answer:Let us analyze the given system of equations: ``` A B C + D E F + G E ------- G E F G E F + D E ------- H F J ``` Our goal is to determine the digits represented by the letters ( A, B, C, D, E, F, G, H, ) and ( J ). 1. **Determining ( C )**: - The first simplification we can make is observing the last digit in each row of the addition: [ C + F + E = F ] - This implies ( C = 0 ), because only then can ( C + F + E = F ) hold true (adding zero retains the digit on the right-hand side). 2. **Find conditions for ( F ) and ( E )**: - Observing the next column, from the right: [ B + E + E = E + 10 ] - Therefore, ( B + E = 10 ). 3. **Considering the second equation**: - Working on the addition: [ 2F text{ is greater than or equal to } 10 ] - Consequently, ( F + F = 10 + (carry) ). We realize that ( E ) contributes to the carry-over, and ( F = E + 1 ). 4. **Exploring possible values of ( F )**: - ( F neq 9 ), as it would force ( E ) to be ( 8 ), making ( J geq 16 ) (which is not a valid digit). - ( F neq 6 ), as it would force ( E ) to be ( 5 ), and ( B = 5 ) leading to duplication. - ( F neq 7 ), as it would force ( E ) to be ( 6 ), making ( B = 4 ) and ( J = 14 ). 5. **Assigning Values**: - ( F = 8 ), ( E = 7 ), then ( B = 3 ) as ( B + 7 = 10 ). - Combining all this: [ H = 5 text{ (as } B + E + E + 2D + A)leq 2H. ] - Solving ( A + 2D + 1 = H = 5 ), we compute: [ A + 2D + 1 = 5 ] - Assign values: ( A = 2 ), ( D = 1 ), and hence ( G = 4 ). 6. **Completing the Solution**: - Revalidate each step and digit substitution to check the consistency of assignments in the original equation setup. # Conclusion This analysis successfully assigns each letter a unique digit satisfying the constraints: [ A = 2, B = 3, C = 0, D = 1, E = 7, F = 8, G = 4, H = 5, J = 6 ] Thus, the correct digit mapping is confirmed to be: [ boxed{A=2, B=3, C=0, D=1, E=7, F=8, G=4, H=5, J=6} ]

answer:The given expression can be rearranged and factored by taking out the common factor: [ (2x+3)(x-6) + (2x+3)(x-8) = (2x+3)((x-6) + (x-8)) ] [ = (2x+3)(2x - 14) = 2(2x+3)(x-7) ] This simplifies to: [ 2(2x+3)(x-7) = 0 ] Setting each factor equal to zero gives: [ 2x+3 = 0 quad text{or} quad x-7 = 0 ] Solving these, we find the roots: [ 2x + 3 = 0 Rightarrow 2x = -3 Rightarrow x = -frac{3}{2} ] [ x - 7 = 0 Rightarrow x = 7 ] Summing the roots: [ -frac{3}{2} + 7 = frac{-3 + 14}{2} = frac{11}{2} ] Thus, the sum of the roots is boxed{frac{11}{2}}.

answer:1. **Setting up the Problem**: Sphere with radius 4, triangle ABC with sides 13, 14, 15 tangential to the sphere. 2. **Triangle Type and Area**: - Using Heron's formula to calculate the triangle's area: - Semiperimeter, s = frac{13 + 14 + 15}{2} = 21. - Area, A = sqrt{s(s-13)(s-14)(s-15)} = sqrt{21 times 8 times 7 times 6} = sqrt{21 times 336} = 84. 3. **Finding Inradius (r)**: - From area formula with inradius: r = frac{A}{s} = frac{84}{21} = 4. 4. **Distance from Sphere's Center (P) to Plane**: - Let d be the distance from P to plane ABC. - Sphere's inradius due to tangency is 4, so closest point of sphere to plane is 4 units away. - Using geometric configurations or offset: [ d^2 + r^2 = R^2 rightarrow d^2 + 4^2 = 4^2 rightarrow d^2 = 0 rightarrow d = 0. ] - **However, this result conflicts** with the problem setup (sphere cannot be completely inside the triangle if it’s only tangent). Let's revise the calculation: [ PA^2 = r^2 + 4^2 rightarrow 4^2 = r^2 + 4^2 rightarrow r^2 = 0 rightarrow r = 4 ] - Here, it appears there was an error in the previous step: As the sphere is tangent to the plane, and r (radius of sphere’s cross-section) is the same as the sphere's radius. The sphere is thus exactly touching the plane. Therefore, d = r - 4 = 0 (assuming tangency at just one point outside). 5. **Conclusion**: - The distance from the center of the sphere to the plane of the triangle is 0. The final answer is boxed{A) 0}

answer:Katie completed 2 problems at the first stop, 3 problems at the second stop, and 1 problem at the third stop. In total, she completed 2 + 3 + 1 = 6 problems on the bus. She started with 9 problems, so after completing 6 on the bus, she had 9 - 6 = boxed{3} problems left to do after she got home.