answer:1. **Identify Triangle and Heights:** Given triangle ABC with heights: AD perp BC , BE perp AC . 2. **Triangle Measurements:** [ AC = 16.25, quad BC = 13.75, quad HE = 6, quad HD = 3, quad text{and} quad b - a = 5 ] 3. **Area Calculation Using Heights:** Since the area of a triangle can be calculated with both heights AD and BE : [ S_{triangle ABC} = frac{1}{2} BC cdot AD = frac{1}{2} AC cdot BE ] Equating the two expressions for the area: [ BC cdot AD = AC cdot BE ] 4. **Substitute the Values:** [ BC = 13.75, quad AC = 16.25, quad AD = 3 + b, quad BE = 6 + a ] Thus: [ 13.75 cdot (3 + b) = 16.25 cdot (a + 6) ] 5. **Substitute the Relation b = a + 5 :** Given b = a + 5 , substitute into the equation: [ 13.75 cdot (3 + a + 5) = 16.25 cdot (a + 6) ] Simplify inside the parentheses: [ 13.75 cdot (a + 8) = 16.25 cdot (a + 6) ] 6. **Expand Both Sides:** [ 13.75a + 13.75 cdot 8 = 16.25a + 16.25 cdot 6 ] [ 13.75a + 110 = 16.25a + 97.5 ] 7. **Rearrange Terms to Isolate a:** [ 110 - 97.5 = 16.25a - 13.75a ] [ 12.5 = 2.5a ] 8. **Solve for a:** [ a = frac{12.5}{2.5} ] [ a = 5 ] 9. **Find b Using b = a + 5:** [ b = 5 + 5 ] [ b = 10 ] 10. **Calculate a + b:** [ a + b = 5 + 10 = 15 ] # Conclusion: [ boxed{D} ]

answer:Solution: (1) When n=1, a_1=S_1=2, When ngeqslant 2, a_n=S_n-S_{n-1} =n^2+n-(n-1)^2-(n-1) =2n, a_1=2 also fits the formula above, Therefore, the general formula for the sequence left{a_nright} is: a_n =2n; (2) From (1), we have: b_n=left( frac{1}{2}right)^{2n}+n=left( frac{1}{4}right)^n+n, Therefore, the sum of the first n terms of the sequence left{b_nright} T_n=left[left( frac{1}{4}right)^1+left( frac{1}{4}right)^2+cdots+left( frac{1}{4}right)^nright]+left(1+2+ldots+nright) = frac{ frac{1}{4}left[1-left( frac{1}{4}right)^nright]}{1- frac{1}{4}} + frac{n(n+1)}{2} = frac{1}{3}left[1-left( frac{1}{4}right)^nright]+ frac{n(n+1)}{2}. Thus, the final answers are: (1) The general formula for the term a_n is boxed{a_n = 2n}. (2) The sum of the first n terms of the sequence left{b_nright}, T_n, is boxed{T_n = frac{1}{3}left[1-left( frac{1}{4}right)^nright]+ frac{n(n+1)}{2}}.

answer:1. **Given:** - Let the complex number ( z = a + b mathrm{i} ) where ( a, b in mathbb{R} ) and ( b neq 0 ). 2. **Magnitude of ( |z| ):** - The magnitude (modulus) of ( z ) is given by: [ |z| = sqrt{a^2 + b^2} ] 3. **Magnitude of ( z^2 ):** - First, compute ( z^2 ): [ z^2 = (a + b mathrm{i})^2 ] Expanding using the distributive property: [ z^2 = a^2 + 2abmathrm{i} + (b mathrm{i})^2 ] Since ( mathrm{i}^2 = -1 ), we get: [ z^2 = a^2 + 2ab mathrm{i} - b^2 = a^2 - b^2 + 2ab mathrm{i} ] 4. **Magnitude of ( z^2 ):** - To find ( |z^2| ), we use the property that for any complex number ( z ), ( |z^2| = |z|^2 ). - We already have: [ |z| = sqrt{a^2 + b^2} ] Therefore: [ |z|^2 = (a^2 + b^2) ] 5. **Magnitude Relationship:** - We know from the modulus property of complex numbers that: [ |z^2| = |z|^2 ] Thus: [ |z^2| = |a^2 - b^2 + 2ab mathrm{i}| = sqrt{(a^2 - b^2)^2 + (2ab)^2} ] By computation: [ sqrt{(a^2 - b^2)^2 + (2ab)^2} = sqrt{a^4 - 2a^2b^2 + b^4 + 4a^2b^2} = sqrt{a^4 + 2a^2b^2 + b^4} = sqrt{(a^2 + b^2)^2} ] Thus: [ |z^2| = a^2 + b^2 ] So, we confirm: [ |z^2| = |z|^2 ] 6. **Conclusion:** - While ( |z^2| = |z|^2 = a^2 + b^2 ), we also found that ( z^2 = a^2 - b^2 + 2abmathrm{i} neq a^2 + b^2 ). - Thus, the correct choice is: [ boxed{text{A}} ]

answer:Let's denote the unknown number as "x". According to the problem, 15% of 40 is greater than 25% of x by 2. We can write this as an equation: 15% of 40 = 25% of x + 2 First, let's calculate 15% of 40: (15/100) * 40 = 0.15 * 40 = 6 Now, we have: 6 = (25/100) * x + 2 To find x, we need to isolate it on one side of the equation. Let's subtract 2 from both sides: 6 - 2 = (25/100) * x 4 = (25/100) * x Now, let's solve for x: x = 4 / (25/100) x = 4 / 0.25 x = 16 Therefore, the number is boxed{16} .