surface area of the outside of a cube shaped closed-top water tank whose edge length is 7.2 ft, can be solved by solving the area of each face of a cube. since the cube has 6 faces and all the faces are congruent because it all has the same edge.
so the formulaA = 6s^2where s is the edge lengthA = 6 ( 7.2^2)A = 311.04 sq ft.
Answer: Feet
Step-by-step explanation:
Answer:
SEE BELOW IN BOLD.
Step-by-step explanation:
a.
h = -16t^2 + 50t
h = 20 t
When the height is the same:
-16t^2 + 50t = 20t
-16t^2 + 30t = 0
t(-16t + 30) = 0
t = 0 or -16t + 30 = 0, so:
t = 0 or -30/-16 = 1.875
So the answer is 1.88 seconds to the nearest hundredth.
b.
For the ball
h = -16t^2 + 50t
Finding the derivative and equating to zero:
dh/dt = -32t + 50 = 0
t = -50/-32 = 1.563
Maximum height after 1.56 seconds to nearest hundredth
c.
When the ball hits the ground h = 0 so
-16t^2 + 50t = 0
-16t(t - 50/16)= 0
T = 3.13 SECONDS TO THE NEAREST HUNDERDTH
Answer:
Lo siento. :(
Step-by-step explanation:
Me preguntaba lo mismo. Desearía poder ayudar, pero también necesito ayuda.
