answer:To calculate the total annual cost of Mrs. Garcia's insurance, we need to find the yearly cost of each type of insurance and then sum them up. 1. Car insurance: 378 quarterly means she pays this amount 4 times a year (since there are 4 quarters in a year). Annual car insurance cost = 378 * 4 = 1512 2. Home insurance: 125 monthly means she pays this amount 12 times a year (since there are 12 months in a year). Annual home insurance cost = 125 * 12 = 1500 3. Health insurance: It is already given as an annual cost. Annual health insurance cost = 5045 4. Life insurance: 850 semi-annually means she pays this amount 2 times a year (since there are 2 semi-annual periods in a year). Annual life insurance cost = 850 * 2 = 1700 Now, we add up all the annual costs to get the total annual insurance cost: Total annual insurance cost = Car insurance + Home insurance + Health insurance + Life insurance Total annual insurance cost = 1512 + 1500 + 5045 + 1700 Total annual insurance cost = 8757 Mrs. Garcia pays a total of boxed{8757} per year for all her insurance.

answer:Using the Remainder Theorem, we find the remainder by setting x = 5. The calculation proceeds as follows: [8 cdot 5^5 - 10 cdot 5^4 + 3 cdot 5^3 + 5 cdot 5^2 - 7 cdot 5 - 35.] 1. Compute 5^5 = 3125, so 8 cdot 3125 = 25000. 2. Compute 5^4 = 625, so 10 cdot 625 = 6250. 3. Compute 5^3 = 125, so 3 cdot 125 = 375. 4. Compute 5^2 = 25, so 5 cdot 25 = 125. 5. Compute -7 cdot 5 = -35. 6. Add up all terms: 25000 - 6250 + 375 + 125 - 35 - 35 = 19180. Thus, the remainder is boxed{19180}.

answer:1. **Apply Menelaus' Theorem:** We start by applying Menelaus' Theorem on triangle ABC with transversal MKL. Menelaus' Theorem states that for a triangle triangle ABC and a transversal line intersecting BC, CA, and AB at points D, E, and F respectively, the following relation holds: [ frac{BD}{DC} cdot frac{CE}{EA} cdot frac{AF}{FB} = 1 ] In our case, the transversal MKL intersects BC at L, CA at K, and AB at M. Therefore, we have: [ frac{BL}{LC} cdot frac{CK}{KA} cdot frac{AM}{MB} = 1 ] Since AMDK is a parallelogram, AM = DK and MK = AD. This implies that K and M divide AC and AB respectively in the same ratio. Hence, frac{CK}{KA} = frac{AM}{MB}. 2. **Simplify the Menelaus' Theorem Equation:** Given that AM = DK and MK = AD, we can simplify the Menelaus' equation: [ frac{BL}{LC} cdot frac{CK}{KA} cdot frac{AM}{MB} = 1 ] Since frac{CK}{KA} = frac{AM}{MB}, we have: [ frac{BL}{LC} cdot left(frac{AM}{MB}right)^2 = 1 ] Let frac{AM}{MB} = k, then: [ frac{BL}{LC} cdot k^2 = 1 implies frac{BL}{LC} = frac{1}{k^2} ] 3. **Consider the Perpendicular from D to BC:** Let X and Y be the intersection points of AB and AC with the perpendicular line from D to BC. Since D is on BC, the perpendicular from D to BC will intersect AB and AC at points X and Y respectively. 4. **Collinearity of W, D, and Z:** Let W be the point where the tangents to the circumcircle omega of triangle ABC at B and C meet. By taking D = B and D = C, we see that W, D, and the circumcenter Z of triangle AXY are collinear. This implies that W, D, and Z are always collinear. 5. **Midpoint L of DE:** Let E be a point on BC such that (BC, DE) = -1. This means that L is the midpoint of DE. 6. **Lemma Proof:** Consider the main lemma: Line ZL is tangent to omega for all positions of D on line BC. Let the tangent from L to omega touch it at P and meet ell at point Z'. Note that PD and PE are the internal and external bisectors of angle BPC. Consequently, A, P, E are collinear. Let U = PB cap AC and V = PC cap AB; then line VW passes through W and Z'. Also, UV is the polar of E in omega; so D lies on it; hence Z' equiv Z as desired. blacksquare 7. **Conclusion:** From the previous lemma, we conclude that ZL = LD + ZA, so (AXY) and mathcal{C}(L, LD) are indeed tangent to each other. blacksquare

answer:1. When -0.5<x<0, the value of the function y=[x]+1 is 0; 2. When 0leq x<1, the value of the function y=[x]+1 is 1; 3. When 1leq x<2, the value of the function y=[x]+1 is 2; 4. When 2leq x<2.5, the value of the function y=[x]+1 is 3; Summarizing the above, the range of the function y=[x]+1 for -0.5<x<2.5 is {0, 1, 2, 3}. Therefore, the answer is boxed{{0, 1, 2, 3}}.