answer:To evaluate the correctness of each statement, we proceed as follows: - **Statement ①**: Unit vectors with the same starting point must have the same endpoint. This statement is incorrect because unit vectors only specify direction and magnitude (which is 1 for unit vectors), but not the specific endpoint. Two unit vectors with the same starting point can have different directions, and thus, different endpoints. - **Statement ②**: Given vectors overrightarrow{AB}∥overrightarrow{CD}, then points A, B, C, D must lie on the same line. This statement is incorrect because overrightarrow{AB}∥overrightarrow{CD} implies that the vectors have the same or opposite direction, but it does not necessarily mean that all four points lie on the exact same line. They could be parallel vectors in different positions in space. - **Statement ③**: If overrightarrow{a}∥overrightarrow{b} and overrightarrow{b}∥overrightarrow{c}, then overrightarrow{a}∥overrightarrow{c}. This statement seems correct at first glance because parallelism is transitive in most cases. However, when considering the special case where overrightarrow{b}=overrightarrow{0} (the zero vector), overrightarrow{a} and overrightarrow{c} do not necessarily have to be parallel. This is because the zero vector is technically parallel to any vector, but this does not imply that any two vectors parallel to the zero vector are parallel to each other. - **Statement ④**: Collinear vectors, if they have different starting points, then their endpoints must be different. This statement is incorrect because if vectors are collinear and have different starting points, it is possible for them to still end at the same point. For example, consider vectors overrightarrow{AC} and overrightarrow{BC} where A, B, and C are collinear with A and B being different starting points but both vectors ending at point C. Given the analysis above, all four statements are incorrect. Therefore, the number of correct statements among the given is 0. Thus, the correct answer is boxed{text{A}}.

answer:First, we find the set M. The inequality x^{2}+x-12leqslant 0 can be factored into (x+4)(x-3)leqslant 0. The solutions to this inequality are -4 leq x leq 3, so M=[-4,3]. Next, we find the set N. Since y=3^{x} and xleqslant 1, we can find the maximum value of y by substituting x=1 into the equation, giving us y=3^{1}=3. The minimum value of y is 0, as x approaches negative infinity. Therefore, N=(0,3]. Now, we want to find the set of all x that are in M but not in N. This is represented as {x|xin M and xnotin N}. Looking at the sets M and N, we can see that the values of x that satisfy this condition are -4 leq x < 0. Therefore, the set {x|xin M and xnotin N} is [-4,0). Hence, the correct answer is: boxed{text{C: } [-4,0)}

answer:1. Let ( ABCD ) be a convex quadrilateral with ( angle BAC = 3angle CAD ), ( AB = CD ), and ( angle ACD = angle CBD ). We need to find the measure of ( angle ACD ). 2. Assume ( AB = AD ). Let ( O ) be the intersection of diagonals ( AC ) and ( BD ). Since ( angle DOC = angle DBC ), we have ( triangle DOC sim triangle DBC ). This implies: [ frac{DC}{DB} = frac{DO}{DC} Rightarrow DC^2 = DO cdot DB ] 3. Let ( x = angle DAC ) and ( y = angle ACD = angle DBC ). Then, ( angle BAC = 3x ). 4. Since ( angle ACD = angle CBD ), we have: [ angle ADB = angle ABD = 90^circ - 2x ] and [ angle AOD = 90^circ + x ] 5. Using the Law of Sines in ( triangle AOD ) and ( triangle ABD ): [ frac{sin x}{sin y} = frac{sin angle ADC}{sin angle ACD} = frac{DC}{AD} ] 6. Since ( AB = AD ), we have: [ left( frac{sin y}{sin x} right)^2 = frac{AD^2}{DC^2} = frac{AD}{DO} cdot frac{AB}{DB} = frac{sin angle AOD}{sin angle DAO} cdot frac{sin angle ADB}{sin angle DAB} ] 7. Substituting the angles: [ left( frac{sin y}{sin x} right)^2 = frac{sin (90^circ + x)}{sin x} cdot frac{sin (90^circ - 2x)}{sin 4x} ] 8. Simplifying the trigonometric expressions: [ frac{sin (90^circ + x)}{sin x} = frac{cos x}{sin x} = cot x ] and [ frac{sin (90^circ - 2x)}{sin 4x} = frac{cos 2x}{4 sin x cos x cos 2x} = frac{cos 2x}{4 sin x cos x cos 2x} = frac{1}{4 sin x} ] 9. Combining these results: [ left( frac{sin y}{sin x} right)^2 = cot x cdot frac{1}{4 sin x} = frac{1}{4 sin^2 x} ] 10. Therefore: [ sin^2 y = frac{1}{4} Rightarrow sin y = frac{1}{2} ] 11. Hence: [ y = 30^circ ] 12. Therefore, ( angle ACD = 30^circ ). The final answer is ( boxed{30^circ} )

answer:Using the identity cos theta = frac{e^{i theta} + e^{-i theta}}{2}, we express cos^7 theta: [ cos^7 theta = left( frac{e^{i theta} + e^{-i theta}}{2} right)^7. ] Expanding this using the binomial theorem: [ cos^7 theta = frac{1}{128} (e^{7i theta} + 7 e^{5i theta} + 21 e^{3i theta} + 35 e^{i theta} + 35 e^{-i theta} + 21 e^{-3i theta} + 7 e^{-5i theta} + e^{-7i theta}). ] Rewriting in terms of cosine: [ cos^7 theta = frac{1}{64} cos 7theta + frac{7}{64} cos 5theta + frac{21}{64} cos 3theta + frac{35}{64} cos theta. ] Therefore, we find a_1 = frac{35}{64}, a_2 = 0, a_3 = frac{21}{64}, a_4 = 0, a_5 = frac{7}{64}, a_6 = 0, a_7 = frac{1}{64}. Now, we compute: [ a_1^2 + a_2^2 + a_3^2 + a_4^2 + a_5^2 + a_6^2 + a_7^2 = left( frac{35}{64} right)^2 + left( frac{21}{64} right)^2 + left( frac{7}{64} right)^2 + left( frac{1}{64} right)^2. ] [ = frac{1225}{4096} + frac{441}{4096} + frac{49}{4096} + frac{1}{4096} = frac{1716}{4096}. ] The final answer is (boxed{frac{1716}{4096}}).