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question:Ellina has twelve blocks, two each of red (textbf{R}), blue (textbf{B}), yellow (textbf{Y}), green (textbf{G}), orange (textbf{O}), and purple (textbf{P}). Call an arrangement of blocks textit{even} if there is an even number of blocks between each pair of blocks of the same color. For example, the arrangement[textbf{R B B Y G G Y R O P P O}]is even. Ellina arranges her blocks in a row in random order. The probability that her arrangement is even is frac{m}{n}, where m and n are relatively prime positive integers. Find m+n.
answer:Consider this position chart: [textbf{1 2 3 4 5 6 7 8 9 10 11 12}]Since there has to be an even number of spaces between each pair of the same color, spots 1, 3, 5, 7, 9, and 11 contain some permutation of all 6 colored balls. Likewise, so do the even spots, so the number of even configurations is 6! cdot 6! (after putting every pair of colored balls in opposite parity positions, the configuration can be shown to be even). This is out of frac{12!}{(2!)^6} possible arrangements, so the probability is: [frac{6!cdot6!}{frac{12!}{(2!)^6}} = frac{6!cdot2^6}{7cdot8cdot9cdot10cdot11cdot12} = frac{2^4}{7cdot11cdot3} = frac{16}{231},]which is in simplest form. So, m + n = 16 + 231 = boxed{247}.~Oxymoronic15
question:The distance from a focus of the hyperbola dfrac {y^{2}}{a^{2}} - dfrac {x^{2}}{b^{2}} = 1 (a > 0, b > 0) to one of its asymptotes is dfrac {2 sqrt {5}}{5}a. Find the eccentricity of the hyperbola.A: dfrac {sqrt {5}}{5}B: dfrac {2 sqrt {5}}{5}C: dfrac {3 sqrt {5}}{5}D: dfrac {4 sqrt {5}}{5}
answer:Given the hyperbola dfrac {y^{2}}{a^{2}} - dfrac {x^{2}}{b^{2}} = 1 (a > 0, b > 0),The equations of the asymptotes of the hyperbola are y = pm dfrac {a}{b}x, or equivalently, ax pm by = 0.Given that the distance from a focus of the hyperbola to one of its asymptotes is dfrac {2 sqrt {5}}{5}a,The distance d from the right focus F(0,c) to the asymptote ax pm by = 0 is d = dfrac {bc}{sqrt {a^{2} + b^{2}}} = dfrac {2 sqrt {5}}{5}a.Solving this equation yields b = dfrac {2 sqrt {5}}{5}a, which implies b^{2} = dfrac {4}{5}a^{2}. Simplifying further, we find c^{2} = dfrac {9}{5}a^{2}.Therefore, the standard eccentricity of the hyperbola is e = dfrac {c}{a} = dfrac {3 sqrt {5}}{5}.Hence, the correct choice is boxed{C}.The problem involves using the distance formula from a point to a line, the standard equation of a hyperbola, and basic geometric properties. It is a foundational question.
question:Which of the following represents the force corresponding to the given potential?[asy]// Code by ribensize(400);picture pic;// Rectangledraw(pic,(0,0)--(22,0)--(22,12)--(0,12)--cycle);label(pic,"-15",(2,0),S);label(pic,"-10",(5,0),S);label(pic,"-5",(8,0),S);label(pic,"0",(11,0),S);label(pic,"5",(14,0),S);label(pic,"10",(17,0),S);label(pic,"15",(20,0),S);label(pic,"-2",(0,2),W);label(pic,"-1",(0,4),W);label(pic,"0",(0,6),W);label(pic,"1",(0,8),W);label(pic,"2",(0,10),W);label(pic,rotate(90)*"F (N)",(-2,6),W);label(pic,"x (m)",(11,-2),S);// Tick Marksdraw(pic,(2,0)--(2,0.3));draw(pic,(5,0)--(5,0.3));draw(pic,(8,0)--(8,0.3));draw(pic,(11,0)--(11,0.3));draw(pic,(14,0)--(14,0.3));draw(pic,(17,0)--(17,0.3));draw(pic,(20,0)--(20,0.3));draw(pic,(0,2)--(0.3,2));draw(pic,(0,4)--(0.3,4));draw(pic,(0,6)--(0.3,6));draw(pic,(0,8)--(0.3,8));draw(pic,(0,10)--(0.3,10));draw(pic,(2,12)--(2,11.7));draw(pic,(5,12)--(5,11.7));draw(pic,(8,12)--(8,11.7));draw(pic,(11,12)--(11,11.7));draw(pic,(14,12)--(14,11.7));draw(pic,(17,12)--(17,11.7));draw(pic,(20,12)--(20,11.7));draw(pic,(22,2)--(21.7,2));draw(pic,(22,4)--(21.7,4));draw(pic,(22,6)--(21.7,6));draw(pic,(22,8)--(21.7,8));draw(pic,(22,10)--(21.7,10));// Pathspath A=(0,6)--(5,6)--(5,4)--(11,4)--(11,8)--(17,8)--(17,6)--(22,6);path B=(0,6)--(5,6)--(5,2)--(11,2)--(11,10)--(17,10)--(17,6)--(22,6);path C=(0,6)--(5,6)--(5,5)--(11,5)--(11,7)--(17,7)--(17,6)--(22,6);path D=(0,6)--(5,6)--(5,7)--(11,7)--(11,5)--(17,5)--(17,6)--(22,6);path E=(0,6)--(5,6)--(5,8)--(11,8)--(11,4)--(17,4)--(17,6)--(22,6);draw(A);label("(A)",(9.5,-3),4*S);draw(shift(35*right)*B);label("(B)",(45.5,-3),4*S);draw(shift(20*down)*C);label("(C)",(9.5,-23),4*S);draw(shift(35*right)*shift(20*down)*D);label("(D)",(45.5,-23),4*S);draw(shift(40*down)*E);label("(E)",(9.5,-43),4*S);add(pic);picture pic2=shift(35*right)*pic;picture pic3=shift(20*down)*pic;picture pic4=shift(35*right)*shift(20*down)*pic;picture pic5=shift(40*down)*pic;add(pic2);add(pic3);add(pic4);add(pic5);[/asy]
answer:To determine the force corresponding to the given potential, we use the relationship between force and potential energy. The force ( F ) in the ( x )-direction is given by the negative gradient of the potential energy ( U ):[ F_x = -frac{dU}{dx} ]We need to analyze the given potential energy graph and compute its derivative to find the force.1. Identify the segments of the potential energy graph: - From ( x = -15 ) to ( x = -10 ), the potential energy ( U ) is constant. - From ( x = -10 ) to ( x = 0 ), the potential energy ( U ) decreases linearly. - From ( x = 0 ) to ( x = 10 ), the potential energy ( U ) increases linearly. - From ( x = 10 ) to ( x = 15 ), the potential energy ( U ) is constant.2. Compute the derivative of ( U ) with respect to ( x ): - For ( x ) in the range ( -15 leq x < -10 ) and ( 10 < x leq 15 ), ( U ) is constant, so ( frac{dU}{dx} = 0 ). - For ( x ) in the range ( -10 leq x < 0 ), ( U ) decreases linearly. The slope of this segment is negative. Let’s denote the slope as ( m_1 ). Since the potential decreases, ( m_1 < 0 ). - For ( x ) in the range ( 0 leq x leq 10 ), ( U ) increases linearly. The slope of this segment is positive. Let’s denote the slope as ( m_2 ). Since the potential increases, ( m_2 > 0 ).3. Determine the slopes ( m_1 ) and ( m_2 ): - The slope ( m_1 ) from ( x = -10 ) to ( x = 0 ) is calculated as: [ m_1 = frac{Delta U}{Delta x} = frac{U(0) - U(-10)}{0 - (-10)} = frac{U(0) - U(-10)}{10} ] - The slope ( m_2 ) from ( x = 0 ) to ( x = 10 ) is calculated as: [ m_2 = frac{Delta U}{Delta x} = frac{U(10) - U(0)}{10 - 0} = frac{U(10) - U(0)}{10} ]4. Compute the force ( F_x ): - For ( x ) in the range ( -15 leq x < -10 ) and ( 10 < x leq 15 ), ( F_x = 0 ). - For ( x ) in the range ( -10 leq x < 0 ), ( F_x = -m_1 ). Since ( m_1 ) is negative, ( F_x ) is positive. - For ( x ) in the range ( 0 leq x leq 10 ), ( F_x = -m_2 ). Since ( m_2 ) is positive, ( F_x ) is negative.5. Match the force graph with the given options: - The force graph should show ( F_x = 0 ) for ( x ) in the ranges ( -15 leq x < -10 ) and ( 10 < x leq 15 ). - The force graph should show a positive constant value for ( x ) in the range ( -10 leq x < 0 ). - The force graph should show a negative constant value for ( x ) in the range ( 0 leq x leq 10 ).By examining the given options, we find that option ( E ) matches these criteria. Therefore, the correct answer is:[boxed{E}]
question:Given an ellipse C:frac{x^2}{25}+frac{y^2}{9}=1, a line l:(m+2)x-(m+4)y+2-m=0 (min R), the relationship between the line l and the ellipse C is ( ).A: intersectB: tangentC: separateD: uncertain
answer:To analyze the relationship between the line l and the ellipse C, we first simplify the equation of the line l given by (m+2)x-(m+4)y+2-m=0. This simplifies to:[2x - 4y + 2 + m(x - y - 1) = 0.]Next, we solve the system of equations formed by setting m(x - y - 1) = 0, which gives us the simpler line equation 2x - 4y + 2 = 0, and the condition x - y - 1 = 0. Solving this system:[begin{aligned}&left{begin{array}{l}2x - 4y + 2 = 0 x - y - 1 = 0end{array}right. &Rightarrowleft{begin{array}{l}x = 3 y = 2end{array}right.end{aligned}]This means the line l always passes through the fixed point P(3, 2). To determine the relationship between this line and the ellipse C: frac{x^2}{25} + frac{y^2}{9} = 1, we substitute the coordinates of point P into the equation of the ellipse:[frac{3^2}{25} + frac{2^2}{9} = frac{9}{25} + frac{4}{9} = frac{181}{225}.]Since frac{181}{225} < 1, this indicates that the point P(3, 2) lies inside the ellipse C. Therefore, any line passing through this point must intersect the ellipse, as it cannot be tangent or separate without crossing the boundary of the ellipse.Thus, the relationship between the line l and the ellipse C is that they intersect.Therefore, the answer is: boxed{text{A}}.
question:Given a=2^{frac{1}{2}}, b=(2^{log_{2}{3}})^{-frac{1}{2}}, c=cos 50^{circ}cos 10^{circ}+cos 140^{circ}sin 170^{circ}, then the relationship among the real numbers a, b, c is ( ).A: a > c > bB: b > a > cC: a > b > cD: c > b > a
answer:We have a= sqrt{2} > 1, b=3^{-frac{1}{2}}= frac{sqrt{3}}{3}in(frac{1}{2},1), c=cos 50^{circ}cos 10^{circ}-sin 50^{circ}sin 10^{circ}=cos (50^{circ}+10^{circ})=cos 60^{circ}= frac{1}{2}.Therefore, a > b > c.Hence, the correct answer is boxed{C}.By using the sum-to-product formulas and the properties of exponents, we can find c, and then use the properties of exponents to find a and b. This question tests the understanding of sum-to-product formulas and the properties of exponents, as well as reasoning and computational skills, and is considered a basic question.
question:How many four-digit integers abcd, with a neq 0, have the property that the three two-digit integers ab<bc<cd form an increasing arithmetic sequence? One such number is 4692, where a=4, b=6, c=9, and d=2.textbf{(A)} 9qquadtextbf{(B)} 15qquadtextbf{(C)} 16qquadtextbf{(D)} 17qquadtextbf{(E)} 20
answer:The numbers are 10a+b, 10b+c, and 10c+d. Note that only d can be zero, the numbers ab, bc, and cd cannot start with a zero, and ale ble c.To form the sequence, we need (10c+d)-(10b+c)=(10b+c)-(10a+b). This can be rearranged as 10(c-2b+a)=2c-b-d. Notice that since the left-hand side is a multiple of 10, the right-hand side can only be 0 or 10. (A value of -10 would contradict ale ble c.) Therefore we have two cases: a+c-2b=1 and a+c-2b=0.Case 1If c=9, then b+d=8, 2b-a=8, so 5le ble 8. This gives 2593, 4692, 6791, 8890.If c=8, then b+d=6, 2b-a=7, so 4le ble 6. This gives 1482, 3581, 5680.If c=7, then b+d=4, 2b-a=6, so b=4, giving 2470.There is no solution for c=6.Added together, this gives us 8 answers for Case 1.Case 2This means that the digits themselves are in an arithmetic sequence. This gives us 9 answers,[1234, 1357, 2345, 2468, 3456, 3579, 4567, 5678, 6789.]Adding the two cases together, we find the answer to be 8+9= boxed{textbf{(D) }17}.