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# HELP!

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1. The total area of fictional Puget Loud, near Seapple, Washington, can be roughly estimated using a rectangular shape that is 110 miles long and 15 miles wide. In Puget Loud, orcas require approximately 5 mi^2 of territory per animal. Based on this information, how many orcas could live in Puget Loud? Show your work.

2. The mass of a ball is 128g. It has a density of 0.5g/cm^3. What is the volume of the ball? Bonus Point: what is the radius of the ball?

May 22, 2019

#1
+92
+1

1. The total area of Puget Loud is 110 mi x 15 mi = 110*15 = 1650 mi^2. Because every orca needs 5 mi^2 of teritory, the number of orcas that can live in Puget Loud is 1650 mi^2 / 5 mi^2=1650/5 = $$\boxed{330}$$

2. $$density = \frac{mass}{volume}$$ so $$.5 =\frac{128}{v}$$, where v is volume.

$$.5v=128$$

$$v=\boxed{256 cm^3}$$

The eaqution for the volume of a sphere is $$v=\frac{4}{3} \pi r^3$$ where v is the volume and r is the radius so $$256=\frac{4}{3} \pi r^3$$

Dividing by $$\frac{4}{3} \pi$$ yields $$r^3=78842.81$$. Taking the cube root of both sides gets $$r = \boxed{42.88}$$

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May 22, 2019

#1
+92
+1

1. The total area of Puget Loud is 110 mi x 15 mi = 110*15 = 1650 mi^2. Because every orca needs 5 mi^2 of teritory, the number of orcas that can live in Puget Loud is 1650 mi^2 / 5 mi^2=1650/5 = $$\boxed{330}$$

2. $$density = \frac{mass}{volume}$$ so $$.5 =\frac{128}{v}$$, where v is volume.

$$.5v=128$$

$$v=\boxed{256 cm^3}$$

The eaqution for the volume of a sphere is $$v=\frac{4}{3} \pi r^3$$ where v is the volume and r is the radius so $$256=\frac{4}{3} \pi r^3$$

Dividing by $$\frac{4}{3} \pi$$ yields $$r^3=78842.81$$. Taking the cube root of both sides gets $$r = \boxed{42.88}$$

power27 May 22, 2019
#2
+101086
+1

CPhill  May 22, 2019
#3
+2

thank you so much! The explanations helped a lot!

Guest May 22, 2019