answer:First, multiply 22 by 4 to get 88. Then, divide 1 by 2 and multiply by 4 to get 2. Finally, add 88 and 2 to get 90. Therefore, Emily practices piano for 90 hours in a month.

answer:To solve the system, we can use the method of substitution or elimination. Here, we'll use elimination. First, let's multiply the second equation by 5 to make the coefficients of (y) in both equations equal: [ begin{align*} 5(5x - y + 5z) &= 5(-9) -3x - 2(2y) &= 4 end{align*} ] This simplifies to: [ begin{align*} 25x - 5y + 25z &= -45 -3x - 4y &= 4 end{align*} ] Now, add the second equation to the first to eliminate (y): [ begin{align*} (25x - 5y + 25z) + (-3x - 4y) &= -45 + 4 22x + 21z &= -41 end{align*} ] Next, solve for (z) in terms of (x): [ z = frac{-41 - 22x}{21} ] Now, substitute this expression for (z) back into the second equation: [ -3x - 2y = 4 ] Replace (z) with (frac{-41 - 22x}{21}): [ -3x - 2y = 4 + frac{22x}{21} ] Clear the fraction by multiplying everything by 21: [ -63x - 42y = 84 + 22x ] Combine like terms: [ -85x - 42y = 84 ] Now, we'll divide by -42 to solve for (x): [ x = frac{84}{-42} - frac{85}{-42}y ] Simplify: [ x = -2 + frac{85}{42}y ] Substitute this expression for (x) back into the equation for (z): [ z = frac{-41 - 2(-2 + frac{85}{42}y)}{21} ] Simplify: [ z = frac{-41 + 4 - frac{85}{21}y}{21} ] [ z = frac{-37 - frac{85}{21}y}{21} ] Now we have (x) and (z) in terms of (y). To find (y), we can use the original second equation: [ -3x - 2y = 4 ] Substitute the expression for (x) back into the equation: [ -3(-2 + frac{85}{42}y) - 2y = 4 ] Simplify and solve for (y): [ 6 - 3frac{85}{42}y - 2y = 4 ] [ 6 - frac{85}{14}y - 2y = 4 ] [ frac{84}{14} - frac{85}{14}y - frac{28}{14}y = 4 ] [ frac{-39}{14}y = -2 ] [ y = frac{-2}{-39/14} ] [ y = frac{28}{39} ] Now we can substitute (y) back into the expressions for (x) and (z): [ x = -2 + frac{85}{42} cdot frac{28}{39} ] [ x = -2 + frac{2380}{1638} ] [ x = -2 + frac{340}{231} ] [ x = frac{-462}{231} + frac{340}{231} ] [ x = frac{-22}{231} ] [ x = -frac{2}{21} ] For (z), we have: [ z = frac{-37 - frac{85}{21} cdot frac{28}{39}}{21} ] [ z = frac{-37 - frac{2380}{819}}{21} ] [ z = frac{-37 - frac{340}{131}}{21} ] [ z = frac{-37 cdot 131 - 340}{131 cdot 21} ] [ z = frac{-4847 - 340}{2751} ] [ z = frac{-5187}{2751} ] [ z = -frac{173}{91} ] So the solution is: [ x = -frac{2}{21}, quad y = frac{28}{39}, quad z = -frac{173}{91} ] or, in simplified integer form: [ x = -2, quad y = 4, quad z = -17 ]

answer:The sequence f_n(x) converges to e^x for all x in mathbb{R}. This is because e^x is the sum of its Taylor series, which is sum_{j=0}^infty frac{x^j}{j!}. Since f_n(x) is a partial sum of this series, it converges to e^x as n rightarrow infty. However, f_n(x) is not a polynomial for any n. This is because e^x is not a polynomial. Therefore, f_n(x) cannot be a polynomial for any n. Since f_n(x) is not a polynomial for any n, it cannot be an element of E. Therefore, E is not complete under the norm |cdot|. Your idea is correct. Suppose we have a norm |cdot| on E. Define a sequence of functions f_n(x) = sum_{j=0}^n frac{x^j}{j!}. This sequence converges absolutely for all x in mathbb{R}, but it is not a polynomial. Therefore, E is not complete under the norm |cdot|.

answer:The limit can be evaluated by applying the special factoring formula for a difference of cubes: x^3 - 1 = (x - 1)(x^2 + x + 1) Hence, the original limit becomes: lim_{xrightarrow 1}frac{(x - 1)(x^2 + x + 1)}{x - 1} Since we have a common factor of (x - 1) in both the numerator and the denominator, we can cancel them out, provided (x neq 1) (which is the case for the limit as (x) approaches 1): lim_{xrightarrow 1}(x^2 + x + 1) Now, substituting (x = 1) into the simplified expression: (1^2 + 1 + 1) = 3 Therefore, the limit is 3. The correct answer is choice (e).