answer:Given the function: [ g(x) = (alpha + |x|)^2 e^{(5 - |x|)^2} ] We need to find: i) All values of (alpha) for which (g(x)) is continuous for all (x in mathbb{R}). ii) All values of (alpha) for which (g(x)) is differentiable for all (x in mathbb{R}). First, let's rewrite (g(x)) in piecewise form: [ g(x) = begin{cases} (alpha + x)^2 e^{(5 - x)^2} & text{if } x > 0 (alpha - x)^2 e^{(5 + x)^2} & text{if } x < 0 alpha^2 e^{5^2} & text{if } x = 0 end{cases} ] # Part i) Continuity To check the continuity of (g(x)) at (x = 0), we need to ensure that: [ lim_{x to 0^+} g(x) = g(0) quad text{and} quad lim_{x to 0^-} g(x) = g(0) ] 1. **Right-hand limit**: [ lim_{x to 0^+} g(x) = lim_{x to 0^+} (alpha + x)^2 e^{(5 - x)^2} ] As (x to 0^+): [ (alpha + x)^2 to alpha^2 ] [ e^{(5 - x)^2} to e^{5^2} ] Therefore: [ lim_{x to 0^+} g(x) = alpha^2 e^{5^2} ] 2. **Left-hand limit**: [ lim_{x to 0^-} g(x) = lim_{x to 0^-} (alpha - x)^2 e^{(5 + x)^2} ] As (x to 0^-): [ (alpha - x)^2 to alpha^2 ] [ e^{(5 + x)^2} to e^{5^2} ] Therefore: [ lim_{x to 0^-} g(x) = alpha^2 e^{5^2} ] 3. **Value at (x = 0)**: [ g(0) = alpha^2 e^{5^2} ] Since: [ lim_{x to 0^+} g(x) = lim_{x to 0^-} g(x) = g(0) ] (g(x)) is continuous at (x = 0) for all (alpha in mathbb{R}). # Part ii) Differentiability To check the differentiability of (g(x)) at (x = 0), we need to ensure that: [ g'(0^+) = g'(0^-) ] 1. **Right-hand derivative**: [ g'(x) = frac{d}{dx} left[ (alpha + x)^2 e^{(5 - x)^2} right] ] Using the product rule: [ g'(x) = 2(alpha + x) e^{(5 - x)^2} + (alpha + x)^2 cdot frac{d}{dx} e^{(5 - x)^2} ] [ frac{d}{dx} e^{(5 - x)^2} = e^{(5 - x)^2} cdot (-2(5 - x)) = -2(5 - x) e^{(5 - x)^2} ] Therefore: [ g'(x) = 2(alpha + x) e^{(5 - x)^2} - 2(alpha + x)^2 (5 - x) e^{(5 - x)^2} ] As (x to 0^+): [ g'(0^+) = 2alpha e^{5^2} - 10alpha^2 e^{5^2} ] 2. **Left-hand derivative**: [ g'(x) = frac{d}{dx} left[ (alpha - x)^2 e^{(5 + x)^2} right] ] Using the product rule: [ g'(x) = 2(alpha - x) e^{(5 + x)^2} + (alpha - x)^2 cdot frac{d}{dx} e^{(5 + x)^2} ] [ frac{d}{dx} e^{(5 + x)^2} = e^{(5 + x)^2} cdot 2(5 + x) = 2(5 + x) e^{(5 + x)^2} ] Therefore: [ g'(x) = 2(alpha - x) e^{(5 + x)^2} + 2(alpha - x)^2 (5 + x) e^{(5 + x)^2} ] As (x to 0^-): [ g'(0^-) = 2alpha e^{5^2} + 10alpha^2 e^{5^2} ] For (g'(0)) to exist: [ g'(0^+) = g'(0^-) ] [ 2alpha e^{5^2} - 10alpha^2 e^{5^2} = 2alpha e^{5^2} + 10alpha^2 e^{5^2} ] [ -10alpha^2 e^{5^2} = 10alpha^2 e^{5^2} ] [ -10alpha^2 = 10alpha^2 ] [ -20alpha^2 = 0 ] [ alpha = 0 ] Therefore, (g(x)) is differentiable for all (x in mathbb{R}) if and only if (alpha = 0). The final answer is ( boxed{ alpha in mathbb{R} } ) for continuity and (alpha = 0) for differentiability.

answer:Let's denote the sequence of ten natural numbers as (a_1, a_2, a_3, ldots, a_{10}). According to the problem, - (a_3 = a_1 + a_2) - (a_4 = a_2 + a_3) - (a_5 = a_3 + a_4) - ... - (a_n = a_{n-1} + a_{n-2}) for (n geq 3). It is given: - (a_7 = 42) - (a_9 = 110) 1. **Find (a_8):** [ a_8 = a_9 - a_7 = 110 - 42 = 68 ] 2. **Find (a_6):** [ a_6 = a_8 - a_7 = 68 - 42 = 26 ] 3. **Find (a_5):** [ a_5 = a_7 - a_6 = 42 - 26 = 16 ] 4. **Find (a_4):** [ a_4 = a_6 - a_5 = 26 - 16 = 10 ] Thus, the fourth number in the sequence is (10). # Conclusion: [ boxed{10} ]

answer:To solve the problem {(frac{sqrt{3}}{2}+frac{i}{2})}^{2023}, we first simplify the expression inside the parentheses to understand its behavior when raised to a power. Step 1: Square the expression to find a pattern. [ begin{align*} left(frac{sqrt{3}}{2}+frac{i}{2}right)^2 &= left(frac{sqrt{3}}{2}right)^2 + 2cdotfrac{sqrt{3}}{2}cdotfrac{i}{2} + left(frac{i}{2}right)^2 &= frac{3}{4} + frac{sqrt{3}}{2}i - frac{1}{4} &= frac{2}{4} + frac{sqrt{3}}{2}i &= frac{1}{2} + frac{sqrt{3}}{2}i. end{align*} ] Step 2: Multiply by i to simplify further. [ begin{align*} ileft(frac{sqrt{3}}{2}+frac{i}{2}right)^2 &= ileft(frac{1}{2} + frac{sqrt{3}}{2}iright) &= frac{i}{2} - frac{sqrt{3}}{2} &= -frac{sqrt{3}}{2} + frac{i}{2}. end{align*} ] This shows that left(frac{sqrt{3}}{2}+frac{i}{2}right)^2 = frac{1}{i}left(-frac{sqrt{3}}{2} + frac{i}{2}right). Step 3: Calculate the cube of the original expression. [ begin{align*} left(frac{sqrt{3}}{2}+frac{i}{2}right)^3 &= frac{1}{i}left(-frac{sqrt{3}}{2} + frac{i}{2}right)left(frac{sqrt{3}}{2}+frac{i}{2}right) &= frac{1}{i}left(-frac{3}{4} - frac{1}{4}right) &= -frac{1}{i}. end{align*} ] Step 4: Use the cube to find the 2023rd power. [ begin{align*} left(frac{sqrt{3}}{2}+frac{i}{2}right)^{2023} &= left[left(frac{sqrt{3}}{2}+frac{i}{2}right)^3right]^{674} times left(frac{sqrt{3}}{2}+frac{i}{2}right) &= left(-frac{1}{i}right)^{674} times left(frac{sqrt{3}}{2}+frac{i}{2}right) &= left[left(frac{1}{-i}right)^2right]^{337} times left(frac{sqrt{3}}{2}+frac{i}{2}right) &= (-1)^{337} times left(frac{sqrt{3}}{2}+frac{i}{2}right) &= -left(frac{sqrt{3}}{2}+frac{i}{2}right) &= -frac{sqrt{3}}{2}-frac{i}{2}. end{align*} ] Therefore, the correct answer is boxed{text{B: } -frac{sqrt{3}}{2}-frac{i}{2}}.

answer:1. **Identify Sides and Hypotenuse:** Since angle B = 90^circ, AB and BC are the legs, AC (hypotenuse) can be calculated using AC = sqrt{AB^2 + BC^2} = sqrt{6^2 + 8^2} = sqrt{36 + 64} = sqrt{100} = 10. 2. **Calculate the Inradius r:** The inradius r of a right triangle is given by [ r = frac{A}{s} ] where A is the area of the triangle and s is the semi-perimeter. The area A = frac{1}{2}ab = frac{1}{2} times 6 times 8 = 24 and the semi-perimeter s = frac{a+b+c}{2} = frac{6+8+10}{2} = 12. Thus, [ r = frac{24}{12} = 2 ] 3. **Determine the Length of Segment BI:** In a right triangle, the incenter is always located at r distance vertically upward from the hypotenuse and horizontally from the right-angled corner. Thus we must determine the length of BI. Due to the formation of an isosceles right triangle triangle BID (where D is the point where the inradius meets AC), this length is rsqrt{2} = 2sqrt{2}. Hence, BI = boxed{2sqrt{2}}.