i}}{partial t_{j}}, j=1,2, ldots, mwhere all partial derivatives frac{partial f}{partial x_{i}} are computed at the point P_{0}, and all partial derivatives frac{partial x_{i}}{partial t_{j}} - at the point M_{0}.From formula (3.12), some important formulas for differentiating composite functions of several variables follow.1) u=f(x, y, z, t), where x=x(t), y=y(t), z=z(t). In this case, by setting in (3.12) n=4, m=1, x_{1}=x, x_{2}=y, x_{3}= =z, x_{4}=t, we obtain the following formula for the total (or material) derivative of the function f(x, y, z, t) :frac{d f}{d t}=frac{partial f}{partial t}+frac{partial f}{partial x} frac{d x}{d t}+frac{partial f}{partial y} frac{d y}{d t}+frac{partial f}{partial z} frac{d z}{d t}2) z=f(u, v), where u=u(x, y), v=v(x, y). The formulas for the partial derivatives f_{x}^{prime}, f_{y}^{prime} are obtained from (3.12) by settingbegin{gathered}x_{1}=u, x_{2}=v, t_{1}=x, t_{2}=y n=2, m=2: frac{partial f}{partial x}=frac{partial f}{partial u} frac{partial u}{partial x}+frac{partial f}{partial v} frac{partial v}{partial x} frac{partial f}{partial y}=frac{partial f}{partial u} frac{partial u}{partial y}+frac{partial f}{partial v} frac{partial v}{partial y}end{gathered}3) f=f(u, v, w), u=u(x, y, z), v=v(x, y, z), w=w(x, y, z). The formulas for the partial derivatives f_{x}^{prime}, f_{y}^{prime} f_{z}^{prime} are obtained from (3.12) similarly to case 2, setting n=3, m=3 and adding the corresponding variables x_{3}=w, t_{3}=z :begin{aligned}& frac{partial f}{partial x}=frac{partial f}{partial u} frac{partial u}{partial x}+frac{partial f}{partial v} frac{partial v}{partial x}+frac{partial f}{partial w} frac{partial w}{partial x} & frac{partial f}{partial y}=frac{partial f}{partial u} frac{partial u}{partial y}