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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
  • Circle A has a center at (5 ,1 ) and an area of 4 pi. Circle . . . - Socratic
    Circle A has a center at # (5 ,1 )# and an area of #4 pi# Circle B has a center at # (2 ,8 )# and an area of #16 pi# Do the circles overlap? If not, what is the shortest distance between them? GeometryAnalytic GeometryDistance between Points
  • Question #4e20b - Socratic
    0,0200 M Since MM of AgI is 235 g mol , 2,35 g contains 0,0100 mol and for the reaction AgNO_3 + HI = AgI + HNO_3 you have also 0,0100 mol of silver nitrate before the reaction the Molarity is given by M= (mol) (V(L)) = (0,0100 mol) (0,0500 L)= 0,200 M pay attention because: 1)a too high excess of hydrogen iodide forms AgI_2^- 2) moreover iodine can be oxidated by Oxygen of air and by nitrate
  • Circle A has a center at # (3 ,2 )# and an area of #96 pi . . . - Socratic
    Circle A has a center at # (3 ,2 )# and an area of #96 pi# Circle B has a center at # (12 ,7 )# and an area of #48 pi# Do the circles overlap? GeometryAnalytic GeometryDistance between Points





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