Answer: your answer is 32/51
Step-by-step explanation:
Answer:
19
Step-by-step explanation:
4(4)+3
16+3=19
Answer:
The volume is ![12000 in^3](https://tex.z-dn.net/?f=12000%20in%5E3)
Step-by-step explanation:
From the question ,
and ![H=10\:in.](https://tex.z-dn.net/?f=H%3D10%5C%3Ain.)
The volume of a pyramid is given as
![V=\frac{1}{3} \times area\:of\:the\:base \times height](https://tex.z-dn.net/?f=V%3D%5Cfrac%7B1%7D%7B3%7D%20%5Ctimes%20area%5C%3Aof%5C%3Athe%5C%3Abase%20%5Ctimes%20height)
From the question, the base of the pyramid is a square.
Hence base area is given by ![L^{2}](https://tex.z-dn.net/?f=L%5E%7B2%7D)
This implies that the volume of the pyramid is
![=\frac{1}{3} \times L^{2}\times H](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B3%7D%20%5Ctimes%20L%5E%7B2%7D%5Ctimes%20H)
![=\frac{1}{3}\times 60^{2} \times 10](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B3%7D%5Ctimes%2060%5E%7B2%7D%20%5Ctimes%2010)
![=\frac{36000}{3}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B36000%7D%7B3%7D)
![=12000 in^3](https://tex.z-dn.net/?f=%3D12000%20in%5E3)
Answer:
It's supposed to be Adjustments, but it's not on here so it's D.
Answer:
a) 30 kangaroos in 2030
b) decreasing 8% per year
c) large t results in fractional kangaroos: P(100) ≈ 1/55 kangaroo
Step-by-step explanation:
We assume your equation is supposed to be ...
P(t) = 76(0.92^t)
__
a) P(10) = 76(0.92^10) = 76(0.4344) = 30.01 ≈ 30
In the year 2030, the population of kangaroos in the province is modeled to be 30.
__
b) The population is decreasing. The base 0.92 of the exponent t is the cause. The population is changing by 0.92 -1 = -0.08 = -8% each year.
The population is decreasing by 8% each year.
__
c) The model loses its value once the population drops below 1/2 kangaroo. For large values of t, it predicts only fractional kangaroos, hence is not realistic.
P(100) = 75(0.92^100) = 76(0.0002392)
P(100) ≈ 0.0182, about 1/55th of a kangaroo