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Answers by Herb K.

let area of rectangular field be A = xy, where x = length, y = width;  then: A = 3,000,000 = xy ---> y = 3,000,000/x = 3,000,000 x^(-1); ---> L = length of fencing = 2x + 2y + y = 2x + 3y = 2x + 3(3,000,000)[x^(-1)] = 2x + (9,000,000)[x^(-1)];   compute...

x - y^2 = 0 ---> x = y^2 ( parabola opening to the right, vertex at  (0,0)) ;  let point on parabola that minimizes distance from (0,3) be (a,b); then, slope of line joining (0,3) and (a,b) = m =  -(3 - b)/a = (b - 3)/a;  using a = b^2, we have the following:  m = (b...

I agree with Kenneth S. that since the integrand is an odd function of x, and the integral, call it I, is between symmetrically placed limits, -a and a, the result should be, even without any detailed calculations, I = 0.

perhaps the two curves were f(x) = 2sin((π/2)x), and g(x) = cos((π/2)x);  if so, then that would seem to imply that the area in question is:                         b             ∫ [(f(x) -...

actually, the work done on a body, W = (component of force in direction of motion)(distance)                                                       = increase...

let x = measure of first angle;  let y = measure of second angle;  let z = measure of third angle;   then:  x = 2y;  z = y + 96;   and x + y + z = 180 ---> 2y + y + y + 96 = 180 ---> 4y + 96 = 180 ---> 4y = 180 - 96   --->...

let x = cost of one donut; and let y = cost of one large coffee;  then                  2x + 4y = 5.10; and ---> 2x + 4y = 5.10                  6x + 3y = 6.39 --------->...

write sin(7*pi/12) = sin[(3*pi/12 + 4*pi/12) = sin(pi/4 + pi/3) = sin(pi/4)cos(pi/3) + sin(pi/3)cos(pi/4)   = [sqrt(2)/2](1/2) + [sqrt(3)/2][sqrt(2)/2], where we have used sum formula:  sin(x + y) = sin(x)cos(y) + sin(y)cos(x)

let x = number of color copies; and let y = number of black and white copies; then,    x + y = 90 -------------------> 9x + 9y = 9(90) = 810   0.49x + 0.09y = 12.90 ----> 49x + 9y = 1290   subtracting the top equation from the bottom...

looks to me like Mark M. is right --->     least amount of time:  number of Field Events = 0; number of Track Events = 100;  time involved = 100(15) = 1500 min   most amount of time:  number of Track Events = 0; number of Field Events = 250;  time...