MATH SOLVE

3 months ago

Q:
# 42. What is the surface area of a sphere with a circumference of 50 feet round the answer to the nearest 10th. 43. The volume of a sphere is 2254 pi m^3. What is the surface of the sphere to the nearest 10th?44. What is the scale factor of a cube with a volume of 729 m^3 to a cube with a volume of 6859?

Accepted Solution

A:

Answer:Part 42) The surface area of the sphere is [tex]SA=795.8\ ft^{2}[/tex]Part 43) The surface area of the sphere is [tex]SA=1,781.6\ m^{2}[/tex]Part 44) The scale factor is [tex]\frac{19}{9}[/tex]Step-by-step explanation:Part 42) What is the surface area of a sphere with a circumference of 50 feet round the answer to the nearest 10thstep 1Find the radius of the sphereThe circumference is equal to[tex]C=2\pi r[/tex]we have[tex]C=50\ ft[/tex]assume[tex]\pi =3.14[/tex]substitute and solve for r[tex]50=2(3.14)r[/tex][tex]r=7.96\ ft[/tex]step 2Find the surface area of the sphereThe surface area of the sphere is equal to[tex]SA=4\pi r^{2}[/tex]substitute the value of r[tex]SA=4(3.14)(7.96)^{2}[/tex][tex]SA=795.82\ ft^{2}[/tex]round to the nearest 10th[tex]795.82=795.8\ ft^{2}[/tex] Part 43) The volume of a sphere is 2254 pi m^3. What is the surface of the sphere to the nearest 10th?step 1Find the radius of the sphereThe volume of the sphere is equal to[tex]V=\frac{4}{3}\pi r^{3}[/tex]we have[tex]V=2,254\pi\ m^{3}[/tex]substitute and solve for r[tex]2,254\pi=\frac{4}{3}\pi r^{3}[/tex]Simplify[tex]1,690.5=r^{3}[/tex][tex]r=11.91\ m[/tex]step 2Find the surface area of the sphereThe surface area of the sphere is equal to[tex]SA=4\pi r^{2}[/tex]substitute the value of r[tex]SA=4(3.14)(11.91)^{2}[/tex][tex]SA=1,781.6\ m^{2}[/tex]Part 44) What is the scale factor of a cube with a volume of 729 m^3 to a cube with a volume of 6859?we know thatIf two figures are similar, then the ratio of its volumes is equal to the scale factor elevated to the cubesoLetz -----> the scale factorx ----> the volume of the larger cubey ----> the volume of the smaller cube[tex]z^{3}=\frac{x}{y}[/tex]we have[tex]x=6,859\ m^{3}[/tex][tex]y=729\ m^{3}[/tex]substitute[tex]z^{3}=\frac{6,859}{729}[/tex][tex]z=\frac{19}{9}[/tex][tex](\frac{6,859}{729})[/tex]