answer:(1) Since v=60 implies d=2.66l, we have k= frac {2.66l- frac {1}{2}l}{60^{2}l}= frac {2.16}{60^{2}}=0.0006. Thus, the relationship between d and v is d=0.0024v^{2}+2. (2) Let Q be the number of vehicles passing through the bridge per hour. To maximize Q, we need to minimize frac {1000v}{d+4}, or equivalently, minimize Q= frac {v}{0.0024v^{2}+6}= frac {1}{0.0024v+ frac {6}{v}}. By the arithmetic mean-geometric mean inequality, we have 0.0024v+ frac {6}{v}geqslant 2 sqrt {0.0024vtimes frac {6}{v}}=0.24, with equality if and only if 0.0024v= frac {6}{v}, which happens when v=50. Therefore, Q attains its maximum value boxed{frac {12500}{3}} when v=boxed{50(km/h)}.

answer:The molecular weight of a substance is the sum of the atomic weights of all the atoms in its chemical formula. The chemical formula for HBrO3 is: H (hydrogen) = 1 atom Br (bromine) = 1 atom O (oxygen) = 3 atoms Using the atomic weights from the periodic table: H = 1.01 g/mol Br = 79.90 g/mol O = 16.00 g/mol The molecular weight of HBrO3 is calculated as follows: (1 x 1.01 g/mol) + (1 x 79.90 g/mol) + (3 x 16.00 g/mol) = 1.01 g/mol + 79.90 g/mol + 48.00 g/mol = 128.91 g/mol Now, to find the weight of 3 moles of HBrO3: 3 moles x 128.91 g/mol = 386.73 g Therefore, the molecular weight of 3 moles of HBrO3 is boxed{386.73} grams.

answer:Let's denote the fixed monthly charge for internet service as ( F ) and the charge for the calls Elvin made in January as ( C_J ). The total telephone bill for January is the sum of the fixed charge and the charge for the calls, so we have: [ F + C_J = 48 ] (Equation 1) For February, the charge for the calls Elvin made is twice the charge for the calls he made in January, so we can denote it as ( 2C_J ). The total telephone bill for February is then: [ F + 2C_J = 90 ] (Equation 2) Now we have a system of two equations with two unknowns. We can solve for ( F ) by subtracting Equation 1 from Equation 2: [ (F + 2C_J) - (F + C_J) = 90 - 48 ] [ F + 2C_J - F - C_J = 42 ] [ 2C_J - C_J = 42 ] [ C_J = 42 ] Now that we have the value for ( C_J ), we can substitute it back into Equation 1 to find ( F ): [ F + 42 = 48 ] [ F = 48 - 42 ] [ F = 6 ] Therefore, Elvin's fixed monthly charge for internet service is boxed{6} .

answer:First, let's represent the points and direction vectors: [bold{a} = begin{pmatrix} 5 1 end{pmatrix}, quad bold{b} = begin{pmatrix} 6 -2 end{pmatrix}, quad bold{d} = begin{pmatrix} 2 -4 end{pmatrix}.] Then, calculate the vector (bold{v}) from a point on the first line to a point on the second line: [bold{v} = bold{b} - bold{a} = begin{pmatrix} 6 -2 end{pmatrix} - begin{pmatrix} 5 1 end{pmatrix} = begin{pmatrix} 1 -3 end{pmatrix}.] Next, compute the projection of (bold{v}) onto (bold{d}): [text{proj}_{bold{d}} bold{v} = frac{bold{v} cdot bold{d}}{bold{d} cdot bold{d}} bold{d} = frac{begin{pmatrix} 1 -3 end{pmatrix} cdot begin{pmatrix} 2 -4 end{pmatrix}}{begin{pmatrix} 2 -4 end{pmatrix} cdot begin{pmatrix} 2 -4 end{pmatrix}} begin{pmatrix} 2 -4 end{pmatrix}.] Calculating the dot products: [bold{v} cdot bold{d} = 1 cdot 2 + (-3) cdot (-4) = 2 + 12 = 14,] [bold{d} cdot bold{d} = 2 cdot 2 + (-4) cdot (-4) = 4 + 16 = 20,] [text{proj}_{bold{d}} bold{v} = frac{14}{20} begin{pmatrix} 2 -4 end{pmatrix} = frac{7}{10} begin{pmatrix} 2 -4 end{pmatrix} = begin{pmatrix} frac{14}{10} -frac{28}{10} end{pmatrix} = begin{pmatrix} 1.4 -2.8 end{pmatrix}.] The perpendicular vector (bold{p}) between the lines is: [bold{p} = bold{v} - text{proj}_{bold{d}} bold{v} = begin{pmatrix} 1 -3 end{pmatrix} - begin{pmatrix} 1.4 -2.8 end{pmatrix} = begin{pmatrix} -0.4 -0.2 end{pmatrix}.] Finally, the distance between the lines is the magnitude of (bold{p}): [|bold{p}| = sqrt{(-0.4)^2 + (-0.2)^2} = sqrt{0.16 + 0.04} = sqrt{0.20} = boxed{frac{sqrt{5}}{5}}.]