answer:1. We are given that ( f(x) ) is a first-degree polynomial (linear function), and ( f^{(10)}(x) = 1024x + 1023 ). 2. Let's assume the form of the function: ( f(x) = ax + b ), where ( a ) and ( b ) are constants. 3. We need to analyze the given derivative ( f^{(10)}(x) ). Here, ( f^{(10)}(x) ) denotes the 10th derivative of ( f(x) ). Given ( f(x) = ax + b ): [ f'(x) = a ] [ f''(x) = 0 ] 4. Notice that any higher-order derivative of a linear function past the first derivative is zero: ( f^{(n)}(x) = 0 ) for ( n geq 2 ). Accordingly, the expression ( f^{(10)}(x) = 1024x + 1023 ) must come from some transformation applied to a higher degree polynomial masquerading as a composition derivative in this context. 5. Thus, we need to equate the derivative form provided within a suitable conceptual framework since if we consider initially: [ f(x) = (ax + b)^{10} + (ax + b) + c ] [ f^{(10)}(x) = (10!a^{10}) times (ax + b)^{10-x-x^{10}} = 1024x + 1023 ] By looking at this context, checking equivalence constraints initially: [ a^{10}/1024 ] where such iterations lead Given ( a^{10}(x - frac{b}{1-a}) + frac{b}{1-a} = 1024x + 1023 ), Matching coefficients [ 1024 = a^{10} ] [ 10a = 1024 rightarrow a = 2 rightarrow resulting, simpler approach: - Matching x variable term and constant: eg [ b ] Comparing coefficients of x and constant term: [ begin{cases} a^{10} = 1024 -frac{a^{10}b}{1-a} + frac{b}{1-a}=1023 rightarrow based constraints] (-frac{1024b}{1-2} + frac{b}{1-2}=1024x) [ a=2x, b1=-3Then consistent finding derivatives breakdown component Thus Any likely (a=2,b=1) + thus matching constraints completing linear scenarios such transformation functionality option In conclusion leading valid results: So, the possible linear functions are: [ f(x) = 2x + 1 leading: f(x) = -2x -3 adding multiple constraints used valid transformation approach: Such Validity boxed: Let's simplified completion equation constraints boxed{f(x) = 2x + 1 bm slash Conclusion Additional valid Consistent Functional LinearTerm: Since multiline transformation Deivtaive ensured function conversion: f(x)=-2x-3

answer:Let's calculate Erwan's total spending step by step. **First store:** - Pair of shoes: 200 with a 30% discount Discount on shoes = 30% of 200 = 0.30 * 200 = 60 Price after discount = 200 - 60 = 140 **Second store:** - Two shirts at 80 each and one pair of pants for 150 with a 20% discount on the entire purchase Total before discount = 2 * 80 (shirts) + 150 (pants) = 160 + 150 = 310 Discount on the entire purchase = 20% of 310 = 0.20 * 310 = 62 Price after discount = 310 - 62 = 248 **Third store:** - Jacket: 250 - Tie: 40 (half price because of the deal) - Hat: 60 (full price, the tie is the cheaper accessory and gets the discount) Discount on the tie = 50% of 40 = 0.50 * 40 = 20 Price after discount for the tie = 40 - 20 = 20 Total for the third store = 250 (jacket) + 20 (tie) + 60 (hat) = 330 **Total before the special 5% discount:** Total = 140 (shoes) + 248 (clothes from the second store) + 330 (items from the third store) Total = 140 + 248 + 330 = 718 **Special 5% discount on the overall purchase:** Special discount = 5% of 718 = 0.05 * 718 = 35.90 Total after special discount = 718 - 35.90 = 682.10 **Sales tax of 8%:** Sales tax = 8% of 682.10 = 0.08 * 682.10 = 54.568 Total after sales tax = 682.10 + 54.568 = 736.668 Erwan's total spending after accounting for all discounts, special offers, and sales tax is approximately boxed{736.67} (rounded to the nearest cent).

answer:1. **Identify the given information and draw the diagram:** - Triangle (ABC) is equilateral, so (angle BAC = angle ABC = angle BCA = 60^circ). - Ray (BE) intersects segment (AC) at (D) such that (angle CBE = 20^circ). - (DE = AB). 2. **Determine the angles in the triangle:** - Since (angle CBE = 20^circ), we have (angle ABE = 60^circ - 20^circ = 40^circ). - Since (ABC) is equilateral, (angle BAC = 60^circ). 3. **Calculate the angles involving point (D):** - (angle ABD = 40^circ) (as calculated above). - (angle BAD = 60^circ) (since (ABC) is equilateral). - Therefore, (angle ADB = 180^circ - angle ABD - angle BAD = 180^circ - 40^circ - 60^circ = 80^circ). 4. **Determine the angles involving point (E):** - Since (DE = AB) and (AB = AC) (as (ABC) is equilateral), triangle (ADE) is isosceles with (DE = AD). - (angle ADE = angle ADB = 80^circ). - Therefore, (angle EDC = 180^circ - angle ADE = 180^circ - 80^circ = 100^circ). 5. **Calculate the angle (angle BEC):** - Since (angle EDC = 100^circ) and (angle CBE = 20^circ), we can use the exterior angle theorem in triangle (BEC). - (angle BEC = angle EDC - angle CBE = 100^circ - 20^circ = 80^circ). The final answer is (boxed{80^circ}).

answer:Let h be the probability of getting a head in one flip. The equations for the probabilities are: {6choose2}h^2(1-h)^4 = {6choose3}h^3(1-h)^3 Cancelling and simplifying gives: {6 choose 2} = 15, {6 choose 3} = 20 15h^2(1-h)^4 = 20h^3(1-h)^3 Rightarrow frac{15}{20} = frac{h}{1-h} frac{3}{4} = frac{h}{1-h} Rightarrow 3(1-h) = 4h Rightarrow 3 = 7h Rightarrow h = frac{3}{7}. Now, calculate the probability of getting exactly four heads: {6choose4}h^4(1-h)^2 = 15left(frac{3}{7}right)^4left(frac{4}{7}right)^2 = 15 cdot frac{81}{2401} cdot frac{16}{49} = 15 cdot frac{1296}{117649} = frac{19440}{117649}. Since both the numerator and denominator are no common factors other than 1: frac{19440}{117649} Thus i+j = 19440 + 117649 = boxed{137089}.