💡Okay, so I have this problem about an isosceles trapezoid. The area is given as S, and the angle between its diagonals opposite to the lateral side is α. I need to find the height of the trapezoid. Hmm, okay, let's break this down step by step.First, let me recall what an isosceles trapezoid is. It's a quadrilateral with a pair of parallel sides (called the bases) and the non-parallel sides (the legs) being equal in length. Also, the base angles are equal. So, in this case, the trapezoid is isosceles, meaning the legs are equal, and the base angles are equal.The area of a trapezoid is given by the formula:[ S = frac{(a + b)}{2} times h ]where ( a ) and ( b ) are the lengths of the two bases, and ( h ) is the height (the perpendicular distance between the bases). So, if I can find ( h ), that's what I need.But in this problem, instead of giving me the lengths of the bases or the legs, they give me the area ( S ) and the angle ( alpha ) between the diagonals opposite to the lateral side. Hmm, okay, so I need to relate this angle to the height.Let me visualize the trapezoid. Let's label it ( ABCD ) with ( AB ) and ( CD ) as the bases, and ( AD ) and ( BC ) as the legs, which are equal. The diagonals are ( AC ) and ( BD ). The angle between the diagonals opposite to the lateral side is ( alpha ). So, I think this angle is at the intersection point of the diagonals, which is point ( O ) somewhere inside the trapezoid.Wait, actually, in an isosceles trapezoid, the diagonals are equal in length and they intersect each other at the same angle. So, the angle between the diagonals is given as ( alpha ). Since the trapezoid is isosceles, the diagonals are congruent, and they intersect at point ( O ), creating four triangles inside the trapezoid.I think the angle ( alpha ) is the angle between the diagonals at their intersection point ( O ). So, if I consider triangle ( AOB ), triangle ( BOC ), triangle ( COD ), and triangle ( DOA ), each of these triangles has some relationship with the angle ( alpha ).But the problem specifies that the angle is opposite to the lateral side. Hmm, so maybe it's the angle at point ( O ) that's opposite to one of the legs. Let me think. If I consider the lateral side ( AD ), then the angle opposite to it would be the angle at ( O ) that's not adjacent to ( AD ). So, perhaps the angle ( alpha ) is the angle between the diagonals at point ( O ) that's opposite to the lateral side.Alternatively, maybe it's the angle between the diagonals when looking from the side opposite to the lateral side. Hmm, I might need to draw a diagram to visualize this better.Since I can't draw here, let me try to imagine it. The trapezoid ( ABCD ) with bases ( AB ) and ( CD ), and legs ( AD ) and ( BC ). Diagonals ( AC ) and ( BD ) intersect at point ( O ). The angle between the diagonals at ( O ) is ( alpha ), and it's opposite to the lateral side. So, if I consider the lateral side ( AD ), then the angle opposite to it would be at point ( O ), which is the angle between the diagonals.Alternatively, maybe it's the angle between the extensions of the diagonals beyond the intersection point. Hmm, not sure. Maybe I need to clarify.Wait, in an isosceles trapezoid, the diagonals are equal and they intersect at the same angle. So, the angle between the diagonals is ( alpha ), which is given. So, perhaps the angle between the diagonals is ( alpha ), and that's the angle we need to use.Now, how can I relate this angle ( alpha ) to the height ( h ) of the trapezoid? Let's think about the properties of the diagonals in an isosceles trapezoid.In an isosceles trapezoid, the diagonals are equal in length and they intersect each other at the same ratio. That is, the ratio of the segments of one diagonal is equal to the ratio of the segments of the other diagonal.Let me denote the point of intersection as ( O ). So, the diagonals ( AC ) and ( BD ) intersect at ( O ). Let me denote ( AO = OC ) and ( BO = OD ) since the trapezoid is isosceles. Wait, no, actually, in an isosceles trapezoid, the diagonals are equal but they are not necessarily bisecting each other unless it's a rectangle or something. Wait, no, in a rectangle, which is a special case of a trapezoid, the diagonals bisect each other, but in a general isosceles trapezoid, the diagonals are equal but they don't necessarily bisect each other.Wait, actually, in an isosceles trapezoid, the diagonals are equal in length but they do bisect each other proportionally. That is, the ratio of the segments of one diagonal is equal to the ratio of the segments of the other diagonal. So, if ( AO = x ) and ( OC = y ), then ( BO = x ) and ( OD = y ), maintaining the same ratio.But I'm not sure if that's directly helpful here. Maybe I need to use trigonometry with the angle ( alpha ).Let me consider triangle ( AOB ). In this triangle, the sides are ( AO ), ( BO ), and ( AB ). The angle between ( AO ) and ( BO ) is ( alpha ). If I can find the lengths of ( AO ) and ( BO ), maybe I can relate them to the height.But wait, I don't know the lengths of the diagonals. Hmm. Maybe I can express the area in terms of the diagonals and the angle between them.I recall that the area of a quadrilateral can be expressed as:[ S = frac{1}{2} times d_1 times d_2 times sin theta ]where ( d_1 ) and ( d_2 ) are the lengths of the diagonals, and ( theta ) is the angle between them. But in this case, the trapezoid is isosceles, so the diagonals are equal, meaning ( d_1 = d_2 = d ). So, the area becomes:[ S = frac{1}{2} times d times d times sin alpha = frac{1}{2} d^2 sin alpha ]So, from this, I can solve for ( d ):[ d^2 = frac{2S}{sin alpha} ][ d = sqrt{frac{2S}{sin alpha}} ]Okay, so now I have the length of the diagonals in terms of ( S ) and ( alpha ). Now, how can I relate this to the height ( h )?Let me think about the relationship between the diagonals and the height. In an isosceles trapezoid, the length of the diagonal can be expressed in terms of the bases and the legs. Let me denote the lengths of the bases as ( a ) and ( b ), with ( a > b ), and the legs as ( c ).The length of the diagonal ( d ) can be found using the formula:[ d = sqrt{c^2 + ab} ]Wait, no, that doesn't sound right. Let me recall the correct formula for the diagonal in a trapezoid.In a trapezoid, the length of a diagonal can be found using the formula:[ d = sqrt{a^2 + c^2 - 2ac cos theta} ]Wait, no, that's for a triangle. Maybe I need to use the law of cosines in the triangles formed by the diagonals.Alternatively, in an isosceles trapezoid, the length of the diagonal can be expressed in terms of the bases and the height. Let me consider the right triangles formed by the legs, the height, and the difference in the bases.Let me denote the height as ( h ), and the difference between the bases as ( a - b ). Then, the projection of the leg onto the base is ( frac{a - b}{2} ). So, the leg ( c ) can be expressed as:[ c = sqrt{h^2 + left( frac{a - b}{2} right)^2} ]But I don't know ( a ) or ( b ), so maybe that's not directly helpful.Wait, but I have the area ( S = frac{(a + b)}{2} times h ). So, if I can express ( a + b ) in terms of other variables, maybe I can relate it to the diagonals.Also, I have the expression for the diagonal ( d ) in terms of ( S ) and ( alpha ):[ d = sqrt{frac{2S}{sin alpha}} ]So, maybe I can find another expression for ( d ) in terms of ( a ), ( b ), and ( h ), and then set them equal.Let me try that. The diagonal ( d ) in an isosceles trapezoid can be found using the formula:[ d = sqrt{c^2 + ab} ]Wait, no, that doesn't seem right. Let me think again.In an isosceles trapezoid, the diagonal can be calculated using the formula:[ d = sqrt{a^2 + c^2 - 2ac cos theta} ]Wait, no, that's for a triangle. Maybe I need to consider the coordinates.Let me place the trapezoid on a coordinate system. Let me set the bases ( AB ) and ( CD ) horizontally, with ( A ) at the origin. So, point ( A ) is at ( (0, 0) ), point ( B ) is at ( (a, 0) ), point ( D ) is at ( (d, h) ), and point ( C ) is at ( (d + b, h) ). Wait, but since it's isosceles, the legs ( AD ) and ( BC ) are equal, so the distances from ( A ) to ( D ) and from ( B ) to ( C ) should be equal.Therefore, the coordinates would be:- ( A = (0, 0) )- ( B = (a, 0) )- ( D = (x, h) )- ( C = (a - x, h) )Because the trapezoid is isosceles, the distances from ( A ) to ( D ) and from ( B ) to ( C ) must be equal, so ( x = a - x - b ) or something? Wait, maybe I need to adjust.Wait, actually, the top base ( CD ) has length ( b ), so the coordinates should be:- ( A = (0, 0) )- ( B = (a, 0) )- ( D = (d, h) )- ( C = (d + b, h) )But since the trapezoid is isosceles, the legs ( AD ) and ( BC ) must be equal in length. So, the distance from ( A ) to ( D ) is equal to the distance from ( B ) to ( C ).Calculating ( AD ):[ AD = sqrt{(d - 0)^2 + (h - 0)^2} = sqrt{d^2 + h^2} ]Calculating ( BC ):[ BC = sqrt{(d + b - a)^2 + (h - 0)^2} = sqrt{(d + b - a)^2 + h^2} ]Since ( AD = BC ), we have:[ sqrt{d^2 + h^2} = sqrt{(d + b - a)^2 + h^2} ]Squaring both sides:[ d^2 + h^2 = (d + b - a)^2 + h^2 ][ d^2 = (d + b - a)^2 ][ d^2 = d^2 + 2d(b - a) + (b - a)^2 ][ 0 = 2d(b - a) + (b - a)^2 ][ 2d(b - a) = -(b - a)^2 ][ 2d = -(b - a) ][ d = frac{a - b}{2} ]Okay, so the x-coordinate of point ( D ) is ( frac{a - b}{2} ). That makes sense because in an isosceles trapezoid, the legs are symmetric with respect to the vertical axis through the midpoint of the bases.So, now, the coordinates are:- ( A = (0, 0) )- ( B = (a, 0) )- ( D = left( frac{a - b}{2}, h right) )- ( C = left( frac{a - b}{2} + b, h right) = left( frac{a + b}{2}, h right) )Okay, so now, the diagonals are ( AC ) and ( BD ). Let's find their coordinates.Diagonal ( AC ) connects ( A(0, 0) ) to ( Cleft( frac{a + b}{2}, h right) ).Diagonal ( BD ) connects ( B(a, 0) ) to ( Dleft( frac{a - b}{2}, h right) ).The angle between these diagonals is ( alpha ). So, we can use the dot product formula to find the angle between vectors ( AC ) and ( BD ).First, let's find the vectors:Vector ( AC ) is ( left( frac{a + b}{2}, h right) ).Vector ( BD ) is ( left( frac{a - b}{2} - a, h - 0 right) = left( -frac{a + b}{2}, h right) ).Wait, is that correct? Let me double-check.Point ( B ) is at ( (a, 0) ), and point ( D ) is at ( left( frac{a - b}{2}, h right) ). So, the vector ( BD ) is ( D - B = left( frac{a - b}{2} - a, h - 0 right) = left( -frac{a + b}{2}, h right) ). Yes, that's correct.So, vectors ( AC ) and ( BD ) are:- ( AC = left( frac{a + b}{2}, h right) )- ( BD = left( -frac{a + b}{2}, h right) )Now, the angle ( alpha ) between vectors ( AC ) and ( BD ) can be found using the dot product formula:[ cos alpha = frac{AC cdot BD}{|AC| |BD|} ]First, compute the dot product ( AC cdot BD ):[ AC cdot BD = left( frac{a + b}{2} right) left( -frac{a + b}{2} right) + (h)(h) ][ = -left( frac{a + b}{2} right)^2 + h^2 ]Next, compute the magnitudes ( |AC| ) and ( |BD| ):[ |AC| = sqrt{left( frac{a + b}{2} right)^2 + h^2} ][ |BD| = sqrt{left( -frac{a + b}{2} right)^2 + h^2} = sqrt{left( frac{a + b}{2} right)^2 + h^2} ]So, both ( |AC| ) and ( |BD| ) are equal:[ |AC| = |BD| = sqrt{left( frac{a + b}{2} right)^2 + h^2} ]Let me denote this common magnitude as ( d ):[ d = sqrt{left( frac{a + b}{2} right)^2 + h^2} ]So, the dot product becomes:[ AC cdot BD = -left( frac{a + b}{2} right)^2 + h^2 ]And the product of the magnitudes is:[ |AC| |BD| = d times d = d^2 = left( frac{a + b}{2} right)^2 + h^2 ]Therefore, the cosine of the angle ( alpha ) is:[ cos alpha = frac{ -left( frac{a + b}{2} right)^2 + h^2 }{ left( frac{a + b}{2} right)^2 + h^2 } ]Let me simplify this expression:[ cos alpha = frac{ h^2 - left( frac{a + b}{2} right)^2 }{ h^2 + left( frac{a + b}{2} right)^2 } ]Hmm, that's an interesting expression. Let me denote ( k = frac{a + b}{2} ), so the expression becomes:[ cos alpha = frac{ h^2 - k^2 }{ h^2 + k^2 } ]Let me solve for ( h^2 ):Multiply both sides by ( h^2 + k^2 ):[ cos alpha (h^2 + k^2) = h^2 - k^2 ][ cos alpha cdot h^2 + cos alpha cdot k^2 = h^2 - k^2 ][ cos alpha cdot h^2 - h^2 + cos alpha cdot k^2 + k^2 = 0 ][ h^2 (cos alpha - 1) + k^2 (cos alpha + 1) = 0 ][ h^2 (1 - cos alpha) = k^2 (1 + cos alpha) ][ frac{h^2}{k^2} = frac{1 + cos alpha}{1 - cos alpha} ][ left( frac{h}{k} right)^2 = frac{1 + cos alpha}{1 - cos alpha} ][ frac{h}{k} = sqrt{ frac{1 + cos alpha}{1 - cos alpha} } ][ frac{h}{k} = frac{ sqrt{1 + cos alpha} }{ sqrt{1 - cos alpha} } ]I can simplify this using the identity ( sqrt{ frac{1 + cos alpha}{1 - cos alpha} } = frac{ cos frac{alpha}{2} }{ sin frac{alpha}{2} } = cot frac{alpha}{2} ). Let me verify that:Recall that:[ 1 + cos alpha = 2 cos^2 frac{alpha}{2} ][ 1 - cos alpha = 2 sin^2 frac{alpha}{2} ]So,[ sqrt{ frac{1 + cos alpha}{1 - cos alpha} } = sqrt{ frac{2 cos^2 frac{alpha}{2}}{2 sin^2 frac{alpha}{2}} } = sqrt{ frac{ cos^2 frac{alpha}{2} }{ sin^2 frac{alpha}{2} } } = frac{ cos frac{alpha}{2} }{ sin frac{alpha}{2} } = cot frac{alpha}{2} ]Yes, that's correct. So,[ frac{h}{k} = cot frac{alpha}{2} ][ h = k cot frac{alpha}{2} ]But ( k = frac{a + b}{2} ), so:[ h = frac{a + b}{2} cot frac{alpha}{2} ]Now, recall that the area ( S ) of the trapezoid is:[ S = frac{(a + b)}{2} times h ]Substituting ( h ) from above:[ S = frac{(a + b)}{2} times left( frac{a + b}{2} cot frac{alpha}{2} right) ][ S = left( frac{a + b}{2} right)^2 cot frac{alpha}{2} ][ S = k^2 cot frac{alpha}{2} ]But from earlier, we have:[ h = k cot frac{alpha}{2} ][ k = frac{h}{ cot frac{alpha}{2} } = h tan frac{alpha}{2} ]Substituting back into the area equation:[ S = (h tan frac{alpha}{2})^2 cot frac{alpha}{2} ][ S = h^2 tan^2 frac{alpha}{2} cot frac{alpha}{2} ]Simplify ( tan^2 frac{alpha}{2} cot frac{alpha}{2} ):[ tan^2 theta cot theta = tan theta ]So,[ S = h^2 tan frac{alpha}{2} ][ h^2 = frac{S}{ tan frac{alpha}{2} } ][ h = sqrt{ frac{S}{ tan frac{alpha}{2} } } ][ h = sqrt{ S cot frac{alpha}{2} } ]Alternatively, since ( cot frac{alpha}{2} = frac{1}{ tan frac{alpha}{2} } ), we can write:[ h = sqrt{ S cdot cot frac{alpha}{2} } ]But let me check the steps again to make sure I didn't make a mistake.Starting from:[ S = k^2 cot frac{alpha}{2} ]And ( k = h tan frac{alpha}{2} ), so substituting:[ S = (h tan frac{alpha}{2})^2 cot frac{alpha}{2} ][ S = h^2 tan^2 frac{alpha}{2} cot frac{alpha}{2} ][ S = h^2 tan frac{alpha}{2} ]Because ( tan^2 theta cot theta = tan theta ).So,[ h^2 = frac{S}{ tan frac{alpha}{2} } ][ h = sqrt{ frac{S}{ tan frac{alpha}{2} } } ][ h = sqrt{ S cot frac{alpha}{2} } ]Yes, that seems correct.Alternatively, since ( cot frac{alpha}{2} = frac{ cos frac{alpha}{2} }{ sin frac{alpha}{2} } ), we can write:[ h = sqrt{ S cdot frac{ cos frac{alpha}{2} }{ sin frac{alpha}{2} } } ][ h = sqrt{ S cdot frac{ cos frac{alpha}{2} }{ sin frac{alpha}{2} } } ]But I think the simplest form is:[ h = sqrt{ S cot frac{alpha}{2} } ]Alternatively, since ( cot frac{alpha}{2} = frac{1}{ tan frac{alpha}{2} } ), we can also write:[ h = sqrt{ frac{S}{ tan frac{alpha}{2} } } ]But both forms are equivalent.Wait, let me check if this makes sense dimensionally. The area ( S ) has units of length squared, and ( cot frac{alpha}{2} ) is dimensionless, so ( h ) has units of length, which is correct.Also, when ( alpha ) approaches 0, ( cot frac{alpha}{2} ) approaches infinity, which would mean the height becomes very large, which makes sense because the diagonals would be almost overlapping, making the trapezoid very "tall".Conversely, when ( alpha ) approaches 180 degrees, ( cot frac{alpha}{2} ) approaches 0, meaning the height approaches 0, which also makes sense because the diagonals would be almost opposite, making the trapezoid very "flat".So, the formula seems to make sense.Therefore, the height ( h ) of the isosceles trapezoid is:[ h = sqrt{ S cot frac{alpha}{2} } ]Alternatively, using ( cot frac{alpha}{2} = frac{ cos frac{alpha}{2} }{ sin frac{alpha}{2} } ), we can write:[ h = sqrt{ S cdot frac{ cos frac{alpha}{2} }{ sin frac{alpha}{2} } } ]But the first form is probably more straightforward.So, to summarize, the height ( h ) is the square root of the area ( S ) multiplied by the cotangent of half the angle ( alpha ).I think that's the answer.**Final Answer**The height of the trapezoid is boxed{sqrt{S cot frac{alpha}{2}}}.

💡Alright, so I've got this problem here: there's a point A at (0,2), and points B and C are on the parabola y² = x + 4. The condition is that AB is perpendicular to BC. I need to find the range of possible values for the y-coordinate of point C.Hmm, okay. Let me start by understanding the problem. We have a fixed point A, and two other points B and C moving along a parabola. The key condition is that the line segments AB and BC are perpendicular. So, I need to figure out where C can be on the parabola such that this perpendicularity condition holds.First, I should probably parameterize points B and C on the parabola. Since the parabola is given by y² = x + 4, I can express any point on it in terms of a parameter. Let me choose a parameter m for point B. So, if I let m be the y-coordinate of B, then the x-coordinate would be m² - 4. So, point B is (m² - 4, m). Similarly, I can let point C be (n² - 4, n), where n is another parameter.Now, I need to find the condition that AB is perpendicular to BC. To do that, I can use the slopes of AB and BC. If two lines are perpendicular, the product of their slopes is -1.So, let's compute the slope of AB first. The slope formula is (y2 - y1)/(x2 - x1). For AB, the coordinates are A(0,2) and B(m² - 4, m). So, the slope k_AB is (m - 2)/(m² - 4 - 0) = (m - 2)/(m² - 4).Similarly, the slope of BC is (n - m)/(n² - 4 - (m² - 4)) = (n - m)/(n² - m²). That simplifies to (n - m)/[(n - m)(n + m)] = 1/(n + m), provided that n ≠ m. Wait, if n = m, then points B and C would be the same, which probably isn't allowed, so I can assume n ≠ m.So, the slope of BC is 1/(n + m). Now, since AB is perpendicular to BC, the product of their slopes should be -1. Therefore:k_AB * k_BC = -1Substituting the slopes:[(m - 2)/(m² - 4)] * [1/(n + m)] = -1Simplify this equation:(m - 2)/( (m² - 4)(n + m) ) = -1Multiply both sides by (m² - 4)(n + m):m - 2 = - (m² - 4)(n + m)Let me expand the right-hand side:m - 2 = - [m²(n + m) - 4(n + m)]= - [m²n + m³ - 4n - 4m]= -m²n - m³ + 4n + 4mSo, bringing all terms to one side:m - 2 + m²n + m³ - 4n - 4m = 0Combine like terms:m³ + m²n - 3m - 4n - 2 = 0Hmm, this seems a bit complicated. Maybe I can rearrange terms:m³ + m²n - 3m - 4n - 2 = 0Let me factor terms with n:n(m² - 4) + m³ - 3m - 2 = 0So,n(m² - 4) = - (m³ - 3m - 2)Therefore,n = - (m³ - 3m - 2)/(m² - 4)Simplify the numerator:m³ - 3m - 2. Let me see if I can factor this. Maybe m = 1 is a root? Plugging in m=1: 1 - 3 - 2 = -4 ≠ 0. m=2: 8 - 6 - 2 = 0. So, m=2 is a root. Therefore, I can factor (m - 2) out of m³ - 3m - 2.Using polynomial division or synthetic division:Divide m³ - 3m - 2 by (m - 2):Coefficients: 1 (m³), 0 (m²), -3 (m), -2 (constant)Using synthetic division:2 | 1 0 -3 -2 2 4 2 1 2 1 0So, it factors to (m - 2)(m² + 2m + 1). And m² + 2m + 1 is (m + 1)². So,m³ - 3m - 2 = (m - 2)(m + 1)²Therefore, n = - [(m - 2)(m + 1)²]/(m² - 4)Note that m² - 4 factors as (m - 2)(m + 2). So,n = - [(m - 2)(m + 1)²]/[(m - 2)(m + 2)]We can cancel out (m - 2) from numerator and denominator, provided m ≠ 2.So, n = - (m + 1)²/(m + 2)Therefore, n = - (m + 1)²/(m + 2)So, now we have n expressed in terms of m. Since n is the y-coordinate of point C, which lies on the parabola y² = x + 4, we can write the coordinates of C as (n² - 4, n). But we need to find the range of possible n.So, n = - (m + 1)²/(m + 2). Let me denote this as:n = - (m + 1)²/(m + 2)We need to find the range of n as m varies over all real numbers except m = -2 (since denominator becomes zero) and m ≠ 2 (since we canceled out (m - 2)).But wait, m can be any real number except m = -2 and m = 2. However, m = 2 is already handled because when m = 2, the original expression for n would have been undefined, but after factoring, we see that m = 2 is a root, so n would be zero? Wait, let me check.Wait, when m approaches 2, n approaches - (2 + 1)²/(2 + 2) = -9/4. But earlier, we had m = 2 as a root in the numerator, so actually, when m = 2, the original expression for n was 0/0, which is indeterminate. So, we can apply L’Hospital’s Rule or recognize that it's a removable discontinuity.But perhaps it's better to consider m ≠ 2 and m ≠ -2.So, n = - (m + 1)²/(m + 2). Let me write this as:n = - (m² + 2m + 1)/(m + 2)Let me perform polynomial division on the numerator and denominator:Divide m² + 2m + 1 by m + 2.m + 2 ) m² + 2m + 1First term: m² / m = mMultiply (m + 2) by m: m² + 2mSubtract from numerator: (m² + 2m + 1) - (m² + 2m) = 1So, it's m + (1)/(m + 2). Therefore,n = - [m + 1/(m + 2)]Wait, no:Wait, actually, the division gives:(m² + 2m + 1) = (m + 2)(m) + 1So,(m² + 2m + 1)/(m + 2) = m + 1/(m + 2)Therefore,n = - [m + 1/(m + 2)]So,n = -m - 1/(m + 2)Hmm, that might be useful. Alternatively, perhaps I can express n in terms of m and then find its range.Alternatively, maybe I can set t = m + 2, so that m = t - 2.Then,n = - ( (t - 2) + 1 )² / t= - (t - 1)² / t= - (t² - 2t + 1)/t= -t + 2 - 1/tSo,n = -t + 2 - 1/tWhere t ≠ 0 (since m ≠ -2, so t = m + 2 ≠ 0).So, n = -t + 2 - 1/tNow, to find the range of n, we can consider t as a real number except t = 0, and find the possible values of n.Let me denote f(t) = -t + 2 - 1/tWe can analyze this function to find its range.First, let's consider t > 0 and t < 0 separately.Case 1: t > 0Then, f(t) = -t + 2 - 1/tLet me compute the derivative to find extrema.f’(t) = -1 + 0 + 1/t²Set derivative to zero:-1 + 1/t² = 01/t² = 1t² = 1t = 1 or t = -1But since t > 0, t = 1.So, at t = 1, f(t) = -1 + 2 - 1/1 = -1 + 2 -1 = 0Now, let's check the behavior as t approaches 0+:As t approaches 0 from the right, -t approaches 0, -1/t approaches -∞. So, f(t) approaches -∞.As t approaches ∞, -t approaches -∞, -1/t approaches 0. So, f(t) approaches -∞.So, in the case t > 0, the function f(t) has a maximum at t = 1, which is 0, and tends to -∞ as t approaches 0+ and ∞.Therefore, for t > 0, n can take any value less than or equal to 0.Case 2: t < 0So, t is negative, let's set t = -s where s > 0.Then, f(t) = -(-s) + 2 - 1/(-s) = s + 2 + 1/sSo, f(t) = s + 2 + 1/s, where s > 0.Now, let's find the range of this function.Compute derivative:f’(s) = 1 - 1/s²Set derivative to zero:1 - 1/s² = 01 = 1/s²s² = 1s = 1 (since s > 0)So, at s = 1, f(s) = 1 + 2 + 1 = 4Now, check behavior as s approaches 0+:s approaches 0+, 1/s approaches ∞, so f(s) approaches ∞.As s approaches ∞, s approaches ∞, 1/s approaches 0, so f(s) approaches ∞.Therefore, the function f(s) has a minimum at s = 1, which is 4, and tends to ∞ as s approaches 0+ and ∞.Therefore, for t < 0, n can take any value greater than or equal to 4.Putting both cases together, the range of n is (-∞, 0] ∪ [4, ∞).But wait, let me verify this because sometimes when dealing with parametric equations, there might be restrictions.Wait, n is the y-coordinate of point C on the parabola y² = x + 4. So, n can be any real number, but in our case, we've found that n can be at most 0 or at least 4.So, the range of possible values for the y-coordinate of point C is (-∞, 0] ∪ [4, ∞).But let me double-check if there are any constraints I might have missed.Wait, when m approaches -2, t approaches 0, but m ≠ -2, so t ≠ 0. So, in the case t > 0, n can be up to 0, but not including 0? Wait, no, at t =1, n=0, so 0 is included.Similarly, when t <0, n can be as low as 4, but wait, no, when t approaches 0 from the negative side, s approaches 0 from the positive side, so n approaches ∞. When s approaches ∞, n approaches ∞ as well. So, the minimum value is 4, achieved when s=1, t=-1.So, yes, the range is n ≤ 0 or n ≥4.Therefore, the range of possible values for the y-coordinate of point C is (-∞, 0] ∪ [4, ∞).**Final Answer**The range of possible values for the y-coordinate of point ( C ) is boxed{(-infty, 0] cup [4, infty)}.

💡Okay, so I have this geometry problem here. It says: In triangle ABC, where angle B is 90 degrees, we know that the tangent of angle A is 3/7, and the length of side AB is 14. I need to find the length of side BC. Hmm, let me try to visualize this.First, since angle B is 90 degrees, triangle ABC is a right-angled triangle at B. So, sides AB and BC are the legs, and AC is the hypotenuse. Let me label the triangle: point A, point B, and point C, with the right angle at B. So, AB is one leg, BC is the other leg, and AC is the hypotenuse.They gave me that tan A is 3/7. I remember that tangent of an angle in a right triangle is the ratio of the opposite side to the adjacent side. So, tan A = opposite over adjacent. In this case, angle A is at point A, so the side opposite to angle A would be BC, and the side adjacent to angle A would be AB. So, tan A = BC / AB.Wait, let me make sure. If angle A is at point A, then the sides relative to angle A are: opposite side is BC, adjacent side is AB, and hypotenuse is AC. So, yes, tan A = BC / AB.They told me that tan A is 3/7, so that means BC / AB = 3/7. And AB is given as 14. So, I can set up the equation: BC / 14 = 3/7.To find BC, I can solve for it. Multiply both sides by 14: BC = (3/7) * 14. Let's calculate that. 14 divided by 7 is 2, so 3 times 2 is 6. So, BC is 6. Wait, that seems straightforward. But let me double-check because sometimes I might mix up the sides.Alternatively, maybe I should use the Pythagorean theorem to find BC. Since I know AB is 14 and BC is 6, then AC should be the hypotenuse. Let me check: AB squared plus BC squared equals AC squared. So, 14 squared is 196, and 6 squared is 36. Adding them together, 196 + 36 equals 232. So, AC squared is 232, which means AC is the square root of 232. Hmm, that simplifies to sqrt(4*58) which is 2*sqrt(58). Okay, that seems right.But wait, the question only asks for BC, which I found to be 6. Is there something I'm missing here? Maybe I should confirm the initial step where I used the tangent ratio. Let me think again: tan A = opposite over adjacent. Opposite side to angle A is BC, adjacent is AB. So, yes, tan A = BC / AB. Plugging in the numbers: 3/7 = BC / 14, so BC = (3/7)*14 = 6. That seems consistent.Alternatively, if I consider the sides in terms of a 3-7 ratio, maybe I can think of the triangle sides as multiples of 3 and 7. Since AB is 14, which is 2 times 7, then BC should be 2 times 3, which is 6. That also makes sense. So, that reinforces that BC is 6.Wait, but earlier when I calculated AC, I got 2*sqrt(58). Is that necessary for the problem? The problem only asks for BC, so maybe I don't need to calculate AC unless I made a mistake in my initial calculation. Let me see.If BC is 6, then using Pythagoras, AC should be sqrt(14^2 + 6^2) = sqrt(196 + 36) = sqrt(232) = 2*sqrt(58). That seems correct, but since the problem doesn't ask for AC, maybe I don't need to worry about that.Alternatively, maybe I should consider using sine or cosine to find another side, but since I have the tangent, and I can directly find BC from that, it seems more straightforward.Wait, another thought: sometimes, in right triangles, if you know one angle, you can find the other non-right angle, but in this case, since we're given the tangent of angle A, and we don't need the other angle, maybe that's not necessary here.So, to recap: right triangle at B, tan A = 3/7, AB = 14. So, tan A = BC / AB = 3/7. Therefore, BC = (3/7)*14 = 6. That seems solid.But just to be thorough, let me consider if I might have mixed up the sides. If angle A is at point A, then the sides are: opposite is BC, adjacent is AB, hypotenuse is AC. So, yes, tan A = BC / AB. So, no, I didn't mix them up.Alternatively, if I had angle C, then the opposite side would be AB, but since it's angle A, it's BC. So, I think I'm correct.Therefore, I think BC is 6. But wait, in my initial calculation using Pythagoras, I got AC as 2*sqrt(58). Let me check if that's correct. 14 squared is 196, 6 squared is 36, 196 + 36 is 232, sqrt(232) is sqrt(4*58), which is 2*sqrt(58). Yes, that's correct. So, AC is 2*sqrt(58), but since the problem only asks for BC, which is 6, I think that's the answer.Wait, but in the initial problem statement, it's written as "Find BC." So, I think I'm confident that BC is 6. But just to make sure, let me think if there's another way to approach this problem.Another approach could be to use the definition of tangent in terms of sine and cosine. Since tan A = sin A / cos A = 3/7. Then, sin A = 3/sqrt(3^2 + 7^2) = 3/sqrt(58), and cos A = 7/sqrt(58). Then, using the definition of sine as opposite over hypotenuse, sin A = BC / AC = 3/sqrt(58). But since AC is the hypotenuse, which we found earlier as 2*sqrt(58), then BC would be sin A * AC = (3/sqrt(58)) * 2*sqrt(58) = 6. So, that also gives BC = 6. So, that's another way to confirm it.Alternatively, using cosine: cos A = AB / AC = 7/sqrt(58). So, AB = 14, so 14 = (7/sqrt(58)) * AC. Therefore, AC = 14 * sqrt(58)/7 = 2*sqrt(58). Then, using Pythagoras, BC = sqrt(AC^2 - AB^2) = sqrt((2*sqrt(58))^2 - 14^2) = sqrt(4*58 - 196) = sqrt(232 - 196) = sqrt(36) = 6. So, that also gives BC = 6.So, regardless of the method I use—whether directly using the tangent ratio, using sine and cosine, or using the Pythagorean theorem—I consistently get BC = 6. Therefore, I'm confident that the length of BC is 6.Wait, but in the initial problem, the user wrote the same problem twice, once in the title and once in the content. Maybe that's just a formatting thing. Anyway, focusing back on the problem: right triangle ABC, angle B is 90 degrees, tan A = 3/7, AB = 14, find BC.I think I've covered all the bases here. I've used multiple methods to confirm that BC is 6. So, unless I made a calculation error somewhere, which I don't see, I think BC is 6.But just to double-check the calculations one more time:Given tan A = 3/7, which is BC / AB.AB = 14, so BC = (3/7)*14 = 6.Yes, that's straightforward.Alternatively, if I consider the sides as parts of a 3-7-? triangle, scaled by a factor. Since AB is 14, which is 2*7, then the scaling factor is 2. Therefore, BC, which corresponds to 3, would be 2*3 = 6. That's another way to see it.So, yes, BC is 6.**Final Answer**The length of side ( BC ) is boxed{6}.

💡Okay, so I have this problem about two robots, A and B, starting from the same point on a circular track that's 8 meters around. Robot A moves at 5 meters per minute, and Robot B moves at 3 meters per minute. They start moving in opposite directions. When they meet head-on, Robot B turns around, and if Robot A catches up from behind, Robot A turns around. I need to figure out where they meet for the 10th time, measured from the starting point.Alright, let's break this down. First, they start at the same point, moving in opposite directions. Since they're moving towards each other, their relative speed is the sum of their speeds. So, Robot A is going 5 m/min and Robot B is going 3 m/min. So together, they're closing the distance between them at 5 + 3 = 8 m/min.The track is 8 meters around, so the first time they meet, they would have covered the entire circumference together. Since their combined speed is 8 m/min, the time it takes for them to meet the first time is 8 meters divided by 8 m/min, which is 1 minute. So, after 1 minute, they meet for the first time.Now, where do they meet? Since Robot A is moving faster, it would have covered more distance. In 1 minute, Robot A would have gone 5 meters, and Robot B would have gone 3 meters. So, they meet 5 meters from the starting point in the direction Robot A is moving. Alternatively, since the track is circular, it's also 3 meters in the opposite direction from Robot B's starting point.But wait, when they meet head-on, Robot B turns around. So, after the first meeting, Robot B is now moving in the same direction as Robot A. That changes things because now they're moving in the same direction, and Robot A is faster. So, Robot A will eventually catch up to Robot B from behind.When Robot A catches up from behind, Robot A turns around. So, now Robot A is moving in the opposite direction, and Robot B is still moving in the same direction as before. This creates a cycle where they keep turning around each time they meet, either head-on or from behind.I think the key here is to figure out the pattern of their meetings. Each time they meet, one of them turns around, changing their direction. Since the track is circular, their meetings will repeat in a periodic fashion. Maybe I can figure out how often they meet and where.Let me try to calculate the time between meetings. Initially, they meet after 1 minute. Then, after they turn around, they start moving in opposite directions again, so their relative speed is still 8 m/min. But wait, after the first meeting, Robot B turns around, so now Robot A is moving at 5 m/min in one direction, and Robot B is moving at 3 m/min in the opposite direction. So, their relative speed is still 5 + 3 = 8 m/min.But wait, after the first meeting, Robot B turns around, so now they're moving in the same direction as each other? No, wait, no. If they were moving in opposite directions initially, and then Robot B turns around, they're now moving in the same direction. So, their relative speed is the difference between their speeds, which is 5 - 3 = 2 m/min.Wait, that's different. So, after the first meeting, they're moving in the same direction, so the relative speed is 2 m/min. That means the time until they meet again is the circumference divided by the relative speed, which is 8 / 2 = 4 minutes.So, the first meeting is at 1 minute, the second meeting is at 1 + 4 = 5 minutes, the third meeting is at 5 + 4 = 9 minutes, and so on.But wait, when they meet the second time, it's because Robot A has caught up from behind. So, according to the problem, when Robot A catches up from behind, Robot A turns around. So, after the second meeting, Robot A is now moving in the opposite direction, and Robot B is still moving in the same direction.So, now, after the second meeting, they're moving in opposite directions again, so their relative speed is 5 + 3 = 8 m/min. Therefore, the time until the next meeting is 8 / 8 = 1 minute.So, the third meeting is at 9 + 1 = 10 minutes, the fourth meeting is at 10 + 1 = 11 minutes, and so on.Wait, that seems inconsistent. Let me try to list out the meetings with their times and directions.1. First meeting: 1 minute, head-on, Robot B turns around.2. Second meeting: 5 minutes, from behind, Robot A turns around.3. Third meeting: 9 minutes, head-on, Robot B turns around.4. Fourth meeting: 13 minutes, from behind, Robot A turns around.5. Fifth meeting: 17 minutes, head-on, Robot B turns around.6. Sixth meeting: 21 minutes, from behind, Robot A turns around.7. Seventh meeting: 25 minutes, head-on, Robot B turns around.8. Eighth meeting: 29 minutes, from behind, Robot A turns around.9. Ninth meeting: 33 minutes, head-on, Robot B turns around.10. Tenth meeting: 37 minutes, from behind, Robot A turns around.Wait, so the pattern is that every two meetings, the time between them alternates between 4 minutes and 1 minute. So, the first meeting is at 1 minute, then 4 minutes later at 5 minutes, then 4 minutes later at 9 minutes, and so on. But that doesn't seem right because after the second meeting, they should be moving in opposite directions again, so the time between meetings should be 1 minute.But according to my initial calculation, after the first meeting, they meet every 4 minutes when moving in the same direction, and every 1 minute when moving in opposite directions. So, the cycle is 4 + 1 = 5 minutes for two meetings.Wait, let me think again. After the first meeting, they're moving in the same direction, so the time until the next meeting is 8 / (5 - 3) = 4 minutes. Then, after that meeting, they turn around again, so they're moving in opposite directions, and the time until the next meeting is 8 / (5 + 3) = 1 minute. So, the cycle is 4 + 1 = 5 minutes for two meetings.Therefore, every 5 minutes, they have two meetings. So, for the 10th meeting, that's 5 cycles of 5 minutes each, totaling 25 minutes.Wait, but according to my earlier list, the 10th meeting is at 37 minutes. That contradicts. Hmm.Let me try to calculate the time for each meeting step by step.1. Start at time 0, both at position 0.2. First meeting: time = 1 minute, position = 5 meters (from starting point). - Robot B turns around.3. Now, Robot A is moving at 5 m/min clockwise, Robot B is moving at 3 m/min clockwise. - Relative speed = 5 - 3 = 2 m/min. - Time until next meeting: 8 / 2 = 4 minutes. - Second meeting at time = 1 + 4 = 5 minutes. - Position: Robot A has moved 5 m/min * 4 min = 20 meters, which is 20 mod 8 = 4 meters from starting point. - But since they meet from behind, it's the same as 4 meters from starting point in the clockwise direction. - Robot A turns around.4. Now, Robot A is moving counterclockwise at 5 m/min, Robot B is moving clockwise at 3 m/min. - Relative speed = 5 + 3 = 8 m/min. - Time until next meeting: 8 / 8 = 1 minute. - Third meeting at time = 5 + 1 = 6 minutes. - Position: Robot A has moved 5 m/min * 1 min = 5 meters counterclockwise, which is 8 - 5 = 3 meters clockwise from starting point. - Robot B has moved 3 m/min * 1 min = 3 meters clockwise, so from 4 meters, it's now at 4 + 3 = 7 meters, which is 7 meters clockwise. - Wait, that doesn't make sense. If they meet at 3 meters, but Robot B was at 4 meters and moved 3 meters, it should be at 7 meters. Hmm, maybe I need to think differently.Wait, maybe I should calculate the position based on their movements.After the second meeting at 5 minutes, Robot A is at 4 meters (from starting point clockwise). Then, Robot A turns around and starts moving counterclockwise at 5 m/min. Robot B continues clockwise at 3 m/min.So, from time 5 to 6 minutes:- Robot A moves counterclockwise: 5 m/min * 1 min = 5 meters. Starting from 4 meters, moving counterclockwise 5 meters: 4 - 5 = -1 meters, which is equivalent to 7 meters on the circular track (since -1 + 8 = 7).- Robot B moves clockwise: 3 m/min * 1 min = 3 meters. Starting from 4 meters, moving clockwise 3 meters: 4 + 3 = 7 meters.So, they meet at 7 meters from starting point at time 6 minutes. So, third meeting at 6 minutes, position 7 meters.Then, since it's a head-on meeting, Robot B turns around.Now, Robot A is moving counterclockwise at 5 m/min, Robot B is moving counterclockwise at 3 m/min.So, relative speed is 5 - 3 = 2 m/min (since they're moving in the same direction).Time until next meeting: 8 / 2 = 4 minutes.Fourth meeting at time = 6 + 4 = 10 minutes.Position: Robot A has moved 5 m/min * 4 min = 20 meters counterclockwise from 7 meters. 20 mod 8 = 4 meters. So, 7 - 4 = 3 meters (since moving counterclockwise). Wait, 7 - 4 = 3 meters counterclockwise from starting point, which is equivalent to 5 meters clockwise.Wait, maybe I should calculate it differently. Starting from 7 meters, moving counterclockwise 20 meters: 7 - 20 = -13 meters, which is equivalent to -13 + 16 = 3 meters (since 8 meters circumference, so 16 is two laps). So, 3 meters counterclockwise from starting point, which is 5 meters clockwise.But Robot B was moving counterclockwise at 3 m/min for 4 minutes: 3 * 4 = 12 meters. Starting from 7 meters, moving counterclockwise 12 meters: 7 - 12 = -5 meters, which is 3 meters counterclockwise (since -5 + 8 = 3).So, they meet at 3 meters counterclockwise, which is 5 meters clockwise from starting point.Since this is a meeting from behind (Robot A catches up from behind), Robot A turns around.Now, Robot A is moving clockwise at 5 m/min, Robot B is moving counterclockwise at 3 m/min.Relative speed = 5 + 3 = 8 m/min.Time until next meeting: 8 / 8 = 1 minute.Fifth meeting at time = 10 + 1 = 11 minutes.Position: Robot A moves clockwise 5 m/min * 1 min = 5 meters from 5 meters (previous meeting point). So, 5 + 5 = 10 meters, which is 2 meters on the track (10 mod 8 = 2).Robot B moves counterclockwise 3 m/min * 1 min = 3 meters from 3 meters (previous meeting point). So, 3 - 3 = 0 meters, which is the starting point.Wait, that doesn't make sense. If Robot A is at 5 meters and moves 5 meters clockwise, it's at 10 meters, which is 2 meters. Robot B is at 3 meters and moves 3 meters counterclockwise, which is 0 meters. So, they meet at 2 meters and 0 meters? That can't be right.Wait, maybe I made a mistake in tracking their positions. Let me try again.After the fourth meeting at 10 minutes, they meet at 3 meters counterclockwise (5 meters clockwise). Then, Robot A turns around and starts moving clockwise at 5 m/min, while Robot B continues moving counterclockwise at 3 m/min.So, from time 10 to 11 minutes:- Robot A moves clockwise: 5 m/min * 1 min = 5 meters. Starting from 5 meters clockwise, moving 5 meters clockwise: 5 + 5 = 10 meters, which is 2 meters (10 mod 8 = 2).- Robot B moves counterclockwise: 3 m/min * 1 min = 3 meters. Starting from 5 meters clockwise, moving 3 meters counterclockwise: 5 - 3 = 2 meters.So, they meet at 2 meters clockwise from starting point at time 11 minutes.Since it's a head-on meeting, Robot B turns around.Now, Robot A is moving clockwise at 5 m/min, Robot B is moving clockwise at 3 m/min.Relative speed = 5 - 3 = 2 m/min.Time until next meeting: 8 / 2 = 4 minutes.Sixth meeting at time = 11 + 4 = 15 minutes.Position: Robot A moves clockwise 5 m/min * 4 min = 20 meters. Starting from 2 meters, moving 20 meters clockwise: 2 + 20 = 22 meters, which is 22 mod 8 = 6 meters.Robot B moves clockwise 3 m/min * 4 min = 12 meters. Starting from 2 meters, moving 12 meters clockwise: 2 + 12 = 14 meters, which is 14 mod 8 = 6 meters.So, they meet at 6 meters clockwise from starting point at time 15 minutes.Since this is a meeting from behind (Robot A catches up from behind), Robot A turns around.Now, Robot A is moving counterclockwise at 5 m/min, Robot B is moving clockwise at 3 m/min.Relative speed = 5 + 3 = 8 m/min.Time until next meeting: 8 / 8 = 1 minute.Seventh meeting at time = 15 + 1 = 16 minutes.Position: Robot A moves counterclockwise 5 m/min * 1 min = 5 meters. Starting from 6 meters, moving 5 meters counterclockwise: 6 - 5 = 1 meter.Robot B moves clockwise 3 m/min * 1 min = 3 meters. Starting from 6 meters, moving 3 meters clockwise: 6 + 3 = 9 meters, which is 1 meter (9 mod 8 = 1).So, they meet at 1 meter clockwise from starting point at time 16 minutes.Since it's a head-on meeting, Robot B turns around.Now, Robot A is moving counterclockwise at 5 m/min, Robot B is moving counterclockwise at 3 m/min.Relative speed = 5 - 3 = 2 m/min.Time until next meeting: 8 / 2 = 4 minutes.Eighth meeting at time = 16 + 4 = 20 minutes.Position: Robot A moves counterclockwise 5 m/min * 4 min = 20 meters. Starting from 1 meter, moving 20 meters counterclockwise: 1 - 20 = -19 meters, which is equivalent to -19 + 24 = 5 meters (since 8 meters circumference, so 24 is three laps). So, 5 meters counterclockwise from starting point, which is 3 meters clockwise.Robot B moves counterclockwise 3 m/min * 4 min = 12 meters. Starting from 1 meter, moving 12 meters counterclockwise: 1 - 12 = -11 meters, which is equivalent to -11 + 16 = 5 meters (since 8 meters circumference, so 16 is two laps). So, 5 meters counterclockwise from starting point, which is 3 meters clockwise.So, they meet at 5 meters counterclockwise (3 meters clockwise) from starting point at time 20 minutes.Since this is a meeting from behind (Robot A catches up from behind), Robot A turns around.Now, Robot A is moving clockwise at 5 m/min, Robot B is moving counterclockwise at 3 m/min.Relative speed = 5 + 3 = 8 m/min.Time until next meeting: 8 / 8 = 1 minute.Ninth meeting at time = 20 + 1 = 21 minutes.Position: Robot A moves clockwise 5 m/min * 1 min = 5 meters. Starting from 3 meters clockwise, moving 5 meters clockwise: 3 + 5 = 8 meters, which is 0 meters (8 mod 8 = 0).Robot B moves counterclockwise 3 m/min * 1 min = 3 meters. Starting from 3 meters clockwise, moving 3 meters counterclockwise: 3 - 3 = 0 meters.So, they meet at 0 meters, the starting point, at time 21 minutes.Since it's a head-on meeting, Robot B turns around.Now, Robot A is moving clockwise at 5 m/min, Robot B is moving clockwise at 3 m/min.Relative speed = 5 - 3 = 2 m/min.Time until next meeting: 8 / 2 = 4 minutes.Tenth meeting at time = 21 + 4 = 25 minutes.Position: Robot A moves clockwise 5 m/min * 4 min = 20 meters. Starting from 0 meters, moving 20 meters clockwise: 0 + 20 = 20 meters, which is 20 mod 8 = 4 meters.Robot B moves clockwise 3 m/min * 4 min = 12 meters. Starting from 0 meters, moving 12 meters clockwise: 0 + 12 = 12 meters, which is 12 mod 8 = 4 meters.So, they meet at 4 meters clockwise from starting point at time 25 minutes.Therefore, the distance from the starting point to the location where they meet for the 10th time is 4 meters.

💡First, I need to determine the total number of ways to select 3 volunteers from the class without any restrictions. This can be calculated using the combination formula ( C(n, k) ), which represents the number of ways to choose ( k ) items from a set of ( n ) items without regard to order.The total number of students in the class is 4 boys + 3 girls = 7 students. Therefore, the total number of ways to select 3 volunteers is:[C(7, 3) = frac{7!}{3!(7-3)!} = 35]Next, I need to calculate the number of selections that do not meet the criteria of including both boys and girls. These are the selections that consist entirely of boys or entirely of girls.1. **All Boys:** There are 4 boys in the class. The number of ways to select 3 boys from 4 is: [ C(4, 3) = frac{4!}{3!(4-3)!} = 4 ]2. **All Girls:** There are 3 girls in the class. The number of ways to select 3 girls from 3 is: [ C(3, 3) = frac{3!}{3!(3-3)!} = 1 ]Now, subtract the number of all-boy and all-girl selections from the total number of selections to find the number of valid selections that include both boys and girls:[35 - 4 - 1 = 30]Therefore, the total number of different selection methods that include both boys and girls is 30.

💡First, I need to determine the total distance of Ezekiel's three-day hike. I'll start by identifying the distances he covered each day.On the first day, Ezekiel hiked 10 kilometers.On the second day, he covered half of the total hike distance. Let's denote the full hike distance as ( X ). Therefore, the distance he covered on the second day is ( frac{1}{2}X ).On the third day, he hiked 15 kilometers to complete the hike.To find the total hike distance, I'll add up the distances from each day and set the sum equal to the total distance ( X ):[ 10 + frac{1}{2}X + 15 = X ]Combining the constants:[ 25 + frac{1}{2}X = X ]Subtracting ( frac{1}{2}X ) from both sides to isolate ( X ):[ 25 = frac{1}{2}X ]Multiplying both sides by 2 to solve for ( X ):[ 50 = X ]Therefore, the full hike was 50 kilometers long.