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question:Given the quadratic function boxed{y=(x+h)^{2}}, when boxed{x < 3}, y increases as x increases; when boxed{x > 3}, y decreases as x increases. The value of y when boxed{x=0} is (quad). A: boxed{1} B: boxed{9} C: boxed{1} D: boxed{9}
answer:From the given information, we know that the axis of symmetry for the quadratic function boxed{y=-(x+h)^{2}} is boxed{x=-3}. Thus, boxed{h=-3}. Substituting boxed{h=-3} into the quadratic function boxed{y=-(x+h)^{2}}, we get boxed{y=-(x-3)^{2}}. When boxed{x=0}, boxed{y=-9}. Therefore, the answer is boxed{B}. According to the given information, we can find the axis of symmetry of the quadratic function, which helps us determine the value of h. Then, we can obtain the quadratic function's analytical expression: boxed{y=-(x-3)^{2}}. Finally, substituting boxed{x=0} into the analytical expression, we can find the value of y. This problem primarily tests the understanding of the properties of quadratic functions. The key is to master the vertex form of a quadratic function: boxed{y=a(x-h)^{2}+k}, where the axis of symmetry is boxed{x=h}.
question:The order of magnitude for the three numbers 6^{0.5}, 0.5^{6}, log_{0.5}6 is ( ) A: 0.5^{6} < log_{0.5}6 < 6^{0.5} B: 0.5^{6} < 6^{0.5} < log_{0.5}6 C: log_{0.5}6 < 6^{0.5} < 0.5^{6} D: log_{0.5}6 < 0.5^{6} < 6^{0.5}
answer:Since log_{0.5}6 < log_{0.5}1 = 0, 0 < 0.5^{6} < 0.5 = 1, and 1 = 6^0 < 6^{0.5}, Therefore, log_{0.5}6 < 0.5^{6} < 6^{0.5}, Hence, the correct option is boxed{text{D}}.
question:Let triangle ABC have side lengths AB=10, AC=11, and BC=12. Two circles are located inside angle BAC which are tangent to rays overline{AB}, overline{AC}, and segment overline{BC}. Compute the distance between the centers of these two circles.
answer:The two circles described in the problem are the incircle and an excircle of triangle ABC. Let the center of the incircle be I and the center of the excircle opposite to vertex A be E. 1. Compute the semiperimeter ( s ) of triangle ABC. [ s = frac{1}{2}(AB + AC + BC) = frac{1}{2}(10 + 11 + 12) = 16.5 ] 2. Use Heron's formula to find the area ( K ) of triangle ABC. [ K = sqrt{s(s - AB)(s - AC)(s - BC)} = sqrt{16.5 cdot 6.5 cdot 5.5 cdot 4.5} = 36.75 ] 3. Find the inradius ( r ) using ( K = rs ). [ r = frac{K}{s} = frac{36.75}{16.5} = 2.25 ] 4. Label the points of tangency of the incircle and excircle with ray overline{AC} as S and T, respectively. It follows that AS = s - AC = 16.5 - 11 = 5.5 and AT = s = 16.5. 5. Determine AI (distance from A to I) using the Pythagorean theorem in triangle ASI and knowing AS and SI. Assume SI approx AI since I is near the midpoint of AS. [ AI approx sqrt{AS^2 + SI^2} = sqrt{5.5^2 + 2.25^2} = sqrt{30.25 + 5.0625} = sqrt{35.3125} = 5.944 ] 6. Knowing the ratio AI/AE = AS/AT = 5.5 / 16.5 = 1/3, find AE. [ AE = 3 times AI = 3 times 5.944 = 17.832 ] 7. Satisfying the condition IE = AE - AI, compute IE. [ IE = 17.832 - 5.944 approx 11.888 ] Conclusion: The distance between the centers of the incircle and excircle of triangle ABC is boxed{11.888}.
question:The graph of ( r = sin theta ) represents a circle. Determine the smallest value of ( t ) so that when ( r = sin theta ) is plotted for ( 0 le theta le t, ) the resulting graph forms the entire circle.
answer:Let ( f(theta) = sin theta. ) When ( theta = 0, ) ( r = 0, ) so in rectangular coordinates, [ (x, y) = (0 cdot cos theta, 0 cdot sin theta) = (0, 0). ] The function ( f(theta) = sin theta ) is periodic, and we need to find the smallest ( t ) at which the entire circle is completed. This happens when: 1. ( theta = frac{pi}{2} ), ( r = 1 ) (top point of the circle), 2. ( theta = frac{3pi}{2} ), ( r = -1 ) (bottom point of the circle). In polar coordinates, the points ( (r, theta) ) and ( (-r, theta + pi) ) coincide, meaning that the graph repeats after an interval of ( pi. ) However, since ( r = sin theta ) differs from ( cos theta ) by a phase shift of ( frac{pi}{2}, ) we need to ensure that ( t ) covers the shift and completes the cycle. Thus, ( t = frac{3pi}{2} ) is the smallest interval needed to plot the entire circle. [ boxed{frac{3pi}{2}} ]