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
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 #223a7 - Socratic Equation of the circle = x^2 + y^2 = 50 Equation of circle with origin as center is x^2 + y^2 = a^2 where ais the radius of the circle To find the radius we have to calculate the distance from center to a point on the circumference radjus r = sqrt( (0-(-7))^2 + (0-(-1))^2) = sqrt50 Hence equation of the circle is x^2 + y^2 = 50 Verification by finding the distance from center to the other
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
Question #48ef1 - Socratic The answer is 3000 ft^2 To obtain the area of a pavement, the formula: A = l xx w will be used Convert first 100 yd to ft: l = 100 yd xx (3ft) (1yd) = 300ft then substitute the given values l = 300ft w= 10ft A= l xx w = 300ft xx 10ft = 3000ft^2
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