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  • Question #51be1 - Socratic
    A = 2 + pi 4 Since the value of costheta goes from 1 to -1 and back to 1 (as you rotate counterclockwise starting from theta = 0), the value of 1 - costheta will go from 0 to 2 and back to 0 This means that the cardioid r = 1-costheta should intersect the unit circle at two places, and the area we're looking for is between these two values So let's find these two values: r_1 = r_2 1
  • A triangle has corners at # (3 , 2 )#, # (6 ,7 )#, and # (2 . . . - Socratic
    A triangle has corners at # (3 , 2 )#, # (6 ,7 )#, and # (2 ,4 )# What is the radius of the triangle's inscribed circle? GeometryCirclesArea of Inscribed Triangle
  • A solid consists of a cone on top of a cylinder with a . . . - Socratic
    V_T=pir^2 (h_1 3+h_2) We need to calculate r in order to calculate the area of the base of the cylinder, hence we fill in the data given 150pi=pir^2 (39 3+17) We cancel the like term (pi) on each side 150cancelpi=cancelpir^2 (39 3+17) 150=r^2 (13+17) 150=r^2xx30 Divide both sides by 30 150 30=r^2 5=r^2 The formula of area of the base of a
  • Question #08cc1 - Socratic
    See a solution process below: The formula for the circumference of a circle is: C = 2pir Where: C is the circumference of the circle r is the radius of the circle However, we also know; 2r = d Where: r is the radius of the circle d is the diameter of the circle We can rewrite the equation for the circumference of a circle as: C = 2pir = 2rpi = pid Substituting for C and solving for d gives
  • Questions asked by Karla - Socratic
    Questions asked by Karla Back to user's profile A viscous liquid is poured onto a flat surface it forms a circular patch whose area grows at a steady rate of 5cm^2s^-1 find in term of pi (a) the radius of patch 20 sec after pouring has commenced (b) the rate of increase of the radius at this instant? The volume of a spherical balloon is increasing at a constant rate of 0 25m3s−1 Find the
  • Question #5321e - Socratic
    a 1cm^3 b 20,000 04cm^2 c 0 0001cm Per data given, the volume of the gold sheet can be solved directly through the formula: rho=m V; rearrange it to isolate the volume (V) V=m rho; plug in values V= (19 3cancel (g)) ( (19 3cancel (g)) (cm^3)) V=1cm^3 Now, find the thickness (T) of the gold sheet Given the measurements of L=100cm and W=100cm; T can be computed as follows: V=LxxWxxT T= (V
  • Lakia wants to make an enclosed paper basket shaped like a . . . - Socratic
    pirl is the formula for the surface area and pir^2 is the formula for the area of the base 3 14xx2 5xx8=62 8 square inches 3 14xx2 5^2=19 625 She needs 82 425 square inches of paper
  • Thomas C. on Socratic
    B S in Accounting, B A in Liberal Arts Fan of History and Math From Louisiana
  • A chord with a length of 13 runs from pi 12 to pi 2 radians on a circle . . .
    A chord with a length of #13 # runs from #pi 12 # to #pi 2 # radians on a circle What is the area of the circle? GeometryPerimeter, Area, and VolumeCircumference and Area of Circles
  • A triangle has corners at (6 ,8 ), (1 ,2 ), and (3 ,9 ). What is the . . .
    Area of the triangle's circumscribed circle is 48 005 If the sides of a triangle are a, b and c, then the area of the triangle Delta is given by the formula Delta=sqrt(s(s-a)(s-b)(s-c)), where s=1 2(a+b+c) and radius of circumscribed circle is (abc) (4Delta) Hence let us find the sides of triangle formed by (6,8), (1,2) and (3,9) This will be surely distance between pair of points, which is





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