The area is that of two 20 yd squares and one 20 yd circle.
.. A = 2*(20 yd)^2 +(π/4)*(20 yd)^2
.. = (2 +π/4)*(400 yd^2)
.. = (800 +100π) yd^2
.. ≈ 1114.16 yd^2
The perimeter is that of a 20 yd circle and 80 yd more.
.. P = π*20 yd + 80 yd
.. ≈ 142.83 yd
Answer:
Step-by-step explanation:
bottom left is 13a + 2b
bottom right is 15y - 6
Answer:
-2, 1, 2, 3, 4
Step-by-step explanation:
The domain is the x values of the graph.
Remember:
a^m * a^n=a^(m+n)
a^-m=1/a^m
8³ x 8⁻⁵=8³⁻⁵=8⁻²=1/8²=1/64<1
Answer: the student is not correct, because 1/64<1, and 8³ x 8⁻⁵=1/64.
Answer:
2160 cm³/hour
Step-by-step explanation:
By default, we know that the volume of a cube is given as s³
Thus, the Volume function, V = s³
When we differentiate with respect to time we have
dV/dt = 3s² (ds/dt), where ds/dt = 0.2
Then we go ahead and substitute all the given parameters
dV/dt = 3 x 60 x 60 x 0.2
dV/dt = 10800 * 0.2
dV/dt = 2160 cm³/hour
This means that the volume decreases by a rate of 2160 cm³/hour at the instant its edge is 60 cm
